Chapter 5 Miscellaneous (differentiation )

Differentiate w.r.t. x the function in Exercises 1 to 11.(Class 12 ncert solution math chapter 5 miscellaneous)

Question 1: $(3x^2 - 9x + 5)^9$

Solution: Let $y =(3x^2 - 9x + 5)^9$

Differentiate with respect to x

$\frac{dy}{dx}=\frac{d}{dx}(3x^2 - 9x + 5)^9$

$= 9(3x^2 - 9x + 5)^8\frac{d}{dx}(3x^2 - 9x + 5)$

$=9(3x^2 - 9x + 5)^8(6x-9)$

$=27(2x-3)(3x^2 - 9x + 5)^8$

Question 2: $\sin^3 x + \cos^6 x$

Solution: Let $y = \sin^3 x + \cos^6 x$

Differentiate with respect to x

$\frac{dy}{dx}=\frac{d}{dx}( \sin^3 x + \cos^6 x)$

$= 3\sin^2 x\frac{d}{dx}\sin x+6\cos^5 x\frac{d}{dx}\cos x$

$= 3\sin^2x\cos x+6\cos^5 x(-\sin x)$

$= 3\sin^2x\cos x-6\sin x\cos^5 x$

$= 3\sin x\cos x(\sin x-2\cos^4x)$

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

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

Taking ‘log’ both side

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

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

$\frac{d}{dx}\log y = 3\cos 2x\frac{d}{dx}\log 5x+3\log 5x\frac{d}{dx}\cos 2x$

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

$\Rightarrow \frac{dy}{dx}=y(3\cos 2x\frac{5}{5x}-6\log 5x \sin 2x)$

$\Rightarrow \frac{dy}{dx}= (5x)^{3\cos 2x}(\frac{3\cos 2x}{x}-6\log 5x.\sin 2x)$

Question 4: $\sin^{-1}(x\sqrt{x})$

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

Differentiate with respect to x

$\frac{dy}{dx} = \frac{d}{dx}\sin^{-1}(x\sqrt{x})$

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

$=\frac{1}{\sqrt{1-(x)^3}}\frac{d}{dx}x^{3/2}$

$=\frac{1}{\sqrt{1-(x)^3}}\frac{3}{2}x^{1/2}$

$=\frac{3x^{1/2}}{2\sqrt{1-(x)^3}}$

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

Question 5: $\frac{\cos{-1}\frac{x}{2}}{\sqrt{2x+7}}$

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

Differentiate with respect to x

$\frac{dy}{dx}=\frac{d}{dx}\frac{\cos{-1}\frac{x}{2}}{\sqrt{2x+7}}$

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

$=\frac{-\sqrt{2x+7}.\frac{1}{\sqrt{1-(x/2)^2}}.\frac{1}{2}-\cos^{-1}(\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.2}{2x+7}$
$=\frac{-\frac{2\sqrt{2x+7}}{2\sqrt{4-x^2}}-\frac{\cos^{-1}(x/2)}{\sqrt{2x+7}}}{2x+7}$

$=\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}.(2x+7)}-\frac{\cos^{-1}x/2}{\sqrt{2x+7}(2x+7)}$

$=-\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}-\frac{\cos^{-1}x/2}{(2x+7)^{3/2}}$

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

Question6: $\cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right],0<x<\frac{\pi}{2}$

Solution: Let $y=\cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] --(i)$

Then,

$\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$

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

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

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

$=\frac{2(1+\cos x)}{2\sin x}$

$=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}.\cos\frac{x}{2}}$

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

Therefore from equation (i)

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

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

Differentiate with respect to x

$\Rightarrow \frac{dy}{dx}= \frac{1}{2}$

Question 7: $(\log x)^{(\log x)},x>1$

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

Taking log both side, we obtain

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

$\Rightarrow \log y =\log x.\log(\log x)$

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

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

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

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

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

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

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

Solution: Let $y=\cos (a \cos x+b \sin x)$
Using chain rule, we get
$\frac{d y}{d x} =\frac{d}{d x} \cos (a \cos x+b \sin x)$

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

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

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

Solution: Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$
Taking log on both the sides, we obtain

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

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

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 10: Differentiate with respect to $x$ the function $x^x+x^a+a^x+a^a$ , for some fixed $a>0$ and $x>0$ .
Solution: 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,
$\Rightarrow y=u+v+w+s$

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

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

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

$\Rightarrow \frac{d u}{d x} =u\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]$

$=x^x[\log x+1]=x^x(1+\log x)$
Now, $v=x^a$
Hence,
$\frac{d v}{d x} & =\frac{d}{d x}\left(x^a\right)$

$=a x^{a-1}$

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

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

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

Now, $s=a^a$
Since $a$ is constant, $a^a$ is also a constant.
Hence,
$\frac{d s}{d x}=0$
From (1), (2), (3), (4) and (5), we obtain

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

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

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

Solution: Let $y=x^{x^2-3}+(x-3)^{x^2}$
Also, let $u=x^{x^2-3}$ and $v=(x-3)^{x^2}$ Therefore,
$y=u+v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

Now, $u=x^{x^2-3}$ Taking logarithm on both the sides, we obtain
$\log u =\log \left(x^{x^2-3}\right)$

$=\left(x^2-3\right) \log x$

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

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

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

$\Rightarrow \frac{d u}{d x}=x^{x^2-3}\left[\frac{x^2-3}{x}+2 x \log x\right]$

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

Taking logarithm on both the sides, we obtain
$\log v =\log (x-3)^{x^2}$
$=x^2 \log (x-3)$

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

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

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

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

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

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

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

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

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

Hence,

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

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

$=10(1-\cos t)$

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

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

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

$=12 \sin t$

Therefore,

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

$=\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^2 \frac{t}{2}}$

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

Question 13: Find $\frac{d y}{d x}$ , if $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2},0 \leq x \leq 1$ .

Solution: The given function is $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2} --(i)$

Let $I=\sin^{-1}\sqrt{1-x^2}$

Let $x =\cos z\Rightarrow z=\cos^{-1}x$

$I=\sin^{-1}(\sqrt{1-\cos^2z})$

$=\sin^{-1}\sin z =z$

$\Rightarrow I=\cos^{-1}x$

Hence,

From (i)

$y=\sin^{-1}x+\cos^{-1}x = \frac{\pi}{2}$

Differentiate with respect to x

$\frac{dy}{dx}= 0$

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

Solution: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

$x^2(1+y)=y^2(1+x)$

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

$\Rightarrow x^2-y^2=x y^2-x^2 y$

$\Rightarrow x^2-y^2=x y(y-x)$

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

$\Rightarrow x+y=-x y$

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

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

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

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

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

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

Hence proved.

Question 15: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 $b$ .

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

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

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

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

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

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

Therefore,

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

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

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

$=-\left[\frac{(y-b)-(x-a) \cdot\left\{\frac{-(x-a)}{y-b}\right\}}{(y-b)^2}\right] \quad[\text { Using (1)] }$

$=-\left[\frac{\left.(y-b)^2+(x-a)^2\right]}{(y-b)^3}\right]$

Hence,

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

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

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

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

$=-c$

$-c$ is a constant and 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 $\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\sin a}$ .
Solution: The given function is $\cos y=x \cos (a+y)$

Therefore,

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

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

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

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

Since, $\cos y=x\cos(a+y)$

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

Then, equation (1) becomes,

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

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

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

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

Hence proved.

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

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

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

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

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

$=a t \cos t$

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

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

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

$=a t \sin t$

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a t \sin t}{a t \cos t}=\tan t$

$\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tan t)=\sec ^2 t \cdot \frac{d t}{d x}$

$=\sec ^2 t . \frac{1}{a t \cos t} \quad\left[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}\right]$

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

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

Solution: It is known that $|x|=\left\{\begin{array}{l}x, \text { if } x \geq 0 \\ -x, \text { if } x<0\end{array}\right.$

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,

$f^{\prime \prime}(x)=\left\{\begin{array}{l} 6 x, \text { if } x \geq 0 \\ -6 x, \text { if } x<0 \end{array}\right.$

Question 19: Using mathematical induction prove that $\frac{d}{d x}\left(x^n\right)=n x^{n-1}$ for all positive integers $n$ .

Solution:To prove: $P(n): \frac{d}{d x}\left(x^n\right)=n x^{n-1}$ 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, $P(k): \frac{d}{d x}\left(x^k\right)=k x^{k-1}--(i)$

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

$P(k+1):\frac{d}{dx}x^{k+1}=(k+1)x^k$

Taking, L.H.S.

$\frac{d}{d x}\left(x^{k+1}\right) =\frac{d}{d x}\left(x \cdot x^k\right)$

$=x^k \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}\left(x^k\right) \quad \text { [By applying product rule] }$

$=x^k \cdot 1+x \cdot k \cdot x^{k-1} [\text{ from (i)}]$

$\frac{d}{d x}\left(x^{k+1}\right) =x^k+k x^k$

$=(k+1) \cdot x^k$

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

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

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

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

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

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

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

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

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

Question 21: Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer?

Solution: Consider, $\quad y= \begin{cases}|x| & -\infty<x \leq 1 \\ 2-x & 1 \leq x \leq \infty\end{cases}$

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.

Question 22: $y=\left|\begin{array}{ccc} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{array}\right| \text {, prove that } \frac{d y}{d x}=\left|\begin{array}{ccc} f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c \end{array}\right|$

Solution: $y =\begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n \\ a & b & c\end{vmatrix}$

Differentiate with respect to x

$\frac{dy}{dx}=\begin{vmatrix} \frac{d}{dx}f(x) & \frac{d}{dx}g(x) & \frac{d}{dx}h(x) \\ l & m & n\\ a & b & c\end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\\frac{d}{dx}l & \frac{d}{dx}m & \frac{d}{dx}n \\ a & b & c\end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x)\\l & m & n\\\frac{d}{dx}a & \frac{d}{dx}b & \frac{d}{dx}c\end{vmatrix}$

$=\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n\\ a & b & c\end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\ 0 & 0 & 0 \\ a & b & c\end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x)\\l & m & n \\ 0 & 0 & 0\end{vmatrix}$

$=\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n\\ a & b & c\end{vmatrix}+0+0$
$=\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n\\ a & b & c\end{vmatrix}$