NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solution for Class 12 Maths Chapter 5 Solutions: Download PDF

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NCERT Solutions for Class 12 Maths Chapter 5: Exercise Questions

Question 1: Prove that the function $f ( x) = 5 x -3$ is continuous at $x = 0, at\: \: x = - 3$ and at $x = 5$

Answer:

The given function is
$f ( x) = 5 x -3$
$f(0) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) =f(0)$
Hence, the function is continuous at x = 0
$f(-3)= 5(-3)-3=-15-3=-18 \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18$

$ \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = f(-3)$
Hence, the function is continuous at $x = -3$
$f(5)= 5(5)-3=25-3=22 \Rightarrow \lim\limits_{x \rightarrow 5} f(x) = 5(5)-3 = 25-3=-22$

$\Rightarrow \lim\limits_{x \rightarrow 5} f(x) = f(5)$
Hence, the function is continuous at $x = 5$

Question 2: Examine the continuity of the function $f (x) = 2x ^2 - 1 at x = 3.$

Answer:

The given function is
$f(x) = 2x^2-1$
at $x = 3$
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\$

$ \lim\limits_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim\limits_{x\rightarrow 3}f(x) = f(3)$
Hence, the function is continuous at $x = 3$

Question 3: Examine the following functions for continuity.
$(a) f (x) = x - 5$

Answer:

The given function is
$f(x) = x-5$
Our function is defined for every real number, say k
and value at $x = k$, $f(k) = k-5$
And also,
$\lim\limits_{x\rightarrow k} f(x) = k -5\\$

$ \lim\limits_{x\rightarrow k} f(x) = f(k)$
Hence, the function $f(x) = x-5$ is continuous at every real number

Question 3(b): Examine the following functions for continuity.

$f (x) = \frac{1}{x-5} , x \neq 5$

Answer:

The given function is
$f(x ) = \frac{1}{x-5}$
For every real number k, $k \neq 5$
We get,
$f(k) = \frac{1}{k-5}\\$

$ \lim\limits_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\$

$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{1}{x-5}$ continuous for every real value of $x$, $x \neq 5$

Question 3(c): Examine the following functions for continuity.

$f (x) = \frac{x ^2-25}{x+5}, x \neq -5$

Answer:

The given function is
$f(x ) = \frac{x^2-25}{x+5}$
For every real number k, $k \neq -5$
We get,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}$

$= \frac{(k +5)(k-5)}{k+5} = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$

Question 3(d): Examine the following functions for continuity. $f (x) = | x - 5|$

Answer:

The given function is
$f (x) = | x - 5|$
for $x > 5 , f(x) = x – 5$
for $x < 5 , f(x) = 5 – x$
So, there are different cases.
case(i) $x > 5$
for every real number $k > 5$ , $f(x) = x – 5$ is defined
$f(k) = k - 5\\$

$ \lim\limits_{x\rightarrow k }f(x) = k -5\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = x – 5$ is continuous for $x > 5$
case (ii) $x < 5$
for every real number $k < 5$ , $f(x) = 5 – x$ is defined
$f(k) = 5-k\\ \lim\limits_{x\rightarrow k }f(x) = 5 -k\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = 5 – x$ is continuous for $x < 5$
case(iii) $x = 5$
for $x = 5$ , $f(x) = x – 5$ is defined
$f(5) = 5 - 5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = f(5)$
Hence, function $f(x) = x – 5$ is continous for $x = 5$

Hence, the function $f (x) = | x - 5|$ is continuous for every real number.

Question 4: Prove that the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Answer:

The given function is
$f (x) = x^n$
The function $f (x) = x^n$ is defined for all positive integer, n
$f(n) = n^n\\$

$\lim\limits_{x\rightarrow n}f(x) = n^n\\$

$ \lim\limits_{x\rightarrow n}f(x) = f(n)$
Hence, the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Question 5: Is the function f defined by
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
continuous at x = 0? At x = 1? At x = 2?

Answer:

The given function is
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
The function is defined at $x = 0$ and its value is $0$
$f(0) = 0\\$

$ \lim\limits_{x\rightarrow 0}f(x) = f(x) = 0\\$

$ \lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the given function is continuous at $x = 0$
The given function is defined for $x = 1$
Now, for $x = 1$ Right-hand limit and left-hand limit are not equal.
$f(1) = 1\\ \lim\limits_{x\rightarrow 1^-}f(x) = f(x) = 1\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) =f(5) = 5$
R.H.L $\neq$ L.H.L.
Therefore, the given function is not continuous at $x =1$
Given function is defined for $x = 2$ and its value at $x = 2$ is $5$
$f(2) = 2\\ \lim\limits_{x\rightarrow 2}f(x) = f(5) = 5\\\lim\limits_{x\rightarrow 2}f(x) = f(2)$
Hence, the given function is continuous at $x = 2$

Question 6: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$
The given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 2$
$f(k) = 2k-3\\ \lim\limits_{x\rightarrow k}f(x) = 2k-3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$

case(ii) $k < 2$
$f(k) = 2k +3\\ \lim\limits_{x\rightarrow k}f(x) = 2k+3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$

case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ $

$\lim\limits_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
Right hand limit at x= 2 $\neq$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Question 7: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$
The given function is defined for every real number k
There are different cases.
case (i) $k < -3$
$f(k) = -k + 3\\ \lim\limits_{x\rightarrow k}f(x) = -k + 3\\ $

$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k < -3

case(ii) $k = -3$
$f(-3) = -(-3) + 3 = 6\\ \lim\limits_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\$

$ \lim\limits_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\$

$ R.H.L. = L.H.L. = f(-3)$
Hence, the given function is continuous for x = -3

case(iii) $-3 < k < 3$
$f(k) = -2k \\ \lim\limits_{x\rightarrow k}f(x) = -2k\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -3 < k < 3 given function is continuous.

case(iv) $k = 3$
$f(3) = 6x+2 = 6\times3+2 =18+2=20\\$

$ \lim\limits_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\$

$ \lim\limits_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\$

$ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity

case(v) $k > 3$
$f(k) = 6k+2 \\$

$ \lim\limits_{x\rightarrow k}f(x) = 6k+2 \\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 3

Question 8: Find all points of discontinuity of f, where f is defined by

$f (x )= \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$

Answer:

The given function is
$f (x ) \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$
if x > 0 , $f(x)=\frac{x}{x} = 1$
if x < 0 , $f(x)=\frac{-(x)}{x} = -1$
The given function is defined for every real number k
Now,
case(i) k < 0
$f(k) = -1\\ \lim\limits_{x\rightarrow k }f(x) = -1\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim\limits_{x\rightarrow k }f(x) = 1\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0^- }f(x) = -1\\$

$ \lim\limits_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
Hence, 0 is the only point of discontinuity

Question 9: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$
if x < 0 , $f (x) =\frac{x }{|x|} = \frac{x}{-(x)} = -1$
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for every value of x
Hence, no point in discontinuity

Question 10: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right. $

Answer:

The given function is
$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k+1\\ \lim\limits_{x\rightarrow k}f(x) = k+1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$

case(ii) $k < 1$
$f(k) = k^2 +1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\ $

$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$

case(iii) $x = 1$

$\lim\limits_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ $

$R.H.L. = L.H.L. = f(1)$

Hence, at x = 2 given function is continuous.
Therefore, no point of discontinuity

Question 11: Find all points of discontinuity of f, where f is defined by

$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$

Answer:

The given function is
$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 2$
$f(k) = k^2+1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$

case(ii) $k < 2$
$f(k) = k^3 -3\\ \lim\limits_{x\rightarrow k}f(x) = k^3-3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$

case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\$

$ \lim\limits_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\$

$ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.$
Hence, the given function is continuous at $x = 2$
There is no point of discontinuity

Question 12: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$
The given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k^2\\ \lim\limits_{x\rightarrow k}f(x) = k^2\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k > 1

case(ii) $k < 1$
$f(k) = k^{10} -1\\$

$ \lim\limits_{x\rightarrow k}f(x) = k^{10}-1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k < 1

case(iii) x = 1
$\lim\limits_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\$

$ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.$
Hence, x = 1 is the point of discontinuity.

Question 13: Is the function defined by

$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$

A continuous function?

Answer:

The given function is
$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$
The given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k-5\\ \lim\limits_{x\rightarrow k}f(x) = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$

case(ii) $k < 1$
$f(k) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$

case(iii) $x = 1$
$\lim\limits_{x\rightarrow 1^-}f(x) = x+5 = 1 + 5 = 1 + 5 = 6\\ \lim\limits_{x\rightarrow 1^+}f(x) = x-5 = 1-5 = -4\\ f(1) = x+5 =1+5= 6 \\ L.H.L. = f(1) \neq R.H.S.$

Hence, $x = 1$ is the point of discontinuity.

Question 14: Discuss the continuity of the function f, where f is defined by

$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$

Answer:

The given function is
$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$
The given function is defined for every real number k
Different cases are there
case (i) $k < 1$
$f(k) = 3\\ \lim\limits_{x\rightarrow k}f(x) = 3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < 1$

case(ii) $k = 1$
$f(1) = 3 \\ \lim\limits_{x\rightarrow 1^-}f(x) = 3\\ \lim\limits_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)$
Hence, the given function is discontinuous at $x = 1$
Therefore, $x = 1$ is the point of discontinuity.

case(iii) $1 < k < 3$
$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $1 < k < 3$ given function is continuous.

case(iv) $k = 3$
$f(3) =5\\ \lim\limits_{x\rightarrow 3^-}f(x) = 4\\ \lim\limits_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity

case(v) $k > 3$
$f(k) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k > 3$
case(vi) when $k < 3$

$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $k < 3$ given function is continuous

Question 15: Discuss the continuity of the function f, where f is defined by $f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

Answer:
Given function is satisfied for all real values of $x$
case (i) $k < 0$
Hence, the function is continuous for all values of $x < 0$

case (ii) $x = 0$
L.H.L at $x= 0$

R.H.L. at $x = 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous at $x = 0$

case (iii) $k > 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous for all values of x > 0

case (iv) k < 1

Hence, the function is continuous for all values of x < 1

case (v) k > 1

Hence, the function is continuous for all values of x > 1

case (vi) x = 1
Hence, the function is not continuous at x = 1

Question 16: Discuss the continuity of the function f, where f is defined by

$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$

Answer:

The given function is
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
The given function is defined for every real number $k$
Different cases are there
case (i) $k < -1$
$f(k) = -2\\ \lim\limits_{x\rightarrow k}f(x) = -2\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < -1$
case(ii) k = -1
$f(-1) = -2 \\ \lim\limits_{x\rightarrow -1^-}f(x) = -2\\ \lim\limits_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)$
Hence, the given function is continuous at $x = -1$
case(iii) $k > -1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for all values of $x > -1$

case(vi) $-1 < k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $-1 < k < 1$ given function is continuous.

case(v) $k = 1$
$f(1) =2x = 2(1)=2\\ \lim\limits_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim\limits_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.$
Hence, at x =1 function is continuous

case(vi) $k > 1$
$f(k) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 1
case(vii) when $k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in $k < 1$ given function is continuous.

Therefore, continuous at all points

Question 17: Find the relationship between a and b so that the function f is defined by
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
is continuous at x = 3.

Answer:

The given function is
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
For the function to be continuous at $x = 3$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim\limits_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim\limits_{x\rightarrow 3^-}f(x) = \lim\limits_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$

Question 18: For what value of l is the function defined by
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
For the function to be continuous at $x = 0$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim\limits_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, if no value of the function is continuous at $x = 0$

For $x = 1$
$f(1)=4x+1=4(1)+1=5\\ \lim\limits_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim\limits_{x\rightarrow 1}f(x) = f(x)$
Hence, the given function is continuous at $x =1$

Question 19: Show that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x.

Answer:

Given function is
$g (x) = x- [x]$
Given is defined for all real numbers k
$\lim\limits_{x\rightarrow k^-}f(x) = k - (k-1) = k-k+1 =1\\ \lim\limits_{x\rightarrow k^+}f(x) = k – k = 0\\ \lim\limits_{x\rightarrow k^-}f(x) \neq \lim\limits_{x\rightarrow k^+}f(x)$
Hence, by this, we can say that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points

Question 20: Is the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$ ?

Answer:

Given function is
$f (x) = x^2 - sin x + 5$
Clearly, the Given function is defined at x = $\pi$
$f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = f(\pi)$
Hence, the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$

Question 21: Discuss the continuity of the following functions:
a) $f (x) = \sin x + \cos x$

Answer:

Given function is
$f (x) = \sin x + \cos x$
The given function is defined for all real numbers.
We, know that if two function $g(x)$ and $h(x)$ are continuous then $g(x)+h(x)$ , $g(x)-h(x)$ , $g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x)$

$ = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$ \text{We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) =$ $\lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{ We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) + h(x) = \sin x + \cos x$ is also a continuous function

Question 21(b):Discuss the continuity of the following functions:
$f (x) = \sin x - \cos x$

Answer:

Given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions g(x) and h(x) are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$ \text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous discussion of the continuity of the following functions:
$f (x)$ is continuous, we can say that
$f(x) = g(x) - h(x) = \sin x - \cos x$ is also a continuous function.

Question 21(c): Discuss the continuity of the following functions:
$f (x) = \sin x \cdot \cos x$

Answer:

Given function is
$f (x) = \sin x \cdot \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c, \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x are continuous functions.
So, we can say that
$f(x) = g(x).h(x) = \sin x .\cos x$ is also a continuous function

Question 22: Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Answer:

We know that if two functions g(x) and h(x) are continuous, then.
$\frac{g(x)}{h(x)} , h(x) \neq0\text{ is continuous}$

$\frac{1}{h(x)} , h(x) \neq 0\ \text{is continuous} \frac{1}{g(x)} , g(x) \neq0\text{ is continuous}$
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text {We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$

$= \lim\limits_{x\rightarrow c}\cos x $

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h)$

$=\lim\limits_{h \rightarrow 0} (\cos c \cos h+ \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, the function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ a continuous functions.
So, we can say that
cosec x = $\frac{1}{\sin x} = \frac{1}{g(x)}$ is also continuous except at $x=n\pi$
sec x = $\frac{1}{\\cos x} = \frac{1}{h(x)}$ is also continuous except at $x=\frac{(2n+1) \pi}{2}$
cot x = $\frac{\cos x}{\sin x} = \frac{h(x)}{g(x)}$ is also continuous except at $x=n\pi$

Question 23: Find all points of discontinuity of f, where

$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$

Answer:

Given function is
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim\limits_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x)$
Hence, the function is continuous.
Therefore, no point of discontinuity

Question 24: Determine if f is defined by
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
Is it a continuous function?

Answer:

The given function is
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
The given function is defined for all real numbers k
when x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0}f(x)=\lim\limits_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim\limits_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k}f(x)=\lim\limits_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k} = f(k)$
Hence, the given function is continuous for all points

Question 25: Examine the continuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \sin x - \cos x & if x \neq 0 \\ -1 & if x = 0 \end{matrix}\right.$

Answer:

The given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x$

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $ \sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) - h(x) =$ $\sin x$ - $\cos x$ is also a continuous function

When $x = 0$
$f (0) = (-1) \lim\limits_{x\rightarrow 0^-}f(x)$

$= \sin 0 - \cos 0 = -1$

$ \lim\limits_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 $

$ R.H.L. = L.H.L. = f(0)$
Hence, the function is also continuous at $x = 0$

Question 26: Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right. \: \: \: at \: \: x = \pi /2$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} \frac{k\cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right.$
When $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= \lim\limits_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim\limits_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
Therefore, the value of k so that the function f is continuous at 6

Question 27: Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right. \: \: at \: \: x = 2$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right.$
When $x = 2$
For the function to be continuous
$f(2) = R.H.L. = LH.L.$
$f(2) = 4k\\ \lim\limits_{x\rightarrow 2^-}f(x)= 4k\\ \lim\limits_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
Hence, the values of k so that the function f is continuous at x=2 are $\frac{3}{4}$

Question 28: Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right. \: \: at \: \: x = \pi$

Answer:

The given function is
$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right.$
When x = $\pi$
For the function to be continuous
, f( $\pi$ ) = R.H.L. = LH.L.
$f(\pi) = k\pi+1\\ \lim\limits_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim\limits_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim\limits_{x\rightarrow \pi^-}f(x) = \lim\limits_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}$
Hence, the values of k so that the function f is continuous at x= $\pi$ is $\frac{-2}{\pi}$

Question 29: Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right. \: \: at x = 5$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right.$
When $x = 5$
For the function to be continuous
$f(5) = R.H.L. = LH.L.$
$f(5) = 5k+1\\ \lim\limits_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim\limits_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim\limits_{x\rightarrow 5^-}f(x) = \lim\limits_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}$
Hence, the values of k so that the function f is continuous at $x= 5$ is $\frac{9}{5}$

Question 30: Find the values of a and b such that the function defined by
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
It is a continuous function.

Answer:

Given that continuous function is
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
The function is continuous so
$\lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ and\\ \lim\limits_{x\rightarrow 10^-}f(x)=\lim\limits_{x\rightarrow 10^+}f(x)$
$\lim\limits_{x\rightarrow 2^-}f(x) = 5\\ \lim\limits_{x\rightarrow 2^+}f(x)=ax+b=2a+b$

$ 2a+b = 5 \ \ \ \ \ \ -(i)$

$\lim\limits_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim\limits_{x\rightarrow 10^+}f(x)=21$

$ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)$
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by $f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21, & if\: \: x > 10 \end{matrix}\right.$ is a continuous function is 2 and 1 respectively

Question 31 Show that the function defined by $f (x) = \cos (x^2 )$ is a continuous function.

Answer:

Given function is
$f (x) = \cos (x^2 )$
Given function is defined for all real values of x
Let x = k + h
if $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = \lim\limits_{x \rightarrow k}\cos x^2 = \lim\limits_{h \rightarrow 0}\cos (k+h)^2 $

$= \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = f(k)$
Hence, the function $f (x) = \cos (x^2 )$ is a continuous function

Question 32: Show that the function defined by $f (x) = |\cos x |$ is continuous.

Answer:

Given function is
$f (x) = |\cos x |$
Given function is defined for all values of x
f = g o h , $g(x) = \|x\|$ and $h(x) = \cos x$
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when $k < 0$

case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$\lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$

$= \lim\limits_{x\rightarrow c}\cos x$

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that } \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
g(x) is continuous , $h(x)$ is continuous
Therefore, $f(x) = g o h$ is also continuous

Question 33: Examine that sin | x| is a continuous function.

Answer:

The given function is
$f(x) = \sin \|x\|$
f(x) = h o g , h(x) =$\sin x$ and g(x) = |x|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$\lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) =$\sin x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \sin c$

$ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\sin x$

$= \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{ We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \sin x$ is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, $f(x) = h o g$ is also continuous

Question 34: Find all the points of discontinuity of f defined by $f (x) = | x| - | x + 1|.$

Answer:

The given function is
$f (x) = | x| - | x + 1|$
Let g(x) = |x| and h(x) = |x+1|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,
$h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < -1
$h(k) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

case (ii) k > -1
$h(k) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k > -1

case (iii) k = -1
$h(-1) = 0\\ \lim\limits_{x\rightarrow -1^-}h(x) = -(x-1)$

$= 0\\ \lim\limits_{x\rightarrow -1^+}h(x ) = x+1 = 0$

$ \lim\limits_{x\rightarrow -1^-}h(x) = h(0) = \lim\limits_{x\rightarrow -1^+}h(x )$
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x.
g(x) is continuous and h(x) is continuous
Therefore, $f(x) = g(x) - h(x) = |x| - |x+1|$ is also continuous

Question 1: Differentiate the functions with respect to x in

$\sin (x^2 +5 )$

Answer:

The given function is
$f(x)=\sin (x^2 +5 )$
When we differentiate it w.r.t. x.
Let's take $t = x^2+5$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)$
$\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x$
Therefore, the answer is $2x \cos (x^2+5)$

Question 2: Differentiate the functions with respect to x in

$\cos ( \sin x )$

Answer:

The given function is
$f(x)= \cos ( \sin x )$
Let’s take $t = \sin x$ then,
$f(t) = \cos t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ ( By chain rule)
$\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)$
$\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x$
Therefore, the answer is $-\sin(\sin x).\cos x$

Question 3: Differentiate the functions with respect to x in

$\sin (ax +b )$

Answer:

The given function is
$f(x) = \sin (ax +b )$
When we differentiate it w.r.t. x.
Let's take $t = ax+b$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)$
$\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a$
Therefore, the answer is $a \cos (ax+b)$

Question 4: Differentiate the functions with respect to x in

$\sec (\tan (\sqrt x) )$

Answer:

The given function is
$f(x)=\sec (\tan (\sqrt x) )$
When we differentiate it w.r.t. x.
Let's take $t = \sqrt x$. then,
$f(t) = \sec (\tan t)$
take $\tan t = k$ . then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$

$=\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$

Question 5: Differentiate the functions with respect to x in

$\frac{\sin (ax +b )}{\cos (cx + d)}$

Answer:

The given function is
$f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}$
We know that,
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}$
$g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$
Let's take $u = (ax+b) \ and \ v = (cx+d)$
Then,
$\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c$
$g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}$ (By chain rule)
$\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)$
$\frac{du}{dx} = \frac{d(ax+b)}{dx} = a$
$g^{'}(x)=a\cos (ax+b)$ -(i)
Similarly,
$h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}$
$\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))$
$\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c$
$h^{'}(x)=-c\sin(cx+d)$ -(ii)
Now, put (i) and (ii) in
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}$
$= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}$
$= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Therefore, the answer is $a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$

Question 6: Differentiate the functions with respect to x in

$\cos x^3 . \sin ^ 2 ( x ^5 )$

Answer:

The given function is
$f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )$
Differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)$
$g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)$
Lets take $u = x^3 \ and \ v = x^5$
Our functions become,
$\cos x^3 = \cos u$ and $\sin^2(x^5) = \sin^2v$
Now,
$g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)$
$\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2$
$g^{'}(x) = -\sin x^3.3x^2$ -(i)
Similarly,
$h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5$

$ (\because v = x^5)$
$\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4$
$h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5$ -(ii)
Put (i) and (ii) in
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) $

$= -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5 \cos x^5$
Therefore, the answer is $10x^4\sin x^5 \cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$

Question 7: Differentiate the functions with respect to x in

$2 \sqrt { \cot ( x^2 )}$

Answer:

The give function is
$f(x)=2 \sqrt { \cot ( x^2 )}$
Let's take $t = x^2$
Now, take $\cot t = k^2$
$f(k) = 2k$
Differentiation w.r.t. x
$\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ -(By chain rule)
$\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2$
$\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}}(\because t = x^2)$
$\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x$
So,
$\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }$ ( Multiply and divide by $\sqrt 2$ and multiply and divide $\sqrt {\cot x^2}$ by $\sqrt{\sin x^2}$
$(\text{because} \cot x = \frac{\cos x}{\sin x} \ and \csc x = \frac{1}{\sin x })$
$=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}(\because 2\sin x\cos x=\sin2x)$
There, the answer is $\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}$

Question 8: Differentiate the functions with respect to x in

$\cos ( \sqrt x )$

Answer:

Let us assume : $y\ =\ \cos ( \sqrt x )$

Differentiating y with respect to x, we get :

$\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}$

or $=\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}$

or $=\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}$

Question 9: Prove that the function f given by $f (x) = |x-1 |, x \epsilon R$ is not differentiable at x = 1.

Answer:

The given function is
$f (x) = |x-1 | , x \epsilon R$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differentiable at x = 1 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^-}\frac{|h|-0}{h}$
$= \lim\limits_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)$
The right-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^+}\frac{|h|-0}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{h}{h} = 1$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1
Therefore, function $f (x) = |x-1 |$ is not differentiable at x = 1

Question 10: Prove that the greatest integer function defined by $f (x) = [x] , 0 < x < 3$ is not differentiable at

x = 1 and x = 2.

Answer:

The given function is
$f (x) = [x] , 0 < x < 3$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differentiable at x = 1 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{0-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \rightarrow 1+h<1, \therefore [1+h] =0)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{1-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 1+h>1, \therefore [1+h] =1)$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 1
Similarly, for x = 2
The required condition for the function to be differentiable at x = 2 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}$
Now, the Left-hand limit of the function at x = 2 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{1-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \rightarrow 2+h<2, \therefore [2+h] =1)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{2-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 2+h>2, \therefore [2+h] =2)$
Now, it is clear that.
R.H.L. at x= 2 $\neq$ L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2

Question 1: Find $\frac{dy}{dx}$ in the following:

$2 x + 3 y = \sin x$

Answer:

The given function is
$2 x + 3 y = \sin x$
We can rewrite it as
$3y = \sin x - 2x$
Now, differentiation w.r.t. x is
$3\frac{dy}{dx} = \frac{d(\sin x - 2x)}{dx} = \cos x - 2$
$\frac{dy}{dx} = \frac{\cos x-2}{3}$
Therefore, the answer is $\frac{\cos x-2}{3}$

Question 2: Find $\frac{dy}{dx}$ in the following: $2 x + 3y = \sin y$

Answer:

The given function is
$2 x + 3 y = \sin y$
We can rewrite it as
$\sin y - 3y = 2x$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(\sin y - 3y) = \frac{d( 2x)}{dx}$

$(\cos y\frac{dy}{dx} - 3\frac{dy}{dx}) = 2$
$\frac{dy}{dx} = \frac{2}{\cos y -3}$
Therefore, the answer is $\frac{2}{\cos y -3}$

Question 3: Find $\frac{dy}{dx}$ in the following: $ax + by ^2 = \cos y$

Answer:

The given function is
$ax + by ^2 = \cos y$
We can rewrite it as
$by^2-\cos y = -ax$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(2by - (-\sin y)) = \frac{d( -ax)}{dx} = -a$
$\frac{dy}{dx} = \frac{-a}{2b y +\sin y}$
Therefore, the answer is $\frac{-a}{2b y +\sin y}$

Question 4: Find $\frac{dy}{dx}$ in the following:

$xy + y^2 = \tan x + y$

Answer:

The given function is
$xy + y^2 = \tan x + y$
We can rewrite it as
$xy+y^2-y= \tan x$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x$
$\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}$
Therefore, the answer is $\frac{\sec^2 x- y}{x+2y-1}$

Question 5: Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$

Answer:

The given function is
$x^2 + xy + y^2 = 100$
We can rewrite it as
$xy + y^2 = 100 - x^2$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y) = \frac{d( 100-x^2)}{dx} = -2x$
$\frac{dy}{dx} = \frac{-2 x- y}{x+2y}$
Therefore, the answer is $\frac{-2 x- y}{x+2y}$

Question 6: Find $\frac{dy}{dx}$ in the following:

$x ^3 + x^2 y + xy^2 + y^3 = 81$

Answer:

The given function is
$x ^3 + x^2 y + xy^2 + y^3 = 81$
We can rewrite it as
$x^2 y + xy^2 + y^3 = 81 - x^3$
Now, differentiation w.r.t. x is
$\frac{d(x^2 y + xy^2 + y^3)}{dx} = \frac{d(81 - x^3)}{dx}$
$2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2) = -3x^2\\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$
Therefore, the answer is $\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$

Question 7: Find $\frac{dy}{dx}$ in the following: $\sin ^ 2 y + \cos xy = k$

Answer:

The given function is
$\sin ^ 2 y + \cos xy = k$
Now, differentiation w.r.t. x is
$\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}$
$2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0$

$\frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx}$

$= \frac{y\sin xy}{2\sin y \cos y-x\sin xy}$

$= \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)$
Therefore, the answer is $\frac{y\sin xy}{\sin 2y -x\sin xy}$

Question 8: Find $\frac{dy}{dx}$ in the following:

$\sin ^2 x + \cos ^ 2 y = 1$

Answer:

The given function is
$\sin ^2 x + \cos ^ 2 y = 1$
We can rewrite it as
$\cos ^ 2 y = 1-\sin^2x$
Now, differentiation w.r.t. x is
$\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}$
$2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)$
Therefore, the answer is $\frac{\sin 2x}{\sin 2y }$

Question 9: Find $\frac{dy}{dx}$ in the following:

$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$

Answer:

The given function is
$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})$
Our equation reduces to
$y = \sin^{-1}(\sin 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question 10: Find $\frac{dy}{dx}$ in the following:
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }$

Answer:

The given function is
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )$
Our equation reduces to
$y = \tan^{-1}(\tan 3t)$
$y = 3t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}$
Therefore, the answer is $\frac{3}{1+x^2}$

Question 11: Find $\frac{dy}{dx}$ in the following:

$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right ) , 0 < x < 1$

Answer:

The given function is
$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$y = \cos^{-1}(\cos 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question 12: Find $\frac{dy}{dx}$ in the following: $y = \sin ^{-1 } \left ( \frac{1- x ^2 }{1+ x^2} \right ) , 0< x < 1$

Answer:

The given function is
$y = \sin ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
We can rewrite it as
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$\sin y = \cos 2t$
Now, differentiation w.r.t. x is
$\frac{d(\sin y)}{dx} = \frac{d(\cos2t)}{dt}.\frac{dt}{dx}$
$\cos y\frac{dy}{dx} = 2(-\sin 2t).\frac{1}{1+x^2} = \frac{-2\sin2t}{1+x^2}$ $= \frac{-2.\frac{2\tan t}{1+\tan^2t}}{1+x^2} =\frac{-2.\frac{2x}{1+x^2}}{1+x^2} =\frac{-4x}{(1+x^2)^2}$
$(\because \sin 2x = \frac{2\tan x}{1+\tan^2x} \ and \ x = \tan t)$
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )\rightarrow \cos y = \frac{2x}{1+x^2}$
$\frac{2x}{1+x^2}\frac{dy}{dx} = \frac{-4x}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question 13: Find $\frac{dy}{dx}$ in the following:

$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1$

Answer:

The given function is
$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
We can rewrite it as
$\cos y = \left ( \frac{2x}{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )$
Our equation reduces to
$\cos y = \sin 2t$
Now, differentiation w.r.t. x is
$\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}$
$(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2}$ $= \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}$
$(\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)$
$\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\rightarrow \sin y = \frac{1-x^2}{1+x^2}$
$-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question 14: Find $\frac{dy}{dx}$ in the following:

$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }$

Answer:

The given function is
$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )$
Let's take $x = \sin t$
Then,
$\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =\cos t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}$
$(\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )$
And
$2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t}$

$= 2\sin t\cos t =\sin 2t$
$(\because \cos x = \sqrt{1-\sin^2x} \text{ and }\ 2\sin x\cos x = \sin2x )$
Now, our equation reduces to
$y = \sin ^ { -1 } ( \sin 2t )$
$y = 2t$
Now, differentiation with respect to. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{2}{\sqrt{1-x^2}}$

Question 15: Find $\frac{dy}{dx}$ in the following:

$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2$

Answer:

The given function is
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )$
Let's take $x = \cos t$
Then,
$\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )$
And
$\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t$
$(\because \cos 2x = \sqrt{2\cos^2x-1} \text{ and }\cos x = \frac{1}{\sec x} )$

Now, our equation reduces to
$y = \sec ^{-1} \left ( \sec 2t \right )$
$y = 2t$
Now, differentiation with respect to. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{-2}{\sqrt{1-x^2}}$

Question 1: Differentiate the following w.r.t. x:

$\frac{e ^x }{\sin x }$

Answer:

The given function is
$f(x)=\frac{e ^x }{\sin x }$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(e^x)}{dx}.\sin x-e^x.\frac{(\sin x)}{dx} }{\sin^2 x }$
$=\frac{e^x.\sin x-e^x.\cos }{\sin^2 x } = \frac{e^x(\sin x-\cos x)}{\sin^2x}$
Therefore, the answer is $\frac{e^x(\sin x-\cos x)}{\sin^2x}$

Question 2: Differentiate the following w.r.t. x:

$e ^{\sin ^{-1}x}$

Answer:

The given function is
$f(x)=e ^{\sin ^{-1}x}$
Let $g(x)={\sin ^{-1}x}$
Then,
$f(x)=e^{g(x)}$
Now, differentiation with respect to. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = \sin^{-1}x \rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}$
Put this value in our equation (i)
$f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$

Question 3: Differentiate the following w.r.t. x:

$e ^{x^3}$

Answer:

The given function is
$f(x)=e ^{x^3}$
Let $g(x)=x^3$
Then,
$f(x)=e^{g(x)}$
Now, differentiation with respect to. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = x^3 \rightarrow g^{'}(x ) =3x^2$
Put this value in our equation (i)
$f^{'}(x) =3x^2.e^{x^3}$
Therefore, the answer is $3x^2.e^{x^3}$

Question 4: Differentiate the following w.r.t. x:

$\sin ( \tan ^ { -1} e ^{-x })$

Answer:

The given function is
$f(x)=\sin ( \tan ^ { -1} e ^{-x })$
Let's take $g(x ) = \tan^{-1}e^{-x}$
Now, our function reduces to
$f(x) = \sin(g(x))$
Now,
$f^{'}(x) = g^{'}(x)\cos(g(x))$ -(i)
And
$g(x)=\tan^{-1}e^{-x}\\\rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}$
Put this value in our equation (i)
$f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$
Therefore, the answer is $\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$

Question 5: Differentiate the following w.r.t. x:

$\log (\cos e ^x )$

Answer:

The given function is
$f(x)=\log (\cos e ^x )$
Let's take $g(x ) = \cos e^{x}$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\cos e^{x}\\\rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x$
Put this value in our equation (i)
$f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)$
Therefore, the answer is $-e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N$

Question 6: Differentiate the following w.r.t. x:

$e ^x + e ^{x^2} + .....e ^{x^5}$

Answer:

The given function is
$f(x)= e ^x + e ^{x^2} + .....e ^{x^5}$
Now, differentiation w.r.t. x is
$f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}$
$=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4$
$=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
Therefore, answer is $e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$

Question 7: Differentiate the following w.r.t. x:

$\sqrt { e ^{ \sqrt x }} , x > 0$

Answer:

The given function is
$f(x)=\sqrt { e ^{ \sqrt x }}$
Let's take $g(x ) = \sqrt x$
Now, our function reduces to
$f(x) = \sqrt {e^{g(x)}}$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}$ -(i)
And
$g(x)=\sqrt x\\\rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$

Question 8: Differentiate the following w.r.t. x: $\log ( \log x ) , x > 1$

Answer:

The given function is
$f(x)=\log ( \log x )$
Let's take $g(x ) = \log x$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\log x\\\rightarrow g^{'}(x) = \frac{1}{x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}$
Therefore, the answer is $\frac{1}{x\log x}, \ \ x>1$

Question 9: Differentiate the following w.r.t. x:

$\frac{\cos x }{\log x} , x > 0$

Answer:

The given function is
$f(x)=\frac{\cos x }{\log x}$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }$
$=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Therefore, the answer is $\frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$

Question 10: Differentiate the following w.r.t. x:

$\cos ( log x + e ^x ) , x > 0$

Answer:

The given function is
$f(x)=\cos ( log x + e ^x )$
Let's take $g(x) = ( log x + e ^x )$
Then, our function reduces to
$f(x) = \cos (g(x))$
Now, differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x)(-\sin) (g(x))$ -(i)
And
$g(x) = ( log x + e ^x )$
$g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x$
Put this value in our equation (i)
$f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)$
Therefore, the answer is $-\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0$

Question 1: Differentiate the functions w.r.t. x. $\cos x . \cos 2x .\cos 3x$

Answer:

The given function is
$y=\cos x. \cos 2x .\cos 3x$
Now, take a look at both sides.
$\log y=\log (\cos x . \cos 2x .\cos 3x)$

$\log y = \log \cos x + \log \cos 2x + \log \cos 3x$
Now, differentiation with respect to. x
$\log y=\log (\cos x . \cos 2x .\cos 3x)$

$\frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}$

$\frac{1}{y}.\frac{dy}{dx} =$

$(-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}$

$\frac{1}{y} \frac{dy}{dx}= (\tan x+ \tan 2x+ \tan 3x )$

$(\because \frac{\sin x }{\cos x} =\tan x)$

$ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)$

$ \frac{dy}{dx}= -\cos x \cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
There, the answer is $-\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$

Question 2: Differentiate the functions with respect to. x.

$\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$

Answer:

The given function is
$y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
Take logs on both sides.
$\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )$

$\log y =$

$\frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)$

$-\log(x-5))$
Now, differentiation w.r.t. x is
$\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}$

$-\frac{d(\log(x-4))}{dx}- \frac{d(\log(x-5))}{dx})$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}$

$=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

$ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$

$(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
Therefore, the answer is $\frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

Question 3: Differentiate the functions w.r.t. x. $(\log x ) ^{\cos x}$

Answer:

The given function is
$y=(\log x ) ^{\cos x}$
Take logs on both sides.
$\log y=\cos x\log (\log x )$
Now, differentiation w.r.t x is

$\frac{d(\log y)}{dx}= \frac{d(\cos x\log(\log x))}{dx}$

$\frac{1}{y} \frac{dy}{dx}= (-\sin x)(\log(\log x))+\cos x (\frac{1}{\log x} \cdot \frac{1}{x})$

$\frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\$

$\frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$
Therefore, the answer is $(\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$

Question 4: Differentiate the functions w.r.t. x. $x ^x - 2 ^{ \sin x }$

Answer:

The given function is
$y = x ^x - 2 ^{ \sin x }$
Let's take $t = x^x$
Take logs on both sides.
$\log t=x\log x\\$
Now, differentiation w.r.t x is
$\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$

$\frac{1}{t}.\frac{dt}{dx} = \log x +1$

$\frac{dt}{dx} = t(\log x+1)$

$\frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )$
Similarly, take $k = 2^{\sin x}$
Now, take the log on both sides and differentiate with respect to. x
$\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$

$\frac{1}{k}.\frac{dk}{dx} = \cos x \log 2$

$\frac{dk}{dx} = k(\cos x \log 2)$

$\frac{dk}{dx}= 2^{\sin x}(\cos x\log 2)$

$(\because k = 2^{\sin x} )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}$

$\frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)$
Therefore, the answer is $x^x(\log x+1 )- 2^{\sin x}(\cos x \log 2)$

Question 5: Differentiate the functions w.r.t. x. $( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$

Answer:

The given function is
$y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$
Take logs on both sides.
$\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]$

$ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)$
Now, differentiate w.r.t. x we get,
$\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}$

$\frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right ) $

$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )$

$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.$

$\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )$

$ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3$

$(9x^2 + 70x + 133)$
Therefore, the answer is $(x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)$

Question 6: Differentiate the functions w.r.t. x. $( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$

Answer:

The given function is
$y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$
Let's take $t = ( x+ \frac{1}{x} ) ^ x$
Now, take a look at both sides.
$\log t =x \log ( x+ \frac{1}{x} )$
Now, differentiate with respect. x
We get,
$\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )}$

$= \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )$

$ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$

$ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$
Similarly, take $k = x^{1+\frac{1}{x}}$
Now, take a look at both sides.
$\log k = ({1+\frac{1}{x}})\log x$
Now, differentiate with respect. x
We get,
$\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x$

$= \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x$

$ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)$

$\frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+$

$\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Therefore, the answer is $\left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$

Question 7: Differentiate the functions with respect to. x. $(\log x )^x + x ^{\log x }$

Answer:

The given function is
$y = (\log x )^x + x ^{\log x }$
Let's take $t = (\log x)^x$
Now, take a look at both sides.
$\log t = x \log(\log x)$
Now, differentiate with respect. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}$

$= \log (\log x)+\frac{1}{\log x}\\$

$ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\$

$ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}$

$=(\log x)^x(\log (\log x))+ (\log x )^{x-1}$
Similarly, take $k = x^{\log x}$
Now, take a look at both sides.
$\log k = \log x \log x = (\log x)^2$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\$

$ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\$

$ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$
Therefore, the answer is $(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$

Question 8: Differentiate the functions with respect to. x. $(\sin x )^x + \sin ^{-1} \sqrt x$

Answer:

The given function is
$(\sin x )^x + \sin ^{-1} \sqrt x$
Let's take $t = (\sin x)^x$
Now, take a look at both sides.
$\log t = x \log(\sin x)$
Now, differentiate with respect. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}$

$= \log (\sin x)+x.\cot x$

$ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\$

$ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\$

$ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)$
Similarly, take $k = \sin^{-1}\sqrt x$
Now, differentiate with respect. x
We get,
$\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}$

$= \frac{1}{2\sqrt{x-x^2}}\\$

$ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Therefore, the answer is $(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$

Question 9: Differentiate the functions w.r.t. x $y=x^{\sin x}+(\sin x)^{\cos x}$

Answer:

The given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$

Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$

$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$

$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$

$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$

$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $

$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$

$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$

$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$

Question 10: Differentiate the functions with respect to. x. $x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$

Answer:

The given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$

Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$

$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$

$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$

$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$

$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $

$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$

$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$

$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$

Question 11: Differentiate the functions w.r.t. x. $( x \cos x )^ x + ( x \sin x )^{1/ x}$

Answer:

Given function is
$f(x)=( x \cos x)^ x + ( x \sin x )^{1/ x}$
Let's take $t = (x\cos x)^x$
Now, take a look at both sides.
$\log t =x\log (x\cos x) = x(\log x+\log \cos x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\$

$\frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) $

$\ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\$

$\frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\$

$\frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))$
Similarly, take $k = (x\sin x)^{\frac{1}{x}}$
Now, take a look at both sides.
$\log k = \frac{1}{x}(\log x+\log \sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+$

$\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\$

$\frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})$

$\ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\$

$\frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\$

$\frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Now,
$\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+$

$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Therefore, the answer is $(x\cos x)^x(1-x\tan x+\log (x\cos x))+$

$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$

Question 12: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15

$x ^ y + y ^ x = 1$ .

Answer:

The given function is
$f(x)=x ^ y + y ^ x = 1$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
We get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$

$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\$

$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$

$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\$

$ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\$

$\frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ $

$\frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

Therefore, the answer is $\frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

Question 13: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15.

$y^x = x ^y$

Answer:

The given function is
$f(x)\rightarrow x ^ y = y ^ x$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
We get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$

$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})$

$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate with respect. x
We get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$

$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ $

$\frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)\rightarrow \frac{dt}{dx}= \frac{dk}{dx}$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\$

$ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\$

$ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} $

$= \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

Therefore, the answer is $\frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

Question 14: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $(\cos x )^y = ( \cos y )^x$

Answer:

The given function is
$f(x)\rightarrow (\cos x) ^ y = (\cos y) ^ x$
Now, take the log on both sides.
$y\log \cos x = x \log \cos y$
Now, differentiate w.r.t x
$\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{dy}{dx}(\log \cos x)-y\tan x)$

$= (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )$

$= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}$

$= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )}$

$= \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Therefore, the answer is $\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Question 15: Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $xy = e ^{x-y}$

Answer:

The given function is
$f(x)\rightarrow xy = e ^{x-y}$
Now, take a look at both sides.
$\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)$
Now, differentiate w.r.t x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}$
Therefore, the answer is $\frac{y}{x}.\frac{x-1}{y+1}$

Question 16: Find the derivative of the function given by $f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find

f ' (1)

Answer:

The given function is
$y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Take logs on both sides.
$\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)$
NOW, differentiate with respect. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\$

$ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ $

$\frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Therefore, $f^{'}(x)=$

$ (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Now, the value of $f^{'}(1)$ is
$f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\$

$ f^{'}(1)=16.\frac{15}{2} = 120$

Question 17(1): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(i) By using the product rule

Answer:

The given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, we need to differentiate using the product rule.
$f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\$
$= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\$

$ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\$

$ = 5x^4 -20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4 -20x^3+45x^2-52x+11$

Question 17(2): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.

Answer:

The given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Multiply both to obtain a single higher-degree polynomial.
$f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$= x^5-5x^4+15x^3-26x^2+11x+72$
Now, differentiate with respect. x
We get,
$f^{'}(x)=5x^4-20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-52x+11$

Question 17(3): Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?

Answer:

The given function is
$y=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, take a look at both sides.
$\log y = \log (x^2-5x+8)+\log (x^3+7x+9)$
Now, differentiate with respect. x
We get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\$

$ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$

$ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$

$ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-56x+11$
And yes, they all give the same answer.

Question 18: If u, v and w are functions of x, then show that $\frac{d}{dx} ( u,v,w) = \frac{du}{dx} v.. w +u. \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx}$ in two ways - first by repeated application of the product rule, second by logarithmic differentiation.

Answer:

It is given that u, v and w are the functions of x
Let $y = u.v.w$
Now, we differentiate using the product rule with respect to x
First, take $y = u.(vw)$
Now,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u$ -(i)
Now, again, by the product rule
$\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v$
Put this in equation (i)
We get,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, by the product rule, we proved it.

Now, by taking the log
Again take $y = u.v.w$
Now, take a look at both sides.
$\log y = \log u + \log v + \log w$
Now, differentiate with respect. x
We get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} $

$= (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\$
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, we proved it by taking the log.

Question 1: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.

$x = 2at^2, y = at^4$

Answer:

The given equations are
$x = 2at^2, y = at^4$
Now, differentiate both with respect to t
We get,
$\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at$
Similarly,
$\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2$
Therefore, the answer is $\frac{dy}{dx}= t^2$

Question 2: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.

Answer:

The given equations are
$x= a \cos \theta , y = b \cos \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}$
Therefore, answer is $\frac{dy}{dx}= \frac{b}{a}$

Question 3: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$. $x = \sin t , y = \cos 2 t$

Answer:

The given equations are
$x = \sin t , y = \cos 2 t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t$
Similarly,
$\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t $

$\ \ \ \ \ (\because \sin 2x = \sin x\cos x)$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t$
Therefore, the answer is $\frac{dy}{dx} = -4\sin t$

Question 4: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\ \frac {dy}{dx}$

$x = 4t , y = 4/t$

Answer:

The given equations are
$x = 4t , y = 4/t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4$
Similarly,
$\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}$
Therefore, the answer is $\frac{dy}{dx} = \frac{-1}{t^2}$

Question 5: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2 \theta$

Answer:

The given equations are
$x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Therefore, answer is $\frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$

Question 6: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \theta - \sin \theta ), y = a ( 1+ \cos \theta )$

Answer:

Given equations are
$x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})$
Therefore, the answer is $\frac{dy}{dx}=-\cot \frac{\theta}{2}$

Question 7: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = \frac{\sin ^3 t }{\sqrt {\cos 2t }}, y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$

Answer:

Given equations are
$x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$
Now, differentiate both w.r.t
We get,
$\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}$

$=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$

$=\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}$
$=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos }{\sin x}=\cot x)$
Similarly,
$\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}$

$=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$

$=\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}$
$=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}}$

$= \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}$
$= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} $

$= \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}$
$=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}$
$(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)$
$=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}$

$\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t$ $\left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )$

Therefore, the answer is $\frac{dy}{dx} = -\cot 3t$

Question 8: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a ( \cos t + \log \tan t/2 ),y = a \sin t$

Answer:

Given equations are
$x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})$
$= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})$
Similarly,
$\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{dy}{dx} = \tan t$

Question 9: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a \sec \theta, y = b \ tan \theta$

Answer:

Given equations are
$x = a \sec \theta , y = b \ tan \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}$
Therefore, the answer is $\frac{dy}{dx} = \frac{b cosec \theta}{a}$

Question 10: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \cos \theta + \theta \sin \theta ), y = a ( \sin \theta - \theta \cos \theta )$

Answer:

The given equations are
$x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$
Now, differentiate both with respect to $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta$
Therefore, the answer is $\frac{dy}{dx}= \tan \theta$

Question 11: If $x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$ , show that $\frac{dy}{dx}$ = - y /x$

Answer:

The given equations are
$x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$

$xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\sin ^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c$

Differentiating with respect to x

$x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}$

Question 1: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x^2 + 3x+ 2$

Answer:

The given function is
$y=x^2 + 3x+ 2$
Now, differentiation with respect to. x
$\frac{dy}{dx}= 2x+3$
Now, the second-order derivative
$\frac{d^2y}{dx^2}= 2$
Therefore, the second order derivative is $\frac{d^2y}{dx^2}= 2$

Question 2: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x ^{20}$

Answer:

The given function is
$y=x ^{20}$
Now, differentiation with respect to. x
$\frac{dy}{dx}= 20x^{19}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}$
Therefore, second-order derivative is $\frac{d^2y}{dx^2}= 380x^{18}$

Question 3: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x \cos x$

Answer:

The given function is
$y = x \cos x$
Now, differentiation with respect to. x
$\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2}= -2\sin x - x\sin x$

Question 4: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\log x$

Answer:

The given function is
$y=\log x$
Now, differentiation with respect to. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2}= \frac{-1}{x^2}$

Question 5: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x ^3 \log x$

Answer:

The given function is
$y=x^3\log x$
Now, differentiation with respect. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2} = x(6.\log x+5)$

Question 6: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$e ^x \sin5 x$

Answer:

Given function is
$y= e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))$
$= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)$
$=2e^x(5\cos5x-12\sin5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)$

Question 7: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$e ^{6x}\cos 3x$

Answer:

Given function is
$y= e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)$
$= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)$
$e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)$
Therefore, second order derivative is $\frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)$

Question 8: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\tan ^{-1} x$

Answer:

Given function is
$y = \tan^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}$

Question 9: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\log (\log x )$

Answer:

Given function is
$y = \log(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}$

Question 10: Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\sin (\log x )$

Answer:

Given function is
$y = \sin(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}$
Now, the second order derivative is
Using Quotient rule
$\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$

Question 11: If $y = 5 \cos x - 3 \sin x$ prove that $\frac{d^2y}{dx^2}+y = 0$

Answer:

Given function is
$y = 5 \cos x - 3 \sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}$

$=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}$

$=-5\cos x+3\sin x$
Now,
$\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0$
Hence proved

Question 12: If $y = \cos ^{-1} x$ Find $\frac{d ^2 y }{dx^2 }$ in terms of y alone.

Answer:

Given function is
$y = \cos ^{-1} x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^2y}{dx^2}$ in terms of y
$y = \cos ^{-1} x$
$x = \cos y$
Now, put the value of x in (i)
$\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y$
$(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)$
Therefore, answer is $\frac{d^2y}{dx^2} = -2\cot y cosec y$

Question 13: If $y = 3 \cos (\log x) + 4 \sin (\log x)$ , show that $x^2 y_2 + xy_1 + y = 0$

Answer:

The given function is
$y = 3 \cos (\log x) + 4 \sin (\log x)$
Now, differentiation with respect to. x
$y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin(\log x)}{x}$ -(i)
Now, the second-order derivative is
By using the Quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}$
$=\frac{-\sin(\log x)+7\cos (\log x)}{x^2}$ -(ii)
Now, from equation (i) and (ii), we will get $y_1 \ and \ y_2$
Now, we need to show.
$x^2 y_2 + xy_1 + y = 0$
Put the value of $y_1 \ and \ y_2$ from equation (i) and (ii)
$x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x)$ $+4\sin(\log x)$
$-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\cos (\log x)$ $+4\sin(\log x)$
$=0$
Hence proved

Question 14: If $y = A e ^{mx} + Be ^{nx}$ , show that $\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + many = 0$

Answer:

The given function is
$y = A e ^{mx} + Be ^{nx}$
Now, differentiation with respect to. x
$\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}$ -(ii)
Now, we need to show.
$\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Put the value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})$
$m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}$ $+mnBe^{nx}$
$=0$
Hence proved

Question 15: If $y = 500 e ^{7x} + 600 e ^{- 7x }$ , show that $\frac{d^2 y}{dx ^2} = 49 y$
Answer:

The given function is
$y = 500 e ^{7x} + 600 e ^{- 7x }$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}$
$= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show.
$\frac{d^2 y}{dx ^2} = 49 y$
Put the value of $\frac{d^2y}{dx^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$= 24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved

Question 16: If $e ^y (x+1) = 1$ show that $\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$

Answer:

The given function is
$e ^y (x+1) = 1$
We can rewrite it as
$e^y = \frac{1}{x+1}$
Now, differentiation with respect to. x
$\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}$ -(i)
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ -(ii)
Now, we need to show.
$\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Put value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2$
$=\frac{1}{(x+1)^2}$
Hence, L.H.S. = R.H.S.
Hence proved

Question 17: If $y = (\tan^{-1} x)^2$ show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$

Answer:

The given function is
$y = (\tan^{-1} x)^2$
Now, differentiation with respect to. x
$y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}$ -(ii)
Now, we need to show.
$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Put the value from equations (i) and (ii)
$(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2$
Hence, L.H.S. = R.H.S.
Hence proved

Question 1: Differentiate with respect to. x the function in Exercises 1 to 11.

$( 3x^2 - 9x + 5 )^9$

Answer:

The given function is
$f(x)=( 3x^2 - 9x + 5 )^9$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d((3x^2-9x+5)^9)}{dx}= 9(3x^2-9x+5)^8.(6x-9)$
$= 27(2x-3)(3x^2-9x+5)^8$
Therefore, differentiation w.r.t. x is $27(3x^2-9x+5)^8(2x-3)$

Question 2: Differentiation with respect to. x the function in Exercises 1 to 11.

Answer:

The given function is
$f(x)= \sin ^3 x + \cos ^6 x$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}$
$=3\sin^2x.\cos x+6\cos^5x.(-\sin x)$
$=3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)$

Therefore, differentiation w.r.t. x is $3\sin x\cos x(\sin x- 2\cos ^4x)$

Question 3: Differentiate with respect. x the function in Exercises 1 to 11.

$( 5 x) ^{ 3 \cos 2x }$

Answer:

The given function is
$y=( 5 x) ^{ 3 \cos 2x }$
Take a log on both sides.
$\log y = 3\cos 2x\log 5x$
Now, differentiation w.r.t. x is
By using the product rule
$\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5$

$= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} $

$= y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\$

$ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )$

Therefore, differentiation w.r.t. x is $(5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )$

Question 4: Differentiate with respect to. x the function in Exercises 1 to 11.

$\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$

Answer:

The given function is
$f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}$
$=\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )$
$=\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )$
$=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Therefore, differentiation w.r.t. x is $\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Question 5: Differentiate with respect to. x the function in Exercises 1 to 11.

$\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$

Answer:

The given function is
$f(x)=\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$
Now, differentiation w.r.t. x is
By using the Quotient rule
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}$

$=\frac{\frac{d(\cos^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7})^2}\\$

$ f^{'}(x) = \frac{\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{1}{2}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7}\\$

$ f^{'}(x)= -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$
Therefore, differentiation w.r.t. x is $-\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$

Question 6: Differentiate with respect to. x the function in Exercises 1 to 11.

$\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$

Answer:

The given function is
$f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$
Now, rationalise the part.
$\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]$

$= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]$
$=\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2}$

$ \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}$
$(Using \ (a+b)^2=a^2+b^2+2ab)$
$=\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}$
$=\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}$
$=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)$
$=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}$
Given function reduces to
$f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}$

$=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}$
Therefore, differentiation w.r.t. x is $\frac{1}{2}$

Question 7: Differentiate with respect to. x the function in Exercises 1 to 11. $( \log x )^{ \log x } , x > 1$

Answer:

The given function is
$y=( \log x )^{ \log x } , x > 1$
Take logs on both sides.
$\log y=\log x\log( \log x )$
Now, differentiate with respect.
$\frac{1}{y}.\frac{dy}{dx}= \frac{1}{x}.\log (\log x)+\log x.\frac{1}{\log x}.\frac{1}{x} = \frac{\log x+1}{x}$
$\frac{dy}{dx} = y.\left ( \frac{\log x+1}{x} \right )\\$
$\frac{dy}{dx} = (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$
Therefore, differentiation w.r.t x is $(\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$

Question 8: $\cos ( \cos x + b \sin x )$, for some constant a and b. r:

The given function is
$f(x)=\cos ( a \cos x + b \sin x )$
Now, differentiation w.r.t x
$f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}$
$= -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}$
$= -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)$
$= (a\sin x-b\cos x)\sin(a\cos x+b\sin x).$
Therefore, differentiation w.r.t x $(a\sin x-b\cos x)\sin(a\cos x+b\sin x)$

Question 9: $(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$

Answer:

The given function is
$y=(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$
Take logs on both sides.
$\log y=(\sin x - \cos x)\log (\sin x - \cos x)$
Now, differentiate with respect. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{d(\sin x-\cos x)}{dx}.\log(\sin x- \cos x)+(\sin x- \cos x).\frac{d(\log(\sin x- \cos x))}{dx}$
$\frac{1}{y}.\frac{dy}{dx} =(\cos x -(-\sin x)).\log(\sin x-\cos x)+(\sin x- \cos x).\frac{(\cos x -(-\sin x))}{(\sin x- \cos x)}$
$\frac{dy}{dx} =y.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
$\frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
Therefore, differentiation w.r.t x is $(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right ), sinx>cosx$

Question 10: $x ^x + x ^a + a ^x + a ^a$ , for some fixed a > 0 and x > 0

Answer:

The given function is
$f(x)=x ^x + x ^a + a ^x + a ^a$
Let's take
$u = x^x$
Now, take a look at both sides.
$\log u = x \log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1)$ -(i)
Similarly, take $v = x^a$
Take logs on both sides.
$\log v = a\log x$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x}$ -(ii)

Similarly, take $z = a^x$
Take logs on both sides.
$\log z = x\log a$
Now, differentiate w.r.t x
$\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a$ -(iii)

Similarly, take $w = a^a$
Take logs on both sides.
$\log w = a\log a= \ constant$
Now, differentiate w.r.t x
$\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0$ -(iv)
Now,
$f(x)=u+v+z+w$
$f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}$
Put values from equations (i), (ii),(iii) and (iv)
$f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a$
Therefore, differentiation w.r.t. x is $x^x(\log x+1)+ax^{a-1}+a^x\log a$

Question 11: $x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$

Answer:

The given function is
$f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$
take $u=x ^{x^2 -3}$
Now, take a look at both sides.
$\log u=(x^2-3)\log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\$

$\frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\$

$ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\$

$ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$

$ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$ -(i)
Similarly,
take $v=(x-3)^{x^2}\\$
Now, take a look at both sides.
$\log v=x^2\log (x-3)$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\$

$ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\$

$ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\$

$ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\$

$ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\$ -(ii)
Now
$f(x)= u + v$
$f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}$
Put the value from equations (i) and (ii)
$f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$
Therefore, differentiation w.r.t x is $x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$

Question 12: Find $\frac{dy}{dx}$ if $y = 12 (1 - \cos t), x = 10 (t - \sin t),$ $-\frac{\pi }{2} <t< \frac{\pi }{2}$

Answer:

The given equations are
$y = 12 (1 - \cos t), x = 10 (t - \sin t),$
Now, differentiate both y and x w.r.t t independently.
$\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t$
And
$\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t$
Now
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\$
$(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
Therefore, differentiation w.r.t x is $\frac{6}{5}.\cot \frac{t}{2}$

Question 13: Find $\frac{dy}{dx}$ if $y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$

Answer:

The given function is
$y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$
Now, differentiate with respect. x
$\frac{dy}{dx}= \frac{d(sin ^{-1} x + sin^{-1} \sqrt{1- x^2})}{dx} = \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}}.\frac{d(\sqrt{1-x^2})}{dx}\\$

$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-1+x^2}}.\frac{1}{2\sqrt{1-x^2}}.(-2x)\\$

$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\$

$ \frac{dy}{dx}= 0$
Therefore, differentiate w.r.t. x is 0

Question 14: If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Answer:

The given function is
$x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0$
$x\sqrt{1+y} = - y\sqrt{1+x}$
Now, squaring both sides.
$(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\$

$ x^2+x^2y=y^2x+y^2\\$

$ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\$

$ x+y =-xy\\ y = \frac{-x}{1+x}$
Now, differentiate w.r.t. x is
$\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}$
Hence proved

Question 15: If $(x - a)^2 + (y - b)^2 = c^2$ , for some c > 0, prove that $\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\:$ is a constant independent of a and b.

Answer:

The given function is
$(x - a)^2 + (y - b)^2 = c^2$
$(y - b)^2 = c^2-(x - a)^2$ - (i)
Now, differentiate with respect. x
$\frac{d((x-a)^2)}{dx}+\frac{((y-b)^2)}{dx}=\frac{d(c^2)}{dx}\\ \\ 2(x-a)+2(y-b).\frac{dy}{dx}=0\\ \\ \frac{dy}{dx} = \frac{a-x}{y-b}$ -(ii)
Now, the second derivative
$\frac{d^2y}{dx^2} = \frac{\frac{d(a-x)}{dx}.(y-b) -(a-x).\frac{d(y-b)}{dx}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} =\frac{ (-1).(y-b)-(a-x).\frac{dy}{dx}}{(y-b)^2}\\ \\$
Now, put values from equations (i) and (ii)
$\frac{d^2y}{dx^2} =\frac{-(y-b)-(a-x).\frac{a-x}{y-b}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} = \frac{-((y-b)^2+(a-x)^2)}{(y-b)^\frac{3}{2}} = \frac{-c^2}{(y-b)^\frac{3}{2}}$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Now,
$\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}} = \frac{\left ( 1+\left ( \frac{x-a}{y-b} \right )^2 \right )^\frac{3}{2}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{\frac{\left ( (y-b)^2 +(x-a)^2\right )^\frac{3}{2}}{(y-b)^\frac{3}{2}}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{(c^2)^\frac{3}{2}}{-c^2}= \frac{c^3}{-c^2}= c$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Which is independent of a and b
Hence proved

Question 16: If $\cos y = x \cos (a + y)$ , with $\cos a \neq \pm 1$ , prove that $\frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }$

Answer:

The given function is
$\cos y = x \cos (a + y)$
Now, differentiate w.r.t x
$\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\$

$-\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\$

$ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\$

$ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) $

$\ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\$
$ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\$

$ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b)$

$ \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\$

$ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}$
Hence proved

Question 17: If $x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t),$ find $\frac{d^2 y }{dx^2 }$

Answer:

Given functions are
$x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t)$
Now, differentiate both the functions w.r.t. t independently.
We get
$\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)$
$=-a\sin t+a\sin t+at\cos t = at\cos t$
Similarly,
$\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))$
$= a\cos t -a\cos t+at\sin t =at\sin t$
Now,
$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t$
Now, the second derivative
$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}$
$(\because \frac{dx}{dt} = at\cos t \rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})$
Therefore, $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$

Question 18: If $f (x) = |x|^3$, show that f ''(x) exists for all real x and find it.

Answer:

The given function is
$f (x) = |x|^3$
$f(x)\left\{\begin{matrix} -x^3 & x<0\\ x^3 & x>0 \end{matrix}\right.$
Now, differentiate in both cases.
$f(x)= x^3\\ f^{'}(x)=3x^2\\ f^{''}(x)= 6x$
And
$f(x)= -x^3\\ f^{'}(x)=-3x^2\\ f^{''}(x)= -6x$
In both cases, f ''(x) exists.
Hence, we can say that f ''(x) exists for all real x
And values are
$f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.$

Question 19: Using the fact that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and the differentiation,
Obtain the sum formula for cosines.

Answer:

The given function is
$\sin (A + B) = \sin A \cos B + \cos A \sin B$
Now, differentiate with respect. x
$\frac{d(\sin(A+B))}{dx} = \frac{d\sin A}{dx}.\cos B+\sin A.\frac{d\cos B}{dx}+\frac{d\cos A}{dx}.\sin B+\cos A.\frac{d\sin B}{dx}$
$\cos (A+b)\frac{d(A+B)}{dx}$ $=\frac{dA}{dx}(\cos A\cos B-\sin A\cos B)+\frac{dB}{dx}(\cos A \sin B-\sin A\sin B)$
$=(\cos A \sin B-\sin A\sin B).\frac{d(A+B)}{dx}$
$\cos(A+B)= \cos A\sin B-\sin A\cos B$
Hence, we get the formula by differentiation of sin(A + B)

Question 20: Does there exist a function which is continuous everywhere but not differentiable
At exactly two points? Justify your answer.

Answer:

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere, and the sum of two continuous functions is also a continuous function.
Therefore, our function f(x) is continuous.
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0$ $(|h| = - h \ because\ h < 0)$
R.H.L. at x = 0
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{2h}{h}= 2$ $(|h| = h \ because \ h > 0)$
R.H.L. is not equal to L.H.L.
Hence. At x = 0, the function is not differentiable.
Now, Similarly
R.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{0}{h}= 0$ $(|h| = h \ because \ h > 0)$
L.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}$
$=\lim\limits_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{-2h}{h}= -2$ $(|h| = - h \ because\ h < 0)$
L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable

Question 21: If $y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$ , prove that $\frac{dy}{dx}$ = $\begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}$

Answer:

Given that
$y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$
We can rewrite it as
$y = f(x)(mc-bn)-g(x)(lc-an)+h(x)(lb-am)$
Now, differentiate w.r.t x
We will get
$\frac{dy}{dx} = f^{'}(x)(mc-bn)-g^{'}(x)(lc-an)+h^{'}(x)(lb-am) \rightarrow \begin{bmatrix} f^{'}(x) &g^{'}(x) &h^{'}(x) \\ l&m &n \\ a& b &c \end{bmatrix}$
Hence proved

Question 22: If $y=e^{a \cos ^{-1} x},-1 \leq x\leq 1$ , show that $( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$

Answer:

The given function is
$y=e^{a \cos ^{-1} x},-1 \leq 1$

Now, differentiate w.r.t x, we will get.
$\frac{dy}{dx}= \frac{d(e^{a\cos^{-1}x})}{dx}.\frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}$ -(i)
Now, again differentiate with respect to x
$\frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx}= \frac{-ae^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}.\sqrt{1-x^2}+ae^{a\cos^{-1}x}.\frac{1.(-2x)}{2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}$
$=\ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2}$ -(ii)
Now, we need to show that.
$( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$
Put the values from equations (i) and (ii)
$(1-x^2).\left ( \ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} \right )-x.\left ( \frac{-ae^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x}$
$a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}+\left ( \frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x} = 0$
Hence proved

Also, read,

• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.1
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.2
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.3
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.4
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.5
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.6
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.7
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.8
• Continuity And Differentiability Class 12 NCERT Solutions Miscellaneous Exercise

Class 12 Maths NCERT Chapter 5: Extra Question

Question: If $f(x)=x|x|_{\text {, then }} f^{\prime}(x)=$

Solution:

$\begin{aligned} & f(x)=x|x| \\ & f(x)=\left\{\begin{array}{cc}x^2 & x \geq 0 \\ -x^2 & x<0\end{array}\right\} \\ & f^{\prime}(x)=\left\{\begin{array}{cc}2 x & x \geq 0 \\ -2 x & x<0\end{array}\right\}\end{aligned}$

Hence, the answer is $f^{\prime}(x)=\left\{\begin{array}{cc}2 x & x \geq 0 \\ -2 x & x<0\end{array}\right\}$

Continuity and Differentiability Class 12 NCERT Solutions: Topics

Here are the topics that are discussed in the Class 12 Maths NCERT Chapter 5, Continuity and Differentiability.

• Introduction
• Continuity
• Algebra of continuous functions
• Differentiability
• Derivatives of composite functions
• Derivatives of implicit functions
• Derivatives of inverse trigonometric functions
• Exponential and Logarithmic Functions
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms
• Second Order Derivative

Continuity and Differentiability Class 12 NCERT Solutions: Important Formulae

Continuity:

A function $f(x)$ is continuous at a point $x = a$ if:

• $f(a)$ exists (finite, definite, and real).
• $\lim\limits_{x\rightarrow a} f(x)$ exists.
• $\lim\limits_ {x\rightarrow a} f(x) = f(a)$.
• $\lim\limits_{x\rightarrow 0} f(x) =f(0)$
• $\lim\limits _{x \rightarrow a} f(x)$ exists

Discontinuity:

$f(x)$ is discontinuous in an interval if it is discontinuous at any point in that interval.

Algebra of Continuous Functions:

The sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

The derivative of $f(x)$ at $x = a$, denoted as $f’(a)$, represents the slope of the tangent line to the graph.

Chain Rule:

If $f = v o u$, where $t = u(x)$, and if both $\frac{dt}{dx}$ and $\frac{dv}{dx}$ exist, then: $\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$.

Derivatives of Some Standard Functions:

• $\frac{d}{dt}(x^n) = nx^{n-1}$
• $\frac{d}{dt}(\sin x) = \cos x$
• $\frac{d}{dt}(\cos x) = -\sin x$
• $\frac{d}{dt}(\tan x) = \sec^2 x$
• $\frac{d}{dt}(\cot x) = -\csc^2 x$
• $\frac{d}{dt}(\sec x) = \sec x \cdot \tan x$
• $\frac{d}{dt}(\csc x) = -\csc x \cdot \cot x$
• $\frac{d}{dt}(a^x) = a^x \cdot ln(a)$
• $\frac{d}{dt}(e^x) = e^x$
• $\frac{d}{dt}(ln x) = \frac{1}{x}$

Mean Value Theorem:

The mean Value Theorem states that if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists some $c$ in $(a, b)$ such that: $f’(c) = \frac{(f(b) - f(a))}{(b - a)}$.

Rolle's Theorem:

Rolle's Theorem states that if $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists some $c$ in $(a, b)$ such that $f’(c) = 0$.

Lagrange's Mean Value Theorem:

Lagrange's Mean Value Theorem states that if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists some c in $(a, b)$ such that:

$f’(c) = \frac{(f(b) - f(a))}{(b - a)}$.

Approach to Solve Questions of Continuity and Differentiability

1. First of all, observe the question and check whether the limit of the given function exists or not.
2. If the limit of the function exists, then the next step is to check whether the function is continuous or discontinuous at the given limit or not.
3. If we want to check the differentiability the we have to check the left-hand derivative and right-hand derivative of the function. If they are equal, then the function is differentiable.
4. If it is given that the function is differentiable, then its continuity is assured, but the vice versa is not true.
5. If the function is differentiable, then we can use the Chain Rule, Differentiation using parametric form, etc.

What Extra Should Students Study Beyond NCERT for JEE?

NCERT Solutions for Class 12 Maths: Chapter Wise

Students can access all the Maths solutions from the NCERT book from the links below.

Also, read,

• NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability
• NCERT Notes Class 122 Chapter 5 Continuity and Differentiability

NCERT solutions for class 12 subject-wise

For subject-wise solutions, you can refer here

• NCERT Solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions class-wise

For the solution of other classes, you can refer here

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Here, you can refer to the latest syllabus and NCERT Books

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12