Q: Differentiate with respect to $x$ the function $\left(3 x^{2}-9 x+5\right)^{9}$.

Let $y=\left(3 x^{2}-9 x+5\right)^{9}$ Using chain rule, we get $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(3 x^{2}-9 x+5\right)^{9} \\ & =9\left(3 x^{2}-9 x+5\right)^{8} \cdot \frac{d}{d x}\left(3 x^{2}-9 x+5\right) \\ & =9\left(3 x^{2}-9 x+5\right)^{8} \cdot(6 x-9) \\ & =9\left(3 x^{2}-9 x+5\right)^{8} \cdot 3(2 x-3) \\ & =27\left(3 x^{2}-9 x+5\right)^{8}(2 x-3) \end{aligned} $$

Q: If $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5)$.

Given, $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function. Since every differentiable function is a continuous function, we obtain (i) $f$ is continuous on $[-5,5]$ (ii) $f$ is continuous on $(-5,5)$ Thus, by the Mean Value Theorem, there exists $c \in(-5,5)$ such that $\Rightarrow f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}$ $\Rightarrow 10 f^{\prime}(c)=f(5)-f(-5)$ It is also given that $f^{\prime}(x)$ does not vanish anywhere. Therefore, $f^{\prime}(c) \neq 0$ Thus, $\Rightarrow 10 f^{\prime}(c) \neq 0$ $\Rightarrow f(5)-f(-5) \neq 0$ $\Rightarrow f(5) \neq f(-5)$ Hence proved.

Q: Differentiate with respect to $x$ the function $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$.

Let $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$ Then, $$ \begin{aligned} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} & =\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \\ & =\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{(1+\sin x)-(1-\sin x)} \\ & =\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}=\frac{1+\cos x}{\sin x} \\ & =\frac{1+2 \cos ^{2} \frac{x}{2}-1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ & =\cot \frac{x}{2} \end{aligned} $$ Therefore, equation (1) becomes, $y=\cot ^{-1}\left(\cot \frac{x}{2}\right)$ $\Rightarrow y=\frac{x}{2}$ Thus, $\Rightarrow \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)$ $=\frac{1}{2}$

Question 1

Differentiate with respect to $x$ the function $\left(3 x^{2}-9 x+5\right)^{9}$.

Accepted Answer

Let $y=\left(3 x^{2}-9 x+5ight)^{9}$

Using chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(3 x^{2}-9 x+5ight)^{9} \

& =9\left(3 x^{2}-9 x+5ight)^{8} \cdot \frac{d}{d x}\left(3 x^{2}-9 x+5ight) \

& =9\left(3 x^{2}-9 x+5ight)^{8} \cdot(6 x-9) \

& =9\left(3 x^{2}-9 x+5ight)^{8} \cdot 3(2 x-3) \

& =27\left(3 x^{2}-9 x+5ight)^{8}(2 x-3)

\end{aligned}

$$

Question 2

Differentiate with respect to $x$ the function $\sin ^{3} x+\cos ^{6} x$.

Accepted Answer

Let $y=\sin ^{3} x+\cos ^{6} x$

Using chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\sin ^{3} xight)+\frac{d}{d x}\left(\cos ^{6} xight) \

& =3 \sin ^{2} x \cdot \frac{d}{d x}(\sin x)+6 \cos ^{5} x \cdot \frac{d}{d x}(\cos x) \

& =3 \sin ^{2} x \cdot \cos x+6 \cos ^{5} x \cdot(-\sin x) \

& =3 \sin x \cos x\left(\sin x-2 \cos ^{4} xight)

\end{aligned}

$$

Question 3

Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$

Accepted Answer

Given, $f(x)=x^{2}+2 x-8$, being polynomial function is continuous in $[-4,2]$ and also differentiable in $(-4,2)$.

$$

\begin{aligned}

f(-4) & =(-4)^{2}+2 \cdot(-4)-8 \

& =16-8-8 \

& =0

\end{aligned}

$$

\begin{aligned}

f(2) & =(2)^{2}+2 	imes 2-8 \

& =4+4-8 \

& =0

\end{aligned}

$$

Therefore, $f(-4)=f(2)=0$

The value of $f(x)$ at -4 and 2 coincides.

Rolle's Theorem states that there is a point $c \in(-4,2)$ such that $f^{\prime}(c)=0$

$f(x)=x^{2}+2 x-8$

Therefore, $f^{\prime}(x)=2 x+2$

Hence,

$f^{\prime}(c)=0$

$2 c+2=0$

$c=-1$

Thus, $c=-1 \in(-4,2)$

Hence, Rolle's Theorem is verified.

Question 4

Examine if Rolle's Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle's Theorem from these examples?

(i)

$$

f(x)=[x] 	ext { for } x \in[5,9]

$$

(ii) $\quad f(x)=[x]$ for $x \in[-2,2]$

(iii) $\quad f(x)=x^{2}-1$ for $x \in[1,2]$

Accepted Answer

By Rolle's Theorem, $f:[a, b] ightarrow \mathbf{R}$,

If

(a) $f$ is continuous on $[a, b]$

(b) $f$ is continuous on $(a, b)$

(c) $f(a)=f(b)$

Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$

Thus, Rolle's Theorem is not applicable to those functions that do not satisfy any of three conditions of the hypothesis.

(i) $\quad f(x)=[x]$ for $x \in[5,9]$

Since, the given function $f(x)$ is not continuous at every integral point.

In general, $f(x)$ is not continuous at $x=5$ and $x=9$

Therefore, $f(x)$ is not continuous in $[5,9]$

Also, $f(5)=[5]=5$ and $f(9)=[9]=9$

Thus, $f(5) 
eq f(9)$

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $n$ be an integer such that $n \in(5,9)$

The LHD of $f$ at $x=n$ is

$$

\lim _{h ightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h ightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h ightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h ightarrow 0^{-}} \frac{-1}{h}=\infty

$$

The RHD of $f$ at $x=n$ is

$$

\lim _{h ightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{h ightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h ightarrow 0^{+}} \frac{n-n}{h}=\lim _{h ightarrow 0^{+}} 0=0

$$

Since LHD and RHD of $f$ at $x=n$ are not equal, f is not differentiable at $x=n$ Therefore, $f$ is not differentiable in $(5,9)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Thus, Rolle's Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9]$.

(ii) $\quad f(x)=[x]$ for $x \in[-2,2]$

Since, the given function $f(x)$ is not continuous at every integral point.

In general, $f(x)$ is not continuous at $x=-2$ and $x=2$

Therefore, $f(x)$ is not continuous in $[-2,2]$

Also, $f(-2)=[-2]=-2$ and $f(2)=[2]=2$

Thus, $f(-2) 
eq f(2)$

The differentiability of $f$ in $(-2,2)$ is checked as follows.

Let $n$ be an integer such that $n \in(-2,2)$

The LHD of $f$ at $x=n$ is

$\lim _{h ightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h ightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h ightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h ightarrow 0^{-}} \frac{-1}{h}=\infty$

The RHD of $f$ at $x=n$ is

$\lim _{h ightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{h ightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h ightarrow 0^{+}} \frac{n-1-n}{h}=\lim _{h ightarrow 0^{+}} 0=0$

Since LHD and RHD of $f$ at $x=n$ are not equal, f is not differentiable at $x=n$

Therefore, $f$ is not differentiable in (-2,2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Thus, Rolle's Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2]$

(iii) $\quad f(x)=x^{2}-1$ for $x \in[1,2]$

Since, $f$ being a polynomial function is continuous in $[1,2]$ and is differentiable in $(1,2)$

Thus,

$f(1)=(1)^{2}-1=0$

$f(2)=(2)^{2}-1=3$

Therefore, $f(1) 
eq f(2)$

Since, $f$ does not satisfy a condition of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$.

Question 5

If $f:[-5,5] ightarrow \mathbf{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) 
eq f(5)$.

Accepted Answer

Given, $f:[-5,5] ightarrow \mathbf{R}$ is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(i) $f$ is continuous on $[-5,5]$

(ii) $f$ is continuous on $(-5,5)$

Thus, by the Mean Value Theorem, there exists $c \in(-5,5)$ such that

$\Rightarrow f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}$

$\Rightarrow 10 f^{\prime}(c)=f(5)-f(-5)$

It is also given that $f^{\prime}(x)$ does not vanish anywhere.

Therefore, $f^{\prime}(c) 
eq 0$

Thus,

$\Rightarrow 10 f^{\prime}(c) 
eq 0$

$\Rightarrow f(5)-f(-5) 
eq 0$

$\Rightarrow f(5) 
eq f(-5)$

Hence proved.

Question 6

Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the integral $[a, b]$, where $a=1$ and $b=4$.

Accepted Answer

Given, $f(x)=x^{2}-4 x-3$

$f$, being a polynomial function, is continuous in $[1,4]$ and is differentiable in $(1,4)$, whose derivative is $2 x-4$.

Thus,

$f(1)=1^{2}-4 	imes 1-3=-6$

$f(4)=4^{2}-4 	imes 4-3=-3$

Therefore,

$\frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}$

$$

\begin{aligned}

& =\frac{-3-(-6)}{3} \

& =\frac{3}{3} \

& =1

\end{aligned}

$$

Mean Value Theorem states that there is a point $c \in(1,4)$ such that $f^{\prime}(c)=1$ Hence,

$$

\begin{aligned}

& \Rightarrow f^{\prime}(c)=1 \

& \Rightarrow 2 c-4=1 \

& \Rightarrow c=\frac{5}{2} \quad\left[	ext { where } c=\frac{5}{2} \in(1,4)ight]

\end{aligned}

$$

Thus, mean value theorem is verified for the given function.

Question 7

Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b]$ where $a=1$ and $b=3$. Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$.

Accepted Answer

Given, $f$ is $f(x)=x^{3}-5 x^{2}-3 x$

$f$, being a polynomial function, is continuous in $[1,3]$ and is differentiable in $(1,3)$, whose derivative is $3 x^{2}-10 x-3$

Thus,

$f(1)=1^{3}-5 	imes 1^{2}-3 	imes 1=-7$

$f(3)=3^{3}-5 	imes 3^{2}-3 	imes 3=-27$

Therefore,

$$

\begin{aligned}

\frac{f(b)-f(a)}{b-a} & =\frac{f(3)-f(1)}{3-1} \

& =\frac{-27-(-7)}{3-1} \

& =-10

\end{aligned}

$$

Mean Value Theorem states that there exists a point $c \in(1,3)$ such that $f^{\prime}(c)=-10$ Hence,

$\Rightarrow f^{\prime}(c)=-10$

$\Rightarrow 3 c^{2}-10 c-3=-10$

$\Rightarrow 3 c^{2}-10 c+7=0$

$\Rightarrow 3 c^{2}-3 c-7 c+7=0$

$\Rightarrow 3 c(c-1)-7(c-1)=0$

$\Rightarrow(c-1)(3 c-7)=0$

$\Rightarrow c=1, \frac{7}{3} \quad\left[ight.$ where $\left.c=\frac{7}{3} \in(1,3)ight]$

Thus, Mean Value Theorem is verified for the given function and $c=\frac{7}{3} \in(1,3)$ is the only point for which $f^{\prime}(c)=0$.

Question 8

Examine the applicability of Mean Value Theorem for all three functions given

(i) $\quad f(x)=[x]$ for $x \in[5,9]$

(ii) $\quad f(x)=[x]$ for $x \in[-2,2]$

(iii) $\quad f(x)=x^{2}-1$ for $x \in[1,2]$

Accepted Answer

Mean Value Theorem states that for a function $f:[a, b] \rightarrow \mathbf{R}$, if

(a) $f$ is continuous on $[a, b]$

(b) $f$ is continuous on $(a, b)$

Then there exists some $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ Thus, Mean Value Theorem is not applicable to those functions that do not satisfy any of three conditions of the hypothesis.

(i) $\quad f(x)=[x]$ for $x \in[5,9]$

Since, the given function $f(x)$ is not continuous at every integral point.

In general, $f(x)$ is not continuous at $x=5$ and $x=9$

Therefore, $f(x)$ is not continuous in $[5,9]$

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $n$ be an integer such that $n \in(5,9)$

The LHD of $f$ at $x=n$ is

$\lim _{h \rightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$

The RHD of $f$ at $x=n$ is

$\lim _{h \rightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{+}} \frac{n-n}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

Since LHD and RHD of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$

Therefore, $f$ is not differentiable in $(5,9)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9]$

(ii) $\quad f(x)=[x]$ for $x \in[-2,2]$

Since, the given function $f(x)$ is not continuous at every integral point.

In general, $f(x)$ is not continuous at $x=-2$ and $x=2$

Therefore, $f(x)$ is not continuous in $[-2,2]$

The differentiability of $f$ in ( $-2,2$ ) is checked as follows.

Let n be an integer such that $n \in(-2,2)$

The LHD of $f$ at $x=n$ is

$\lim _{h \rightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$

The RHD of $f$ at $x=n$ is

$\lim _{h \rightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{x \rightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{+}} \frac{n-n}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

Since LHD and RHD of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$

Therefore, $f$ is not differentiable in $(-2,2)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2]$.

(iii) $\quad f(x)=x^{2}-1$ for $x \in[1,2]$

Since, $f$ being a polynomial function is continuous in $[1,2]$ and is differentiable in $(1,2)$ It is observed that $f$ satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$.

It can be proved as follows.

We have, $f(x)=x^{2}-1$

Then,

$f(1)=(1)^{2}-1=0$,

$f(2)=(2)^{2}-1=3$

Therefore,

$\frac{f(b)-f(a)}{b-a}=\frac{f(2)-f(1)}{2-1}=\frac{3-0}{1}$

$$

=3

$$

Hence, $f^{\prime}(x)=2 x$

Thus,

$\Rightarrow f^{\prime}(c)=3$

$\Rightarrow 2 c=3$

$\Rightarrow c=\frac{3}{2}$

$\Rightarrow c=1.5 \quad[$ where $1.5 \in[1,2]]$

\section*{MISCELLANEOUS EXERCISE}

Question 9

Differentiate with respect to $x$ the function $(5 x)^{3 \cos 2 x}$.

Accepted Answer

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

Taking logarithm on both the sides, we obtain

$\log y=3 \cos 2 x \log 5 x$

Differentiating both sides with respect to $x$, we get

$$

\begin{aligned}

\frac{1}{y} \frac{d y}{d x} & =3\left[\log 5 x \cdot \frac{d}{d x}(\cos 2 x)+\cos 2 x \cdot \frac{d}{d x}(\log 5 x)ight] \

\frac{d y}{d x} & =3 y\left[\log 5 x \cdot(-\sin 2 x) \cdot \frac{d}{d x}(2 x)+\cos 2 x \cdot \frac{1}{5 x} \cdot \frac{d}{d x}(5 x)ight] \

& =3 y\left[-2 \sin 2 x \cdot \log 5 x+\frac{\cos 2 x}{x}ight] \

& =y\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 xight] \

& =(5 x)^{3 \cos 2 x}\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 xight]

\end{aligned}

$$

Question 10

Differentiate with respect to $x$ the function $\sin ^{-1}(x \sqrt{x}), 0 \leq x \leq 1$.

Accepted Answer

Let $y=\sin ^{-1}(x \sqrt{x})$

Using chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x} \sin ^{-1}(x \sqrt{x}) \

& =\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} 	imes \frac{d}{d x}(x \sqrt{x}) \

& =\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}\left(x^{\frac{3}{2}}ight) \

& =\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{3}{2} \cdot x^{\frac{1}{2}} \

& =\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}} \

& =\frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}

\end{aligned}

$$

Question 11

Differentiate with respect to $x$ the function $\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2$.

Solution:

Let $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$

Using quotient rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\sqrt{2 x+7} \cdot \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}ight)-\left(\cos ^{-1} \frac{x}{2}ight) \cdot \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}} \

& =\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}ight)^{2}}} \cdot \frac{d}{d x}\left(\frac{x}{2}ight)ight]-\left(\cos ^{-1} \frac{x}{2}ight) \cdot \frac{1}{2 \sqrt{2 x+7}} \cdot \frac{d}{d x}(2 x+7)}{\frac{2 x+7}{\sqrt{2 x+7}} \cdot \frac{-1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}ight) \cdot \frac{2}{2 \sqrt{2 x+7}}} \

& =\frac{2 x+7}{\left(\sqrt{4-x^{2}}ight) \cdot(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)} \

& =-\left[\frac{1}{\sqrt{4-x^{2}} \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}ight]

\end{aligned}

$$

Accepted Answer

(No solution provided.)

Question 12

Differentiate with respect to $x$ the function $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$.

Accepted Answer

Let $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}ight]$

Then,

$$

\begin{aligned}

\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})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \

& =\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{(1+\sin x)-(1-\sin x)} \

& =\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}=\frac{1+\cos x}{\sin x} \

& =\frac{1+2 \cos ^{2} \frac{x}{2}-1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \

& =\cot \frac{x}{2}

\end{aligned}

$$

Therefore, equation (1) becomes,

$y=\cot ^{-1}\left(\cot \frac{x}{2}ight)$

$\Rightarrow y=\frac{x}{2}$

Thus,

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)$

$=\frac{1}{2}$

Question 13

Differentiate with respect to $x$ the function $(\log x)^{\log x}, x>1$.

Accepted Answer

Let $y=(\log x)^{\log x}$

Taking logarithm on both the sides, we obtain

$$

\log y=\log x \cdot \log (\log x)

$$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{aligned}

& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \cdot \log (\log x)] \

& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}[\log (\log x)] \

& \Rightarrow \frac{d y}{d x}=y\left[\log (\log x) \cdot \frac{1}{x}+\log x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)ight] \

& \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x} \cdot \log (\log x)+\frac{1}{x}ight] \

& \Rightarrow \frac{d y}{d x}=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}ight]

\end{aligned}

$$

Question 14

Differentiate with respect to $x$ the function $\cos (a \cos x+b \sin x)$, for some constant $a$ and $b$.

Accepted Answer

Let $y=\cos (a \cos x+b \sin x)$

Using chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x} \cos (a \cos x+b \sin x) \

& =-\sin (a \cos x+b \sin x) \cdot \frac{d}{d x}(a \cos x+b \sin x) \

& =-\sin (a \cos x+b \sin x) \cdot[a(-\sin x)+b \cos x] \

& =(a \sin x-b \cos x) \cdot \sin (a \cos x+b \sin x)

\end{aligned}

$$

Question 15

Differentiate with respect to $x$ the function $(\sin x-\cos x)^{(\sin x-\cos x)}, \frac{\pi}{4}<x<\frac{3 \pi}{4}$

Accepted Answer

Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$

Taking log on both the sides, we obtain

$$

\begin{aligned}

\log y & =\log \left[(\sin x-\cos x)^{(\sin x-\cos x)}ight] \

& =(\sin x-\cos x) \log (\sin x-\cos x)

\end{aligned}

$$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \log (\sin x-\cos x)]$

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

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

$\Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x) \cdot \log (\sin x-\cos x)+(\cos x+\sin x)]$

$\Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)[1+\log (\sin x-\cos x)]$

Question 16

If $\quad y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ a & b & c\end{array}ight|$, prove that $\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \ l & m & n \ a & b & c\end{array}ight|$

Accepted Answer

Given, $\quad y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ a & b & c\end{array}ight|$

$\Rightarrow y=(m c-n b) f(x)-(l c-n a) g(x)+(l b-m a) h(x)$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}[(m c-n b) f(x)]-\frac{d}{d x}[(l c-n a) g(x)]+\frac{d}{d x}[(l b-m a) h(x)] \

& =(m c-n b) f^{\prime}(x)-(l c-n a) g^{\prime}(x)+(l b-m a) h^{\prime}(x) \

& =\left|\begin{array}{ccc}

f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \

l & m & n \

a & b & c

\end{array}ight|

\end{aligned}

$$

Thus, $\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \ l & m & n \ a & b & c\end{array}ight|_{	ext {proved. }}$

Question 17

Differentiate with respect to $x$ the function $x^{x}+x^{a}+a^{x}+a^{a}$, for some fixed $a>0$ and $x>0$.

Accepted Answer

Let $y=x^{x}+x^{a}+a^{x}+a^{a}$

Also, let $x^{x}=u, x^{a}=v, a^{x}=w$ and $a^{a}=s$

Therefore,

$\Rightarrow y=u+v+w+s$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x} 	ag{1}

\end{equation*}

$$

Now, $u=x^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log x^{x}$

$\Rightarrow \log u=x \log x$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{align*}

\frac{1}{u} \frac{d u}{d x} & =\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x) \

\frac{d u}{d x} & =u\left[\log x \cdot 1+x \cdot \frac{1}{x}ight] \

& =x^{x}[\log x+1]=x^{x}(1+\log x) 	ag{2}

\end{align*}

$$

Now, $v=x^{a}$

Hence,

$$

\begin{align*}

\frac{d v}{d x} & =\frac{d}{d x}\left(x^{a}ight) \

& =a x^{a-1} 	ag{3}

\end{align*}

$$

Now, $w=a^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log w=\log a^{x}$

$\Rightarrow \log w=x \log a$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{align*}

\frac{1}{w} \frac{d w}{d x} & =\log a \cdot \frac{d}{d x}(x) \

\frac{d w}{d x} & =w \log a \

& =a^{x} \log a 	ag{4}

\end{align*}

$$

Now, $s=a^{a}$

Since $a$ is constant, $a^{a}$ is also a constant.

Hence,

$$

\begin{equation*}

\frac{d s}{d x}=0 	ag{5}

\end{equation*}

$$

From (1), (2), (3), (4) and (5), we obtain

$$

\begin{aligned}

\frac{d y}{d x} & =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a+0 \

& =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a

\end{aligned}

$$

Question 18

Differentiate with respect to $x$ the function $x^{x^{2}-3}+(x-3)^{x^{2}}$, for $x>3$.

Accepted Answer

Let $y=x^{x^{2}-3}+(x-3)^{x^{2}}$

Also, let $u=x^{x^{2}-3}$ and $v=(x-3)^{x^{2}}$

Therefore,

$y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Now, $u=x^{x^{2}-3}$

Taking logarithm on both the sides, we obtain

$$

\begin{aligned}

\log u & =\log \left(x^{x^{2}-3}ight) \

& =\left(x^{2}-3ight) \log x

\end{aligned}

$$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{align*}

& \frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^{2}-3ight)+\left(x^{2}-3ight) \cdot \frac{d}{d x}(\log x) \

& \Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^{2}-3ight) \cdot \frac{1}{x} \

& \Rightarrow \frac{d u}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log xight] 	ag{2}

\end{align*}

$$

Now, $v=(x-3)^{x^{2}}$

Taking logarithm on both the sides, we obtain

$$

\begin{aligned}

\log v & =\log (x-3)^{x^{2}} \

& =x^{2} \log (x-3)

\end{aligned}

$$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{align*}

& \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^{2}ight)+\left(x^{2}ight) \cdot \frac{d}{d x}[\log (x-3)] \

& \Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^{2} \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \

& \Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^{2}}{x-3} \cdot 1ight] \

& \Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2 x \log (x-3)ight] 	ag{3}

\end{align*}

$$

From (1), (2), and (3), we obtain

$\frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log xight]+(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2 x \log (x-3)ight]$

Question 19

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

Accepted Answer

The given function is $y=12(1-\cos t), x=10(t-\sin t)$

Hence,

$$

\begin{aligned}

\frac{d x}{d t} & =\frac{d}{d t}[10(t-\sin t)] \

& =10 \cdot \frac{d}{d t}(t-\sin t) \

& =10(1-\cos t) \

\frac{d y}{d t} & =\frac{d}{d t}[12(1-\cos t)] \

& =12 \cdot \frac{d}{d t}(1-\cos t) \

& =12 \cdot[0-(-\sin t)] \

& =12 \sin t

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{12 \sin t}{10(1-\cos t)} \

& =\frac{12.2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10.2 \sin ^{2} \frac{t}{2}} \

& =\frac{6}{5} \cot \frac{t}{2}

\end{aligned}

$$

Question 20

Find $\frac{d y}{d x}$, if $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}},-1 \leq x \leq 1$.

Accepted Answer

The given function is $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$

Hence,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left[\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}ight] \

& =\frac{d}{d x}\left(\sin ^{-1} xight)+\frac{d}{d x}\left(\sin ^{-1} \sqrt{1-x^{2}}ight) \

\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-\left(\sqrt{1-x^{2}}ight)^{2}}} \cdot \frac{d}{d x}\left(\sqrt{1-x^{2}}ight) \

& =\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{x} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot \frac{d}{d x}\left(1-x^{2}ight) \

& =\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 x \sqrt{1-x^{2}}}(-2 x) \

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

& =0

\end{aligned}

$$

Question 21

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ for $-1<x<1$, prove that $\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}$.

Accepted Answer

The given function is $x \sqrt{1+y}+y \sqrt{1+x}=0$

$\Rightarrow x \sqrt{1+y}=-y \sqrt{1+x}$

Squaring both sides, we obtain

$$

\begin{aligned}

& x^{2}(1+y)=y^{2}(1+x) \

& \Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2} \

& \Rightarrow x^{2}-y^{2}=x y^{2}-x^{2} y \

& \Rightarrow x^{2}-y^{2}=x y(y-x) \

& \Rightarrow(x+y)(x-y)=x y(y-x) \

& \Rightarrow x+y=-x y \

& \Rightarrow(1+x) y=-x \

& \Rightarrow y=\frac{-x}{(1+x)}

\end{aligned}

$$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{aligned}

\frac{d y}{d x} & =-\left[\frac{(1+x) \frac{d}{d x}(x)-(x) \cdot \frac{d}{d x}(1+x)}{(1+x)^{2}}ight] \

& =-\frac{(1+x)-x}{(1+x)^{2}} \

& =-\frac{1}{(1+x)^{2}}

\end{aligned}

$$

Hence proved.

Question 22

If $y=e^{a \cos ^{-1} x},-1 \leq x \leq 1$, show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Accepted Answer

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

Taking logarithm on both the sides, we obtain

$\Rightarrow \log y=a \cos ^{-1} x \log e$

$\Rightarrow \log y=a \cos ^{-1} x$

Differentiating both sides with respect to $x$, we obtain

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=a \cdot \frac{-1}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^{2}}}$

By squaring both the sides, we obtain

$\Rightarrow\left(\frac{d y}{d x}ight)^{2}=\frac{a^{2} y^{2}}{1-x^{2}}$

$\Rightarrow\left(1-x^{2}ight)\left(\frac{d y}{d x}ight)^{2}=a^{2} y^{2}$

Again, differentiating both sides with respect to $x$, we obtain

$\Rightarrow\left(\frac{d y}{d x}ight)^{2} \frac{d}{d x}\left(1-x^{2}ight)+\left(1-x^{2}ight) 	imes \frac{d}{d x}\left[\left(\frac{d y}{d x}ight)^{2}ight]=a^{2} \frac{d}{d x}\left(y^{2}ight)$

$\Rightarrow\left(\frac{d y}{d x}ight)^{2}(-2 x)+\left(1-x^{2}ight) 	imes 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow-x \frac{d y}{d x}+\left(1-x^{2}ight) \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y \quad\left[\frac{d y}{d x} 
eq 0ight]$

$\Rightarrow\left(1-x^{2}ight) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Hence proved.

Question 23

If $(x-a)^{2}+(y-b)^{2}=c^{2}$ for $c>0$, prove that

$$

\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}

$$

is a constant independent of $a$ and $b$.

Accepted Answer

The given function is $(x-a)^{2}+(y-b)^{2}=c^{2}$

Differentiating both sides with respect to $x$, we obtain

$\frac{d}{d x}\left[(x-a)^{2}ight]+\frac{d}{d x}\left[(y-b)^{2}ight]=\frac{d}{d x}\left(c^{2}ight)$

$\Rightarrow 2(x-a) \cdot \frac{d}{d x}(x-a)+2(y-b) \cdot \frac{d}{d x}(y-b)=0$

$\Rightarrow 2(x-a) \cdot 1+2(y-b) \cdot \frac{d y}{d x}=0$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{-(x-a)}{y-b} 	ag{1}

\end{equation*}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[\frac{-(x-a)}{y-b}ight] \

& =-\left[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^{2}}ight] \

& =-\left[\frac{(y-b)-(x-a) \cdot \frac{d y}{d x}}{(y-b)^{2}}ight] \

& =-\left[\frac{(y-b)-(x-a) \cdot\left\{\frac{-(x-a)}{y-b}ight\}}{(y-b)^{2}}ight] \quad[	ext { Using (1)] }] \

& =-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}ight]

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\frac{\left[1+\left(\frac{d y}{d x}ight)^{2}ight]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}= & \frac{\left[1+\frac{(x-a)^{2}}{(y-b)^{2}}ight]^{\frac{3}{2}}}{-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}ight]} \

= & \frac{\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{2}}ight]^{\frac{3}{2}}}{-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}ight]} \

= & \frac{\left[\frac{c^{2}}{(y-b)^{2}}ight]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}} \

= & \frac{c^{3}}{(y-b)^{3}} \

= & -\frac{c^{2}}{(y-b)^{3}} \

= & -c

\end{aligned}

$$

$-c$ is a constant and is independent of $a$ and $b$.

Hence proved.

Question 24

If $\cos y=x \cos (a+y)$ with $\cos a 
eq \pm 1$, prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$.

Accepted Answer

The given function is $\cos y=x \cos (a+y)$

Therefore,

$\Rightarrow \frac{d}{d x}[\cos y]=\frac{d}{d x}[x \cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}[\cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x \cdot[-\sin (a+y)] \frac{d y}{d x}$

$$

\begin{equation*}

\Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y) 	ag{1}

\end{equation*}

$$

Since, $\cos y=x \cos (a+y) \Rightarrow x=\frac{\cos y}{\cos (a+y)}$

Then, equation (1) becomes,

$\left[\frac{\cos y}{\cos (a+y)} \cdot \sin (a+y)-\sin yight] \frac{d y}{d x}=\cos (a+y)$

$\Rightarrow[\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)] \cdot \frac{d y}{d x}=\cos ^{2}(a+y)$

$\Rightarrow \sin (a+y-y) \frac{d y}{d x}=\cos ^{2}(a+y)$

$\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$

Hence proved.

Question 25

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

Accepted Answer

The given function is $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$ Therefore,

$$

\begin{aligned}

\frac{d x}{d t} & =a \cdot \frac{d}{d t}(\cos t+t \sin t) \

& =a\left[-\sin t+\sin t \cdot \frac{d}{d x}(t)+t \cdot \frac{d}{d t}(\sin t)ight] \

& =a[-\sin t+\sin t+t \cos t] \

& =a t \cos t

\end{aligned}

$$

\begin{aligned}

\frac{d y}{d t} & =a \cdot \frac{d}{d t}(\sin t-t \cos t) \

& =a\left[\cos t-\left\{\cos t \cdot \frac{d}{d t}(t)+t \cdot \frac{d}{d t}(\cos t)ight\}ight] \

& =a[\cos t-\{\cos t-t \sin t\}] \

& =a t \sin t

\end{aligned}

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{a t \sin t}{a t \cos t}=	an t \

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(\frac{d y}{d x}ight)=\frac{d}{d x}(	an t)=\sec ^{2} t \cdot \frac{d t}{d x} \

& =\sec ^{2} t \cdot \frac{1}{a t \cos t} \

& =\frac{\sec ^{3} t}{a t}, 0<t<\frac{\pi}{2}

\end{aligned} \quad\left[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}ight]

$$

Question 26

If $f(x)=|x|^{3}$, show that $f^{\prime \prime}(x)$ exists for all real $x$, and find it.

Accepted Answer

It is known that $|x|=\left\{\begin{array}{l}x, 	ext { if } x \geq 0 \ -x, 	ext { if } x<0\end{array}ight.$

Therefore, when $x \geq 0, f(x)=|x|^{3}=x^{3}$

In this case, $f^{\prime}(x)=3 x^{2}$ and hence, $f^{\prime \prime}(x)=6 x$

When $x<0, f(x)=|x|^{3}=(-x)^{3}=-x^{3}$

In this case, $f^{\prime}(x)=-3 x^{2}$ and hence, $f^{\prime \prime}(x)=-6 x$

Thus, for $f(x)=|x|^{3}, f^{\prime \prime}(x)$ exists for all real $x$ and is given by, $f^{\prime \prime}(x)=\left\{\begin{array}{l}6 x, 	ext { if } x \geq 0 \ -6 x, 	ext { if } x<0\end{array}ight.$

Question 27

Using mathematical induction prove that $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ for all positive integers $n$.

Accepted Answer

To prove: $P(n): \frac{d}{d x}\left(x^{n}ight)=n x^{n-1}$ for all positive integers $n$.

For $n=1$,

$P(1): \frac{d}{d x}(x)=1=1 \cdot x^{1-1}$

Therefore, $P(n)$ is true for $n=1$.

Let $P(k)$ is true for some positive integer $k$.

That is, $P(k): \frac{d}{d x}\left(x^{k}ight)=k x^{k-1}$

It has to be proved that $P(k+1)$ is also true.

Consider

$$

\begin{aligned}

\frac{d}{d x}\left(x^{k+1}ight) & =\frac{d}{d x}\left(x \cdot x^{k}ight) \

& =x^{k} \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}\left(x^{k}ight) \quad 	ext { [By applying product rule] } \

& =x^{k} \cdot 1+x \cdot k \cdot x^{k-1} \

\frac{d}{d x}\left(x^{k+1}ight) & =x^{k}+k x^{k} \

& =(k+1) \cdot x^{k} \

& =(k+1) \cdot x^{(k+1)-1}

\end{aligned}

$$

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Therefore, by the principle of mathematical induction, the statement $P(n)$ is true for every positive integer $n$.

Hence, proved.

Question 28

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

Accepted Answer

Given, $\sin (A+B)=\sin A \cos B+\cos A \sin B$

Differentiating both sides with respect to $x$, we obtain

$$

\begin{aligned}

& \frac{d}{d x}[\sin (A+B)]=\frac{d}{d x}(\sin A \cos B)+\frac{d}{d x}(\cos A \sin B) \

& \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \frac{d}{d x}(\sin A)+\sin A \cdot \frac{d}{d x}(\cos B)+\sin B \cdot \frac{d}{d x}(\cos A)+\cos A \cdot \frac{d}{d x}(\sin B) \

& \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \cos A \frac{d A}{d x}+\sin A(-\sin B) \frac{d B}{d x}+\sin B(-\sin A) \cdot \frac{d A}{d x}+\cos A \cos B \frac{d B}{d x} \

& \Rightarrow \cos (A+B) \cdot\left[\frac{d A}{d x}+\frac{d B}{d x}ight]=(\cos A \cos B-\sin A \sin B) \cdot\left[\frac{d A}{d x}+\frac{d B}{d x}ight] \

& \Rightarrow \cos (A+B)=\cos A \cos B-\sin A \sin B

\end{aligned}

$$

Question 29

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer?

Accepted Answer

Consider, $\quad y=\left\{\begin{array}{lc}|x| & -\infty<x \leq 1 \ 2-x & 1 \leq x \leq \infty\end{array}ight.$

![](https://cdn.mathpix.com/cropped/2025_10_31_3490434beff8fbc1043bg-133.jpg?height=527&width=695&top_left_y=276&top_left_x=688)

It can be seen from the above graph that the given function is continuous everywhere but not differentiable at exactly two points which are 0 and 1 .