Exercise 10.1 - Chapter 10 Differentiability and Methods of Differentiation 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Differentiability and Methods of Differentiation
Ex 10.1
Question 1.
Find the derivatives of the following functions using first principle.
(i) $f(x)=6$
Solution:
Given $f(x)=6$
$f(x+h)=6$
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{6-6}{h} \\
& =\lim _{h \rightarrow 0} \frac{0}{h}
\end{aligned}
$
[ $\mathrm{h} \rightarrow 0$ means $\mathrm{h}$ is very nears to zero from left to right but not zero]
(ii) $f(x)=-4 x+7$
Solution:
Given $f(x)=-4 x+7$
$
\begin{aligned}
& \mathrm{f}(\mathrm{x}+\mathrm{h})=-4(\mathrm{x}+\mathrm{h})+7 \\
& =-4 \mathrm{x}-4 \mathrm{~h}+7 \\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{(-4 x-4 h+7)-(-4 x+7)}{h} \\
& =\lim _{h \rightarrow 0} \frac{-4 x-4 h+\not \lambda+A x-\not 7}{h} \\
& =\lim _{h \rightarrow 0} \frac{-4 h}{h} \\
& =\lim _{h \rightarrow 0}(-4) \\
&
\end{aligned}
$
(iii) $f(x)=-x^2+2$
Given $f(x)=-x^2+2$
$
f(x+h)=-(x+h)^2+2
$

$
\begin{aligned}
=-\mathrm{x}^2-\mathrm{h}^2 & -2 \mathrm{xh}+2 \\
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left(-x^2-h^2-2 x h+2\right)-\left(-x^2+2\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+22+x^2-\not 2}{h} \\
& =\lim _{h \rightarrow 0} \frac{-h(h+2 x)}{h} \\
& =-(0+2 x) \\
& =-2 x
\end{aligned}
$
Question 2.
Find the derivatives from the left and from the right at $\mathrm{x}=1$ (if they exist) of the following functions. Are the functions differentiable at $\mathrm{x}=1$ ?
(i) $f(x)=|x-1|$
Solution:

$\begin{aligned}
& f^{\prime}(a)=\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
& f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{|x-1|-|(1)-1|}{x-1} \\
&=\lim _{x \rightarrow 1^{-}} \frac{|x-1|-0}{x-1} \\
&=\lim _{x \rightarrow 1^{-}} \frac{-(x-1)}{x-1} \\
&=-1 \\
& f^{\prime}\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{|x-1|-|(1)-1|}{x-1} \\
&=\lim _{x \rightarrow 1^{+}} \frac{|x-1|-0}{x-1} \\
&=\lim _{x \rightarrow 1^{+}} \frac{+(x-1)}{(x-1)} \\
&=+1 \\
& f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right) \\
& f^{\prime}(1) \text { does not exist } \\
& \therefore \mathrm{S}^{\prime} \text { is not differentiable at } \mathrm{x}=1 .
\end{aligned}$

(ii) $f(x)=\sqrt{1-x^2}$
Solution:
$
\begin{aligned}
f^{\prime}(a) & =\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2-0}}{\sqrt{1-x^2-1}} \\
& =f^{\prime}(x) \rightarrow-\infty \text { as } x \rightarrow 1^{-}
\end{aligned}
$
$\therefore$ ' $\mathrm{f}$ ' is not differentiable at $\mathrm{x}=1$.
(iii)
$
f(x)= \begin{cases}x, & x \leq 1 \\ x^2, & x>1\end{cases}
$
Solution:
$
\begin{aligned}
f^{\prime}(a) & =\lim _{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{-}} \frac{x-1}{x-1} \\
& =1 \\
f^{\prime}\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{(x+1)(x-1)}{(x-1)} \\
& =\lim _{x \rightarrow 1^{+}} x+1 \\
& =1+1 \\
& =2 \\
f^{\prime}\left(1^{-}\right) & \neq f^{\prime}\left(1^{+}\right), f^{\prime}(x) \text { is not differentiate }
\end{aligned}
$
' $\mathrm{f}$ ' is not differentiable at $\mathrm{x}=1$

Question 3.
Determine whether the following functions is differentiable at the indicated values.
(i) $f(x)=x|x|$ at $x=0$
Solution:
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{h|h|-0}{h} \\
& =\lim _{h \rightarrow 0}|h| \\
& =0
\end{aligned}
$

Limits exists
Hence ' $f$ ' is differentiable at $x=0$.
(ii) $f(x)=\left|x^2-1\right|$ at $x=1$
Solution:
$
\begin{aligned}
f(x) & = \begin{cases}1-x^2 & \text { if } x \leq 0 \\
x^2-1 & \text { if } x>1\end{cases} \\
f^{\prime}\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{x^2-1-0}{x-1} \\
& =\lim _{x \rightarrow 1^{+}} \frac{(x+1)(x-1)}{(x-1)} \\
f^{\prime}\left(1^{-}\right) & =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\
& =\lim _{x \rightarrow 1^{-}}-\frac{(x-1)(x+1)}{(x-1)} \\
& =-2 \\
f^{\prime}\left(1^{+}\right) & \neq f^{\prime}\left(1^{-}\right)
\end{aligned}
$
$f(x)$ is not differentiable at $x=1$.
(iii) $f(x)=|x|+|x-1|$ at $x=0,1$
Solution:

$
f(x)=\left\{\begin{array}{ccc}
1-2 x & \text { if } & x<0 \\
1 & \text { if } & 0 \leq x \leq 1 \\
2 x-1 & \text { if } & x>1
\end{array}\right.
$
Clearly $f(x)$ is differentiable at $x$, where $x \neq 0, x \neq 1$.
$
\begin{aligned}
f^{\prime}\left(0^{+}\right) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{1-1}{x}=0 \\
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{1-2 x-1}{x}=-2 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
$
\begin{aligned}
& f\left(1^{+}\right)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}=-2 \\
& f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{1-1}{x-1}=0 \\
& f^{\prime}\left(1^{+}\right) \neq f^{\prime}\left(1^{-}\right) \\
& \therefore \mathrm{f}(\mathrm{x}) \text { is not differentiable at } \mathrm{x}=1 \text {. } \\
& \text { (iv) } f(x)=\sin |x| \text { at } x=0 \\
& \text { Solution: } \\
& f(x)=\left\{\begin{array}{clc}
-\sin x & \text { if } & x<0 \\
0 & \text { if } & x=0 \\
\sin x & \text { if } & x>0
\end{array}\right. \\
& f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{\sin x-0}{x .} \\
& =1 \text {. } \\
&
\end{aligned}
$

$
\begin{aligned}
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{-\sin x-0}{x} \\
& =-1 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$\therefore \mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

Question 4.
Show that the following functions are not differentiable at the indicated value of $x$.
(i)
$
f(x)=\left\{\begin{array}{ll}
-x+2 & ; x \leq 2 \\
2 x-4 & ; x>2
\end{array} ; x=2\right.
$
Solution:
$
\begin{aligned}
f(x) & = \begin{cases}-x+2 \quad ; x \leq 2 \\
2 x-4 \quad ; x>2\end{cases} \\
& =\lim _{x \rightarrow 2^{+}} \frac{2 x-4-0}{x-2} \\
& =\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{x-2} \\
& =2 \\
f^{\prime}\left(2^{-}\right) & =\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \\
& =\lim _{x \rightarrow 2^{-}} \frac{-x+2-0}{x-2} \\
& =\lim _{x \rightarrow 2^{-}} \frac{-(x-2)}{(x-2)}=\lim _{x \rightarrow 2^{-}}-1 \\
& =-1 \\
f^{\prime}\left(2^{+}\right) & \neq f^{\prime}\left(2^{-}\right)
\end{aligned}
$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $x=2$.
(ii)
$
f(x)=\left\{\begin{array}{cl}
3 x & ; x<0 \quad x=0 \\
-4 x & ; x \geq 0
\end{array}\right.
$
Solution:

$
\begin{aligned}
f^{\prime}\left(0^{+}\right) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{-4 x-0}{x-0} \\
& =\lim _{x \rightarrow 0^{+}} \frac{-4 x}{x} \\
& =-4 \\
f^{\prime}\left(0^{-}\right) & =\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{3 x-0}{x-0} \\
& =\lim _{x \rightarrow 0^{-}} \frac{3 x}{x} \\
& =3 \\
f^{\prime}\left(0^{+}\right) & \neq f^{\prime}\left(0^{-}\right)
\end{aligned}
$
$f(x)$ is not differentiable at $x=0$.
Question 5.
The graph off is shown below. State with reasons that $\mathrm{x}$ values (the numbers), at which $\mathrm{f}$ is not differentiable.
Solution:

(i) at $\mathrm{x}=-1$ and has vertical tangent at $x=-1$ and $x=$ 8(also At $x=-1$. The graph has shape edge $v$ ] and at $x=8$;The graph has shape peak ${ }^{\wedge}$ ]
(ii) At $x=4$ : The graph $f$ is not differentiable, since at $\mathrm{x}=4$. The graph $f$ ' is not continuous.
(iii) At $x=11$; The graph $f^{\prime}$ is not differentiable, since at $x=11$. The tangent line of the graph is perpendicular.
Question 6.
If $f(x)=|x+100|+x^2$, test whether $f^{\prime}(-100)$ exists.
Solution:

$\begin{aligned}
& f(x)=|\mathrm{x}+100|+\mathrm{x}^2 \\
& f^{\prime}\left(-100^{-}\right)=\lim _{x \rightarrow-100^{-}} \frac{f(x)-f(-100)}{x+100}=\lim _{x \rightarrow-100^{-}} \frac{-(x+100)+x^2-(0+10000)}{x+100} \\
&=\lim _{x \rightarrow-100^{-}} \frac{-x-100+x^2-10000}{x+100}=\lim _{x \rightarrow-100^{-}} \frac{x^2+x-10100}{x+100}=1 \\
&=\lim _{x \rightarrow-100^{-}} \frac{(x+100)(x-101)}{x+100}=-201 \\
& f^{\prime}\left(-100^{+}\right)=\lim _{x \rightarrow-100^{+}} \frac{f(x)-f(-100)}{x+100} \\
&=\lim _{x \rightarrow-100^{+}} \frac{x+100+x^2-10000}{x+100} \\
&=\lim _{x \rightarrow-100^{+}} \frac{x^2+x-9900}{x+100} \\
& \lim _{x \rightarrow-100^{+}} \frac{(x+100)(x-99)}{x+100}=-199 \\
& \Rightarrow f^{\prime}\left(-100^{-}\right) \neq f^{\prime}\left(100^{+}\right) \\
& f^{\prime}(-100) \text { does not exists }
\end{aligned}$

Question 7.
Examine the differentiability of functions in R by drawing the diagrams.
(i) $|\sin x|$
Solution:
$
|\sin x|=f(x)
$

Limit exist and continuous for all $x \in R$ clearly, differentiable at $R-\{n \pi n \in z$ ) Not differentiable at $x$ $=n \pi, n \in z$.
(ii) $|\cos x|$
Solution:
$
|\cos x|=f(x)
$

Limit exist and continuous for all $x \in R$ clearly, differentiable at $R\{(2 n+1) \pi / 2 / n \in z\}$ Not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$.