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

Let $f(x)=\sin \left(x^{2}+5\right), u(x)=x^{2}+5$ and $v(t)=\sin t$ Then, $(v o u)(x)=v(u(x))=v\left(x^{2}+5\right)=\tan \left(x^{2}+5\right)=f(x)$ Thus, $f$ is a composite of two functions. Put $t=u(x)=x^{2}+5$ Then, we get $\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos \left(x^{2}+5\right)$ $\frac{d t}{d x}=\frac{d}{d x}\left(x^{2}+5\right)=\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(5)=2 x+0=2 x$ By chain rule of derivative, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos \left(x^{2}+5\right) \times 2 x=2 x \cos \left(x^{2}+5\right)$ \section*{Alternate method:} $$ \begin{aligned} \frac{d}{d x}\left[\sin \left(x^{2}+5\right)\right] & =\cos \left(x^{2}+5\right) \cdot \frac{d}{d x}\left(x^{2}+5\right) \\ & =\cos \left(x^{2}+5\right) \cdot\left[\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(5)\right] \\ & =\cos \left(x^{2}+5\right) \cdot[2 x+0] \\ & =2 x \cos \left(x^{2}+5\right) \end{aligned} $$

Q: Differentiate the function with respect to $x$ $$ \sec (\tan (\sqrt{x})) $$

Let $f(x)=\sec (\tan (\sqrt{x})), u(x)=\sqrt{x}, v(t)=\tan t$ and $w(s)=\sec s$ Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(\tan \sqrt{x})=\sec (\tan \sqrt{x})=f(x)$ Here, $f$ is a composite function of three functions $u, v$ and $w$. Put, $s=v(t)=\tan t$ and $t=u(x)=\sqrt{x}$ Then, $$ \begin{aligned} \frac{d w}{d s} & =\frac{d}{d s}(\sec s) \\ & =\sec s \tan s \\ & =\sec (\tan t) \cdot \tan (\tan t) \quad[s=\tan t] \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \quad[t=\sqrt{x}] \end{aligned} $$ Now, $$ \begin{aligned} & \frac{d s}{d t}=\frac{d}{d t}(\tan t)=\sec ^{2} t=\sec ^{2} \sqrt{x} \\ & \frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}\left(x^{\frac{1}{2}}\right)=\frac{1}{2} \cdot x^{\frac{1}{2}-1}=\frac{1}{2 \sqrt{x}} \end{aligned} $$ Hence, by chain rule, we get $$ \begin{aligned} \frac{d}{d x}[\sec (\tan \sqrt{x})] & =\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x} \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2} \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \\ & =\frac{\sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})}{2 \sqrt{x}} \end{aligned} $$ \section*{Alternate method:} $$ \begin{aligned} \frac{d}{d x}[\sec (\tan \sqrt{x})] & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x})}{2 \sqrt{x}} \end{aligned} $$

Q: Differentiate the function with respect to $x \cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$

Given, $\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$ $$ \begin{aligned} \frac{d}{d x}\left[\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)\right] & =\sin ^{2}\left(x^{5}\right) \times \frac{d}{d x}\left(\cos x^{3}\right)+\cos x^{3} \times \frac{d}{d x}\left[\sin ^{2}\left(x^{5}\right)\right] \\ & =\sin ^{2}\left(x^{5}\right) \times\left(-\sin x^{3}\right) \times \frac{d}{d x}\left(x^{3}\right)+\cos x^{3} \times 2 \sin \left(x^{5}\right) \cdot \frac{d}{d x}\left[\sin x^{5}\right] \\ & =-\sin x^{3} \sin ^{2}\left(x^{5}\right) \times 3 x^{2}+2 \sin x^{5} \cos x^{3} \cdot \cos x^{5} \times \frac{d}{d x}\left(x^{5}\right) \\ & =-3 x^{2} \sin x^{3} \cdot \sin ^{2}\left(x^{5}\right)+2 \sin x^{5} \cos x^{5} \cos x^{3} \times 5 x^{4} \\ & =10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2}\left(x^{5}\right) \end{aligned} $$

Q: Prove that the greatest integer function defined by $f(x)=[x], 0<x<3$ is not differentiable at $x=1$ and $x=2$.

Given, $f(x)=[x], 0<x<3$ It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both $\lim _{h \rightarrow 0^{-}} \frac{f(c)-f(c-h)}{h}$ and $\lim _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal. At $x=1$, Consider the LHD at $x=1$ $$ \begin{aligned} \lim _{h \rightarrow 0^{-}} \frac{f(1)-f(1-h)}{h} & =\lim _{h \rightarrow 0^{-}} \frac{[1]-[1-h]}{h} \\ & =\lim _{h \rightarrow 0^{-}} \frac{1-0}{h} \\ & =\lim _{h \rightarrow 0^{-}} \frac{1}{h} \\ & =\infty \end{aligned} $$ Consider RHD at $x=1$ $$ \begin{aligned} \lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} & =\lim _{h \rightarrow 0^{+}} \frac{[1+h]-[1]}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{1-1}{h} \\ & =\lim _{h \rightarrow 0^{+}} 0 \\ & =0 \end{aligned} $$ Since LHD and RHD at $x=1$ are not equal, Hence, $f$ is not differentiable at $x=1$. To check the differentiability of the given function at $x=2$, Consider LHD at $x=2$ $$ \begin{aligned} \lim _{h \rightarrow 0^{-}} \frac{f(2)-f(2-h)}{h} & =\lim _{h \rightarrow 0^{-}} \frac{[2]-[2-h]}{h} \\ & =\lim _{h \rightarrow 0^{-}} \frac{2-1}{h} \\ & =\lim _{h \rightarrow 0^{-}} \frac{1}{h} \\ & =\infty \end{aligned} $$ Now, consider RHD at $x=2$ $$ \begin{aligned} \lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h} & =\lim _{h \rightarrow 0^{+}} \frac{[2+h]-[2]}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{2-2}{h} \\ & =\lim _{h \rightarrow 0^{+}} 0 \\ & =0 \end{aligned} $$ Since, LHD and RHD at $x=2$ are not equal. Hence, $f$ is not differentiable at $x=2$. \section*{EXERCISE 5.3}

Question 1

Differentiate the function with respect to $x \sin (a x+b)$

Accepted Answer

Let $f(x)=\sin (a x+b), u(x)=a x+b$ and $v(t)=\sin t$

Then, $(v o u)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=f(x)$

Here, $f$ is a composite function of two functions $u$ and $v$.

Put, $t=u(x)=a x+b$

Thus,

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Hence, by chain rule, we get

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

\section*{Alternate method:}

$$

\begin{aligned}

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

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

& =\cos (a x+b) \cdot(a+0) \

& =a \cos (a x+b)

\end{aligned}

$$

Question 2

Differentiate the function with respect to $x$.

$\sin \left(x^{2}+5\right)$

Accepted Answer

Let $f(x)=\sin \left(x^{2}+5ight), u(x)=x^{2}+5$ and $v(t)=\sin t$

Then, $(v o u)(x)=v(u(x))=v\left(x^{2}+5ight)=	an \left(x^{2}+5ight)=f(x)$

Thus, $f$ is a composite of two functions.

Put $t=u(x)=x^{2}+5$

Then, we get

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos \left(x^{2}+5ight)$

$\frac{d t}{d x}=\frac{d}{d x}\left(x^{2}+5ight)=\frac{d}{d x}\left(x^{2}ight)+\frac{d}{d x}(5)=2 x+0=2 x$

By chain rule of derivative,

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos \left(x^{2}+5ight) 	imes 2 x=2 x \cos \left(x^{2}+5ight)$

\section*{Alternate method:}

$$

\begin{aligned}

\frac{d}{d x}\left[\sin \left(x^{2}+5ight)ight] & =\cos \left(x^{2}+5ight) \cdot \frac{d}{d x}\left(x^{2}+5ight) \

& =\cos \left(x^{2}+5ight) \cdot\left[\frac{d}{d x}\left(x^{2}ight)+\frac{d}{d x}(5)ight] \

& =\cos \left(x^{2}+5ight) \cdot[2 x+0] \

& =2 x \cos \left(x^{2}+5ight)

\end{aligned}

$$

Question 3

Differentiate the function with respect to $x \cos (\sin x)$

Accepted Answer

Let $f(x)=\cos (\sin x), u(x)=\sin x$ and $v(t)=\cos t$

Then, $(v o u)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$

Here, $f$ is a composite function of two functions.

Put $t=u(x)=\sin x$

$	herefore \frac{d v}{d t}=\frac{d}{d t}[\cos t]=-\sin t=-\sin (\sin x)$

$\frac{d t}{d x}=\frac{d}{d x}(\sin x)=\cos x$

By chain rule,

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

\section*{Alternate method:}

$$

\begin{aligned}

\frac{d}{d x}[\cos (\sin x)] & =-\sin (\sin x) \cdot \frac{d}{d x}(\sin x) \

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

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

\end{aligned}

$$

Question 4

Differentiate the function with respect to $x$

$$

\sec (	an (\sqrt{x}))

$$

Accepted Answer

Let $f(x)=\sec (	an (\sqrt{x})), u(x)=\sqrt{x}, v(t)=	an t$ and $w(s)=\sec s$

Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(	an \sqrt{x})=\sec (	an \sqrt{x})=f(x)$

Here, $f$ is a composite function of three functions $u, v$ and $w$.

Put, $s=v(t)=	an t$ and $t=u(x)=\sqrt{x}$

Then,

$$

\begin{aligned}

\frac{d w}{d s} & =\frac{d}{d s}(\sec s) \

& =\sec s 	an s \

& =\sec (	an t) \cdot 	an (	an t) \quad[s=	an t] \

& =\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \quad[t=\sqrt{x}]

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \frac{d s}{d t}=\frac{d}{d t}(	an t)=\sec ^{2} t=\sec ^{2} \sqrt{x} \

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

\end{aligned}

$$

Hence, by chain rule, we get

$$

\begin{aligned}

\frac{d}{d x}[\sec (	an \sqrt{x})] & =\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x} \

& =\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \cdot \sec ^{2} \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \

& =\frac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sec (	an \sqrt{x}) 	an (	an \sqrt{x}) \

& =\frac{\sec ^{2} \sqrt{x} \sec (	an \sqrt{x}) 	an (	an \sqrt{x})}{2 \sqrt{x}}

\end{aligned}

$$

\section*{Alternate method:}

$$

\begin{aligned}

\frac{d}{d x}[\sec (	an \sqrt{x})] & =\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \cdot \frac{d}{d x}(	an \sqrt{x}) \

& =\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \

& =\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \

& =\frac{\sec (	an \sqrt{x}) \cdot 	an (	an \sqrt{x}) \cdot \sec ^{2}(\sqrt{x})}{2 \sqrt{x}}

\end{aligned}

$$

Question 5

Differentiate the function with respect to $x$

$\frac{\sin (a x+b)}{\cos (c x+d)}$

Accepted Answer

Given, $f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}$, where $g(x)=\sin (a x+b)$ and $h(x)=\cos (c x+d)$

$	herefore f=\frac{g^{\prime} h-g h^{\prime}}{h^{2}}$

Consider $g(x)=\sin (a x+b)$

Let $u(x)=a x+b, v(t)=\sin t$

Then $(v o u)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=g(x)$

$	herefore g$ is a composite function of two functions, $u$ and $v$.

Put, $t=u(x)=a x+b$

$$

\begin{aligned}

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

& \frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a

\end{aligned}

$$

Thus, by chain rule, we get

$$

g^{\prime}=\frac{d g}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)

$$

Consider $h(x)=\cos (c x+d)$

Let $p(x)=c x+d, q(y)=\cos y$

Then, $(q o p)(x)=q(p(x))=q(c x+d)=\cos (c x+d)=h(x)$

$	herefore h$ is a composite function of two functions, $p$ and $q$.

Put, $y=p(x)=c x+d$

$$

\begin{aligned}

& \frac{d q}{d y}=\frac{d}{d y}(\cos y)=-\sin y=-\sin (c x+d) \

& \frac{d y}{d x}=\frac{d}{d x}(c x+d)=\frac{d}{d x}(c x)+\frac{d}{d x}(d)=c

\end{aligned}

$$

Using chain rule, we get

$$

\begin{aligned}

h^{\prime} & =\frac{d h}{d x}=\frac{d q}{d y} \cdot \frac{d y}{d x} \

& =-\sin (c x+d) 	imes c \

& =-c \sin (c x+d)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

f^{\prime} & =\frac{a \cos (a x+b) \cdot \cos (c x+d)-\sin (a x+b)\{-c \sin (c x+d)\}}{[\cos (c x+d)]^{2}} \

& =\frac{a \cos (a x+b)}{\cos (c x+d)}+c \sin (a x+b) \cdot \frac{\sin (c x+d)}{\cos (c x+d)} 	imes \frac{1}{\cos (c x+d)} \

& =a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) 	an (c x+d) \sec (c x+d)

\end{aligned}

$$

Question 6

Differentiate the function with respect to $x \cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$

Accepted Answer

Given, $\cos x^{3} \cdot \sin ^{2}\left(x^{5}ight)$

$$

\begin{aligned}

\frac{d}{d x}\left[\cos x^{3} \cdot \sin ^{2}\left(x^{5}ight)ight] & =\sin ^{2}\left(x^{5}ight) 	imes \frac{d}{d x}\left(\cos x^{3}ight)+\cos x^{3} 	imes \frac{d}{d x}\left[\sin ^{2}\left(x^{5}ight)ight] \

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

& =-\sin x^{3} \sin ^{2}\left(x^{5}ight) 	imes 3 x^{2}+2 \sin x^{5} \cos x^{3} \cdot \cos x^{5} 	imes \frac{d}{d x}\left(x^{5}ight) \

& =-3 x^{2} \sin x^{3} \cdot \sin ^{2}\left(x^{5}ight)+2 \sin x^{5} \cos x^{5} \cos x^{3} 	imes 5 x^{4} \

& =10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2}\left(x^{5}ight)

\end{aligned}

$$

Question 7

Differentiate the function with respect to $x$

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

Accepted Answer

$$

\begin{aligned}

\frac{d}{d x}\left[2 \sqrt{\cot \left(x^{2}ight)}ight] & =2 \cdot \frac{1}{2 \sqrt{\cot \left(x^{2}ight)}} 	imes \frac{d}{d x}\left[\cot \left(x^{2}ight)ight] \

& =\sqrt{\frac{\sin \left(x^{2}ight)}{\cos \left(x^{2}ight)}} 	imes-\operatorname{cosec}\left(x^{2}ight) 	imes \frac{d}{d x}\left(x^{2}ight) \

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

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

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

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

\end{aligned}

$$

Question 8

Differentiate the function with respect to $x \cos (\sqrt{x})$

Accepted Answer

Let $f(x)=\cos (\sqrt{x})$

Also, let $u(x)=\sqrt{x}$ and, $v(t)=\cos t$

Then,

$$

\begin{aligned}

(v o u)(x) & =v(u(x)) \

& =v(\sqrt{x}) \

& =\cos \sqrt{x} \

& =f(x)

\end{aligned}

$$

Since, $f$ is a composite function of $u$ and $v$.

$$

t=u(x)=\sqrt{x}

$$

Then,

$$

\begin{aligned}

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

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

& \frac{d v}{d t}=\frac{d}{d t}(\cos t)=-\sin t

\end{aligned}

$$

And, $=-\sin (\sqrt{x})$

Using chain rule, we get

$$

\begin{aligned}

\frac{d t}{d x} & =\frac{d v}{d t} \cdot \frac{d t}{d x} \

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

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

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

\end{aligned}

$$

\section*{Alternate method:}

$$

\begin{aligned}

\frac{d}{d x}[\cos (\sqrt{x})] & =-\sin (\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \

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

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

& =\frac{-\sin \sqrt{x}}{2 \sqrt{x}}

\end{aligned}

$$

Question 9

Prove that the function $f$ given by

$f(x)=|x-1|, x \in \mathbf{R}$ is not differentiable at $x=1$.

Accepted Answer

Given, $f(x)=|x-1|, x \in \mathbf{R}$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both $\lim _{h ightarrow 0^{-}} \frac{f(c)-f(c-h)}{h}$ and $\lim _{h ightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$,

Consider LHD at $x=1$

$$

\begin{aligned}

\lim _{h ightarrow 0^{-}} \frac{f(1)-f(1-h)}{h} & =\lim _{h ightarrow 0^{-}} \frac{f|1-1|-|1-h-1|}{h} \

& =\lim _{h ightarrow 0^{-}} \frac{0-|h|}{h} \

& =\lim _{h ightarrow 0^{-}} \frac{-h}{h} \quad(h<0 \Rightarrow|h|=-h) \

& =-1

\end{aligned}

$$

Consider RHD at $x=1$

$$

\begin{aligned}

\lim _{h ightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} & =\lim _{h ightarrow 0^{+}} \frac{f|1+h-1|-|1-1|}{h} \

& =\lim _{h ightarrow 0^{+}} \frac{|h|-0}{h} \

& =\lim _{h ightarrow 0^{+}} \frac{h}{h} \quad(h>0 \Rightarrow|h|=h) \

& =1

\end{aligned}

$$

Since LHD and RHD at $x=1$ are not equal,

Therefore, $f$ 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 $x=2$.

Accepted Answer

Given, $f(x)=[x], 0<x<3$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both $\lim _{h ightarrow 0^{-}} \frac{f(c)-f(c-h)}{h}$ and $\lim _{h ightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

At $x=1$,

Consider the LHD at $x=1$

$$

\begin{aligned}

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

& =\lim _{h ightarrow 0^{-}} \frac{1-0}{h} \

& =\lim _{h ightarrow 0^{-}} \frac{1}{h} \

& =\infty

\end{aligned}

$$

Consider RHD at $x=1$

$$

\begin{aligned}

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

& =\lim _{h ightarrow 0^{+}} \frac{1-1}{h} \

& =\lim _{h ightarrow 0^{+}} 0 \

& =0

\end{aligned}

$$

Since LHD and RHD at $x=1$ are not equal, Hence, $f$ is not differentiable at $x=1$.

To check the differentiability of the given function at $x=2$, Consider LHD at $x=2$

$$

\begin{aligned}

\lim _{h ightarrow 0^{-}} \frac{f(2)-f(2-h)}{h} & =\lim _{h ightarrow 0^{-}} \frac{[2]-[2-h]}{h} \

& =\lim _{h ightarrow 0^{-}} \frac{2-1}{h} \

& =\lim _{h ightarrow 0^{-}} \frac{1}{h} \

& =\infty

\end{aligned}

$$

Now, consider RHD at $x=2$

$$

\begin{aligned}

\lim _{h ightarrow 0^{+}} \frac{f(2+h)-f(2)}{h} & =\lim _{h ightarrow 0^{+}} \frac{[2+h]-[2]}{h} \

& =\lim _{h ightarrow 0^{+}} \frac{2-2}{h} \

& =\lim _{h ightarrow 0^{+}} 0 \

& =0

\end{aligned}

$$

Since, LHD and RHD at $x=2$ are not equal.

Hence, $f$ is not differentiable at $x=2$.

\section*{EXERCISE 5.3}