Exercise 12.2 (Revised) - Chapter 13 - Limits & Derivatives - Ncert Solutions class 11 - Maths

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Chapter 12: Limits & Derivatives - NCERT Solutions for Class 11 Maths

Ex 12.2 Question 1.

Find the derivative of $x^2-2$ at $x=10$.

Answer.

$\frac{d}{d x}\left(x^2-2\right)=\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}(2)=2 x-0=2 x \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]$
Therefore, Derivative of $x^2-2$ at $x=10$ is $2 \times 10=20$
Ex 12.2 Question 2.

Find the derivative of $99 x$ at $x=100$.

Answer.

$\frac{d}{d x}(99 x)=99 \frac{d}{d x}(x)=99 \times 1=99$
Therefore, Derivative of $99 x$ at $x=100$ is 99
Ex 12.2 Question 3.

Find the derivative of $x$ at $x=1$.

Answer.

Here $\frac{d}{d x}(x)=1$
Therefore, Derivative of $x$ at $x=1$ is 1
Ex 12.2 Question 4.

Find the derivatives of the following functions from first principle:
(i) $x^3-27$
(ii) $(x-1)(x-2)$

(iii) $\frac{1}{x^2}$
(iv) $\frac{x+1}{x-1}$

Answer.

(i) Given: $f(x)=x^3-27$
$
\therefore f(x+h)=(x+h)^3-27
$

Now, since $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^3-27-x^3+27}{h} \\
& =\lim _{h \rightarrow 0} \frac{x^3+3 x^2 h+3 x h^2+h^3-27-x^3+27}{h} \\
& =\lim _{h \rightarrow 0} \frac{h\left(h^2+3 x^2+3 x h\right)}{h}=3 x^2
\end{aligned}
$
(ii) Given: $f(x)=(x-1)(x-2)=x^2-3 x+2$

Now $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\left[(x+h)^2-3(x+h)+2\right]-\left[x^2-3 x+2\right]}{h} \\
& =\lim _{h \rightarrow 0} \frac{x^2+2 x h+h^2-3 x-3 h+2-x^2+3 x-2}{h}
\end{aligned}
$

$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{2 x h+h^2-3 h}{h} \\
& =\lim _{h \rightarrow 0} \frac{h(h+2 x-3)}{h}=2 x-3
\end{aligned}
$
(iii) Given: $f(x)=\frac{1}{x^2}$
$
\therefore f(x+h)=\frac{1}{(x+h)^2}
$

Now, since $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} \\
& =\lim _{h \rightarrow 0} \frac{x^2-(x+h)^2}{h x^2(x+h)^2}=\lim _{h \rightarrow 0} \frac{x^2-x^2-h^2-2 x h}{h x^2(x+h)^2} \\
& =\lim _{h \rightarrow 0} \frac{h(-h-2 x)}{h x^2(x+h)^2}=\frac{-2 x}{x^2 \times x^2}=\frac{-2}{x^3}
\end{aligned}
$
(iv) $f(x)=\frac{x+1}{x-1}$
$
\therefore f(x+h)=\frac{x+h+1}{x+h-1}
$

Now, since $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}
$

$\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{h(x+h-1)(x-1)} \\
& =\lim _{h \rightarrow 0} \frac{\left(x^2-x+h x-h+x-1\right)+\left(x^2+h x-x+x+h-1\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{x^2-x+x h-h+x-1-x^2-x h+x-x-h+1}{h(x+h-1)(x-1)}
\end{aligned}$

$
=\lim _{h \rightarrow 0} \frac{-2 h}{h(x+h-1)(x-1)}=\frac{-2}{(x-1)^2}
$
Ex 12.2 Question 5.

For the function $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots .+\frac{x^2}{2}+x+1$, prove that $f^{\prime}(1)=100 f^{\prime}(0)$. Ans. Given: $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots .+\frac{x^2}{2}+x+1$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[\frac{x^{100}}{100}+\frac{x^9}{99}+\ldots . .+\frac{x^2}{2}+x+1\right]
$

$
\begin{aligned}
& =\frac{1}{100} \frac{d}{d x}\left(x^{100}\right)+\frac{1}{99} \frac{d}{d x}\left(x^{99}\right)+\ldots \ldots .+\frac{1}{2} \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}(x)+\frac{d}{d x}(1) \\
& {\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]} \\
& =\frac{1}{100} \times 100 x^{99}+\frac{1}{99} 99 x^{98}+\ldots . . .+\frac{1}{2} 2 x+1+0 \\
& =x^{\infty}+x^{98}+\ldots .+x+1
\end{aligned}
$

Now $f^{\prime}(1)=(1)^{98}+(1)^{98}+\ldots \ldots .+1+1=100$

And $f^{\prime}(0)=(0)^{99}+(0)^{98}+\ldots \ldots . .+0+1=1$
Now $f^{\prime}(1)=100 f^{\prime}(0)$
$
\begin{aligned}
& \Rightarrow 100=1 \times 100 \\
& \Rightarrow 100=100 \text { Proved }
\end{aligned}
$
Ex 12.2 Question 6.

Find the derivative of $x^n+a x^{n-1}+a^2 x^{n-2}+\ldots . . a^{n-1} x+a^n$ for some fixed real number a.

Answer.

Given: $f(x)=x^n+a x^{n-1}+a^2 x^{n-2}+a^{n-1} x+a^n$

$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[x^n+a x^{n-1}+a^2 x^{n-2}+\ldots .+a^{n-1} x+a^n\right] \\
& =\frac{d}{d x} x^n+\frac{d}{d x}\left(a x^{n-1}\right)+\frac{d}{d x}\left(a^2 x^{n-2}\right)+\ldots+\frac{d}{d x}\left(a^{n-1} x\right)+\frac{d}{d x} a^n \\
& {\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]}
\end{aligned}
$
$=n x^{n-1}+a(n-1) x^{n-2}+a^2(n-2) x^{n-3}+\ldots \ldots \ldots \ldots .+a^{n-1} \quad\left[\frac{d}{d x}(K)=0\right], \mathrm{K}$ is a constant

Ex 12.2 Question 7.

For some constants $a$ and $b$, find the derivative of:
(i) $(x-a)(x-b)$
(ii) $\left(a x^2+b\right)^2$
(iii) $\frac{x-a}{x-b}$

Answer.

(i) Given: $f(x)=(x-a)(x-b)$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(x-a)(x-b) \\
& \Rightarrow f^{\prime}(x)=(x-a) \frac{d}{d x}(x-b)+(x-b) \frac{d}{d x}(x-a)
\end{aligned}
$
$
\left[\begin{array}{l}
\frac{d}{d x}(K)=0 \\
\frac{d}{d x}(x)=1
\end{array}\right] \mathrm{K}
$
is a constant

$
\begin{aligned}
& =(x-a) \times 1+(x-b) \times 1 \\
& =x-a+x-b=2 x-a-b
\end{aligned}
$
(ii) Given: $f(x)=\left(a x^2+b\right)^2=a^2 x^4+b^2+2 a b x^2$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[a^2 x^4+b^2+2 a b x^2\right]
$

$\begin{array}{ll}
=a^2 \frac{d}{d x}\left(x^4\right)+\frac{d}{d x}\left(b^2\right)+2 a b \frac{d}{d x}\left(x^2\right) & {\left[\frac{d}{d x}(K)=0\right], \quad \mathrm{K} \text { is a constant }} \\
=a^2 \times 4 x^3+0+2 a b \times 2 x=4 a^2 x^3+4 a b x & {\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]}
\end{array}$

$
=4 a x\left(a x^2+b\right)
$
(iii) Given: $f(x)=\frac{x-a}{x-b} \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x-a}{x-b}\right)$
$\Rightarrow f^{\prime}(x)=\frac{(x-b) \frac{d}{d x}(x-a)-(x-a) \frac{d}{d x}(x-b)}{(x-b)^2}$
$\left[\begin{array}{l}\frac{d}{d x}(K)=0 \\ \frac{d}{d x}(x)=1\end{array}\right] \mathrm{K}$ is a
constant

$
\begin{aligned}
& =\frac{(x-b) \times 1-(x-a) \times 1}{(x-b)^2} \\
& =\frac{x-b-x+a}{(x-b)^2}=\frac{a-b}{(x-b)^2}
\end{aligned}
$
Ex 12.2 Question 8.

Find the derivative of $\frac{x^n-a^n}{x-a}$ for some constant $a$.

Answer.

Given: $\frac{x^n-a^n}{x-a}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x^n-a^n}{x-a}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{(x-a) \frac{d}{d x}\left(x^n-a^n\right)-\left(x^n-a^n\right) \frac{d}{d x}(x-a)}{(x-a)^2} \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]
\end{aligned}
$

$
=\frac{(x-a) n x^{n-1}-\left(x^n-a^n\right) \times 1}{(x-a)^2}
$
$\left[\frac{d}{d x}(K)=0\right], \mathrm{K}$ is a constant

$
=\frac{n x^n-a n x^{n-1}-x^n+a^n}{(x-a)^2}
$
Ex 12.2 Question 9.

Find the derivative of:
(i) $2 x-\frac{3}{4}$
(ii) $\left(5 x^3+3 x-1\right)(x-1)$
(iii) $x^3(5+3 x)$
(iv) $x^5\left(3-6 x^{-9}\right)$
(v) $x^{-4}\left(3-4 x^{-5}\right)$
(vi) $\frac{2}{x+1}-\frac{x^2}{3 x-1}$

Answer.

(i) Given: $f(x)=2 x-\frac{3}{4}$

$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(2 x-\frac{3}{4}\right) \\
& \Rightarrow f^{\prime}(x)=2 \frac{d}{d x} x-\frac{d}{d x}\left(\frac{3}{4}\right)=2 \times 1-0=2
\end{aligned}
$
$\left[\frac{d}{d x}(K)=0\right], \mathrm{K}$ is a constant
(ii) Given: $f(x)=\left(5 x^3+3 x-1\right)(x-1)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[\left(5 x^3+3 x-1\right)(x-1)\right]
$
[Product rule]

$
\begin{aligned}
& =\left(5 x^3+3 x-1\right) \frac{d}{d x}(x-1)+(x-1) \frac{d}{d x}\left(5 x^3+3 x-1\right) \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& =\left(5 x^3+3 x-1\right) \times 1+(x-1)\left(15 x^2+3\right) \\
& =5 x^3+3 x-1+15 x^3+3 x-15 x^2-3 \\
& =20 x^3-15 x^2+6 x-4
\end{aligned}
$
(iv) Given: $f(x)=x^5\left(3-6 x^{-9}\right)$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left[x^5\left(3-6 x^{-9}\right)\right]$
[Product rule]

$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=x^{-3} \frac{d}{d x}(5+3 x)+(5+3 x) \frac{d}{d x}\left(x^{-3}\right) \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& =x^{-3} \times 3+(5+3 x)\left(-3 x^{-4}\right)=\frac{3}{x^3}-\frac{3}{x^4}(5+3 x) \\
& =\frac{3}{x^3}\left(1-\frac{5+3 x}{x}\right)=\frac{3}{x^3}\left(\frac{x-5-3 x}{x}\right)=\frac{3}{x^3}\left(\frac{-5-2 x}{x}\right)=\frac{-3}{x^4}(5+2 x) \\
&
\end{aligned}
$
(iv) Given: $f(x)=x^5\left(3-6 x^{-9}\right)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[x^5\left(3-6 x^{-9}\right)\right]
$
[Product rule]

$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=x^5 \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x} x^5 \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& =x^5\left(54 x^{-10}\right)+\left(3-6 x^{-9}\right) \times 5 x^4=54 x^{-5}+15 x^4-30 x^{-5} \\
& =24 x^{-5}+15 x^4=\frac{24}{x^5}+15 x^4
\end{aligned}
$
(v) Given: $f(x)=x^{-4}\left(3-4 x^{-5}\right)$

$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[x^{-4}\left(3-4 x^{-5}\right)\right]
$
[Product rule]

$\begin{aligned}
& \Rightarrow f^{\prime}(x)=x^{-4} \frac{d}{d x}\left(3-4 x^{-5}\right)+\left(3-4 x^{-5}\right) \frac{d}{d x} x^{-4} \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& =x^{-4}\left(20 x^{-6}\right)+\left(3-4 x^{-5}\right)\left(-4 x^{-5}\right)=20 x^{-10}-12 x^{-5}+16 x^{-10} \\
& =36 x^{-10}-12 x^{-5}=\frac{36}{x^{10}}-\frac{12}{x^5}
\end{aligned}$

(vi) Given: $f(x)=\frac{2}{x+1}-\frac{x^2}{3 x-1}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[\frac{2}{x+1}-\frac{x^2}{3 x-1}\right] \\
& \Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\frac{2}{x+1}\right)-\frac{d}{d x}\left(\frac{x^2}{3 x-1}\right)
\end{aligned}
$
$
=\frac{(x+1) \frac{d}{d x}(2)-2 \frac{d}{d x}(x+1)}{(x+1)^2}-\frac{(3 x-1) \frac{d}{d x} x^2-x^2 \frac{d}{d x}(3 x-1)}{(3 x-1)^2}
$
(Quotient rule)
$
\begin{aligned}
& =\frac{(x+1) \times 0-2 \times 1}{(x+1)^2}-\frac{(3 x-1)(2 x)-x^2(3)}{(3 x-1)^2} \\
& =\frac{-2}{(x+1)^2}-\frac{6 x^2-2 x-3 x^2}{(3 x-1)^2}
\end{aligned}
$

$\begin{aligned}
& =\frac{-2}{(x+1)^2}-\frac{3 x^2-2 x}{(3 x-1)^2} \\
& =\frac{-2}{(x+1)^2}-\frac{x(3 x-2)}{(3 x-1)^2}
\end{aligned}$

Ex 12.2 Question 10.

Find the derivative of $\cos x$ from first principle.

Answer.

Given: $f(x)=\cos x$,
then $f(x+h)=\cos (x+h)$
Since, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h} \\
& {\left[\cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]} \\
& =\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h} \\
& =\lim _{h \rightarrow 0}\left[-\sin \left(\frac{2 x+h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right]
\end{aligned}
$

$
\begin{array}{ll}
=\lim _{h \rightarrow 0}-\sin \left(x+\frac{h}{2}\right) \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right) & {\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]} \\
{\left[h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\right]} & \\
=-\sin x \times 1=-\sin x &
\end{array}
$
Ex 12.2 Question 11.

Find the derivative of the following functions:
(i) $\sin x \cos x$
(ii) $\sec x$

(iii) $5 \sec x+4 \cos x$
(iv) $\operatorname{cosec} x$
(v) $3 \cot x+5 \operatorname{cosec} x$
(vi) $5 \sin x-6 \cos x+7$
(vii) $2 \tan x-7 \sec x$

Answer.

(i) Given: $f(x)=\sin x \cos x$
$
\therefore f^{\prime}(x)=\frac{d}{d x}(\sin x \cos x)
$
[Product rule]

$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\sin x \frac{d}{d x} \cos x+\cos x \frac{d}{d x} \sin x \\
& =\sin x(-\sin x)+\cos x \cos x \\
& =\cos ^2 x-\sin ^2 x=\cos 2 x
\end{aligned}
$
(ii) Given: $f(x)=\sec x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x} \sec x \\
& \Rightarrow f^{\prime}(x)=\sec x \tan x
\end{aligned}
$
(iii) Given: $f(x)=5 \sec x+4 \cos x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(5 \sec x+4 \cos x) \\
& \Rightarrow f^{\prime}(x)=5 \frac{d}{d x} \sec x+4 \frac{d}{d x} \cos x \\
& =5 \sec x \tan x-4 \sin x
\end{aligned}
$

(iv) Given: $f(x)=\operatorname{cosec} x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x} \operatorname{cosec} x \\
& \Rightarrow f^{\prime}(x)=-\operatorname{cosec} x \cot x
\end{aligned}
$
$
\begin{aligned}
& \text { (v) Given: } f(x)=3 \cot x+5 \operatorname{cosec} x \\
& \therefore f^{\prime}(x)=\frac{d}{d x}(3 \cot x+5 \operatorname{cosec} x) \\
& \Rightarrow f^{\prime}(x)=3 \frac{d}{d x} \cot x+5 \frac{d}{d x} \operatorname{cosec} x \\
& =-3 \operatorname{cosec} 2 x-5 \operatorname{cosec} x \cot x
\end{aligned}
$
(vi) Given: $f(x)=5 \sin x-6 \cos x+7$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(5 \sin x-6 \cos x+7) \\
& \Rightarrow f^{\prime}(x)=5 \frac{d}{d x} \sin x-6 \frac{d}{d x} \cos x+\frac{d}{d x} 7 \\
& =5 \cos x-6(-\sin x)+0
\end{aligned}
$

$
=5 \cos x+6 \sin x+0=5 \cos x+6 \sin x
$
(vii) Given: $f(x)=2 \tan x-7 \sec x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(2 \tan x-7 \sec x) \\
& \Rightarrow f^{\prime}(x)=2 \frac{d}{d x} \tan x-7 \frac{d}{d x} \sec x \\
& =2 \sec ^2 x-7 \sec x \tan x
\end{aligned}
$