KC Sinha: Exercise 11.4- Mathematics Solution Class 12 Chapter 11 अवकलन

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Question 1

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following functions w.r.t x]

(i) $e^{x^{3}}$
Sol :
$y=e^{x^{3}}$

Taking log both sides

$logy=loge^{x^{3}}$

$[\because \log _{e} m^{n}=n \cdot \log _{e} m ]$

$\log y=x^{3} \cdot \log e$

[∵ loge=1]

log y=x³

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}=3x^2$

$\frac{d y}{d x}=3 x^{2} \cdot y$

$\frac{dy}{dx}=3 x^{2} \cdot e^{x^{3}}$

(ii) $e^{-x}$
Sol :
Let y=$e^{-x}$

Taking log both sides

log y=log e^-x

log y=-x log e

log y= -x

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{dx}=-1$

$\frac{d y}{dx}=-y$

-y=-e^-x

(iii) $e^{\cos x}$
Sol :
Let y=$e^{\cos x}$

Taking log both sides

log y=log e^{cos x}

log y=cos x. log e

log y=cos x

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=-\sin x$

$\frac{d y}{d x}=-\sin x \cdot y$

$\frac{dy}{d x}=-\sin x \cdot e^{\log x}$

(iv) sin(tan^-1 e^x)
Sol :
Let y=sin(tan^-1 e^x)

Differentiating w.r.t.x

$\frac{dy}{dx}=\cos \left(\tan ^{-1} e^{x}\right) \times \frac{1}{1+\left(e^{x}\right)^{2}} \times e^{x}$

$=\frac{e^{x}}{1+e^{2} x} \cdot \cos \left(\tan ^{-1} e^{x}\right)$

(v) $\sqrt{e^{\sqrt{x}}}, x>0$
Sol :
Let y=$\sqrt{e^{\sqrt{x}}}$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{1}{2 \sqrt{e^{\sqrt{x}}}} \times e^{\sqrt x} \times \frac{1}{2 \sqrt{x}}$

$=\frac{e^{\sqrt{x}}}{4 \sqrt{x} \cdot \sqrt{e^{\sqrt{x}}}}$

(vi) $a^{x^{2}}$
Sol :
y=$a^{x^{2}}$

Differentiating w.r.t.x

$\frac{d y}{d x}=a^{x^{2}} \cdot \log a \times 2 x$

$=2 x a^{x^{2}} \cdot \log _{a}$

(vii) $\frac{e^{x} \tan x+1}{\tan x}$
Sol :
Let y$=\frac{e^{x} \tan x+1}{\tan x}$

$y=\frac{e^{x} \tan x}{\tan x}+\frac{1}{\tan x}$

y=e^x+cotx

Differentiating w.r.t.x

$\frac{dy}{d x}=e^{x}-\operatorname{cosec}^{2} x$

Question 2

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following functions w.r.t. x]

(i) $\tan ^{-1}(\log x)$
Sol :
Let y=$\tan ^{-1}(\log x)$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{1}{1+(\log x)^{2}} \times \frac{1}{x}$

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

(ii) cos(sin(log x))
Sol :
Let y=cos(sin(log x))

Differentiating w.r.t.x

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

$=-\dfrac{1}{x} \cos \left(\log {x}\right) \cdot \sin (\sin (\log x))$

(iii) $\log \left(x^{2} \sqrt{x^{2}+1}\right)$
Sol :
Let y=$\log \left(x^{2} \sqrt{x^{2}+1}\right)$

$y=\log x^{2}+\log \sqrt{x^{2}+1}$

Differentiating w.r.t.x

$\frac{d y}{dx}=2 \times \frac{1}{x}+\frac{1}{2} \times \frac{1}{x^{2}+1} \times 2x$

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

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

$=\frac{3 x^{2}+2}{x\left(x^{2}+1\right)}$

(iv) $\log \left(\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}-x}}\right)$
Sol :
Let y=$\log \left(\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}-x}}\right)$

$y=\log (\sqrt{x^{2}+a^{2}}+x)-\log (\sqrt{x^{2}+a^{2}}-x)$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{1}{(\sqrt{x^{2}+a^{2}}+x)} \times\left(\frac{1}{2 \sqrt{x^{2}+a^{2}}} \times 2 x+1\right)-\frac{1}{(\sqrt{x^{2}+a^{2}-x})} \times\left(\frac{1}{2 \sqrt{x^{2}+a^{2}}} \times 2x-1\right)$

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

$=\frac{2}{\sqrt{x^{2}+a^{2}}}$

(v) $\log _{3}(\log x)$
Sol :
Let y=$\log _{3}(\log x)$

$[\because \log _{n} m=\frac{\log _{e} m}{\log _{e} n}]$

$y=\frac{\log (\log x)}{\log 3}$

$y=\frac{1}{\log 3} \cdot \log (\log x)$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{1}{\log 3} \cdot \frac{1}{\log x} \times \frac{1}{x}$

$=\frac{1}{\log 3 \cdot x \log x}$

(vi) $\sin \left(e^{x} \log x\right)$
Sol :
Let y=$\sin \left(e^{x} \log x\right)$

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

$=\cos \left(e^{x} \log {x}\right)\left[e^{x} \log{x}+\frac{e^{x}}{x}\right]$

Question 3

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following functions w.r.t. x]

(i) $e^{\sqrt{x}} \log (\cos x)$

Sol :

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

Differentiating w.r.t.x

$\frac{dy}{d x}=e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} \cdot \log \left(\cos x+e^{\sqrt{x}} \cdot \frac{1}{\cos x} \times(\sin x)\right.$

$=\frac{e \sqrt{x}}{2 \sqrt{x}} \log (\cos x)-e^{\sqrt{2}} \tan x$

(ii) $\frac{\cos x}{\log x}, x>0$

Sol :
Let y=$\frac{\cos x}{\log x}$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{-\sin x \cdot \log x-\cos x \cdot \frac{1}{x}}{(\log x)^{2}}$

$=\frac{\frac{-2 \sin x \log x-\cos x}{x}}{(\log x)^{2}}$

$=-\frac{x \sin x \log x+\cos x}{x\left(\log x\right)^{2}}$

(iii) $e^{\sec ^{2} x}+3 \cos ^{-1} x$
Sol :
Let y=$e^{\sec ^{2} x}+3 \cos ^{-1} x$

Differentiating w.r.t.x

$\frac{d y}{d x}=e^{s e c^{2} x} \cdot 2 \sec 2 \sec x \tan x+3\left(\frac{-1}{\sqrt{1-x^{2}}}\right)$

$=e^{\sec ^{2} x} \cdot 2 \sec ^{2} x \tan x-\frac{3}{\sqrt{1-x^{2}}}$

Question 4

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following w.r.t. x]

(i) xlog x-x
Sol :
Let y=xlog x-x

Differentiating w.r.t.x

$\frac{d y}{d x}=1 . \log x+x \cdot \frac{1}{x}-1$

=log x+1-1

$\frac{d y}{d x}=\log x$

(ii) $\frac{\log x}{x}$
Sol :
Let y=$\frac{\log x}{x}$

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{\frac{1}{x} \cdot x-\log x \cdot 1}{x^{2}}$

$=\frac{1-\log x}{x^{2}}$

(iii) $x^{4} \cdot \log x$
Sol :
Let y=$x^{4} \cdot \log x$

Differentiating w.r.t.x

$\frac{d y}{d x}=4 x^{3} \cdot \log x+x^{4} \cdot \frac{1}{x}$

$=4 x^{3} \log x+x^{3}$

(iv) $\log x \cdot \sin e^{x}$
Sol :
Let y=$\log x \cdot \sin e^{x}$

Differentiating w.r.t.x

$\frac{dy}{d x}=\frac{1}{2} \cdot \sin^{2} x+\log x \cos e^{x} \times e^{x}$

$=\frac{\sin e^{x}}{x}+e^{x} \cos e^{x} \cdot \log x$

Question 5

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें।

[Differentiate the following w.r.t. x]

(i) $e^{x} \sin x \log x$
Sol :
Let y=$e^{x} \sin x \log x$

Differentiating w.r.t.x

$\frac{d y}{d x}=e^{x} \sin x \log x+e^{x} \cos x \log x+e^{x} \sin x \cdot \frac{1}{x}$

$=e^{x}\left(\sin {x}+\cos x\right) \log x+\frac{e^{x}}{x} \sin x$

(ii) $\frac{e^{x}+\sin x}{1+\log x}$
Sol :
Let y=$\frac{e^{x}+\sin x}{1+\log x}$

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{\left(e^{x}+\cos x \cdot(1+\log x)-\left(e^{x}+\sin x\right) \frac{1}{x}\right.}{(1+\log x)^{2}}$

$=\frac{\frac{\left(1+\log x\right) \cdot x\left(e^{x}+\cos x\right)-e^{x}-\sin x}{x}}{(1+\log x)^{2}}$

$\frac{d y}{d x}=\frac{(1+\log x) \cdot x\left(e^{x}+\cos x \right)-e^x-\sin x}{x \cdot\left(1+\log x^{2}\right)}$

(iii) $\log \left[e^{x}\left(\frac{x-2}{x+2}\right)^{3 / 4}\right]$
Sol :
Let y=$\log \left[e^{x}\left(\frac{x-2}{x+2}\right)^{3 / 4}\right]$

$y=\log e^ x+\log \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}$

$y=x \cdot \log e+\frac{3}{4} \cdot \log \left(\frac{x-2}{x+2}\right)$

$y=x+\frac{3}{4}[\log (x-2)-\log (x+2)]$

Differentiating w.r.t.x

$\frac{d y}{d x}=1+\frac{3}{4}\left[\frac{1}{x-2}-\frac{1}{x+2}\right]$

$=1+\frac{3}{4}\left[\frac{(x+2)-(x-2)}{(x-2)(x+2)}\right]$

$=1+\frac{3}{4}\left[\frac{x+2-x+2}{x^{2}-2^{2}}\right]$

$=1+\frac{3}{4}\left[\frac{4}{x^{2}-4}\right]$

$\frac{dy}{dx}=\frac{x^{2}-4+3}{x^{2}-4}$

$=\frac{x^{2}-1}{x^{2}-4}$

(iv) $\frac{x \log x}{e^{x} \tan x}$
Sol :
Let y=$\frac{x \log x}{e^{x} \tan x}$

Differentiating w.r.t.x

$\frac{dy}{d x}=\frac{\left[1 \cdot \log x+x \cdot \frac{1}{x}\right] e^{x} \tan x-x \log x\left[e^{x} \tan x+e^{x}\sec ^2 x\right] }{\left(e^{x} \cdot \tan x\right)^{2}}$

$=\frac{e^{x}\left[(\log x+1) \tan x-x \log x\left(\tan x+\sec ^{2} x\right)\right]}{\left(e^{x}\right)^{2} \cdot \tan ^{2} x}$

$=\frac{\tan x\left( \log{x}+1\right)-x \log x\left(\tan x+\sec ^{2} x\right)}{e^{x}-\tan ^{2} x}$

Question 6

यदि (If) $y=a^{x}+\sqrt{\frac{1+x}{1-x}}$ ,निकाले ( find )$\frac{d y}{d x}$ at x=0

Sol :
y=$y=a^{x}+\sqrt{\frac{1+x}{1-x}}$

Differentiating w.r.t.x

$\frac{d y}{d x}=a^{x} \cdot \log _{e} a+\frac{1}{2 \sqrt{\frac{1+x}{1-x}}} \times \frac{1 \cdot(1-x)-(1+x)(-1)}{(1-x)^{2}}$

$=a^{x} \cdot \log {e^{a}}+\frac{1}{2} \sqrt{\frac{1-x}{1+x}} \cdot \frac{1-x+1+x}{(1-x)^{2}}$

$\frac{dy}{dx}=a^{x} \cdot \log _{e} a+\frac{1}{2} \sqrt{\frac{1-x}{1+x}} \cdot \frac{2}{(1-x)^{2}}$

$\frac{d y}{d x}=a^{x} \cdot \log e^{a}+\sqrt{\frac{1-x}{1+x}} \frac{1}{(1-x)^{2}}$

At x=0
$\left(\frac{d y}{d x}\right)_{x=0}=a^{0} \cdot \log _{e} a+\sqrt{\frac{1-0}{1+0}} \cdot \frac{1}{(1-0)^{2}}$

$=\log _{e} a+1$

Question 7

यदि (If) x=ylog (xy) दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{y(x-y)}{x(x+y)}$
Sol :
x=ylog (xy)

x=y[logx+logy]

$\log x+\log y=\frac{x}{y}$

Differentiating w.r.t.x

$1=\frac{dy}{dx}\left[\log x+\log y\right]+y \cdot\left[\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x}\right]$

$1=\frac{x}{y} \frac{d y}{d x}+\frac{y}{x}+\frac{d y}{dx}$

$1-\frac{y}{x}=\left(\frac{x}{y}+1\right) \frac{d y}{dx}$

$\frac{x-y}{x}=\left(\frac{x+y}{y}\right) \frac{d y}{d x}$

$\frac{dy}{dx}=\frac{y(x-y)}{x(x+y)}$

Question 8

$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\text{to }\infty}}}$ सिद्ध करें कि ( prove that ) $\frac{d y}{d x}=\frac{1}{x(2 y-1)}$
Sol :
$y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\text{to }\infty}}}$

$y=\sqrt{\log x+y}$

Squaring both sides

$y^{2}=\log x+y$

Differentiating w.r.t.x

$2 y \cdot \frac{d y}{d x}=\frac{1}{x}+\frac{d y}{d x}$

$2 y \frac{d y}{d x}-\frac{dy}{dx}=\frac{1}{x}$

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

$\frac{dy}{dx}=\frac{1}{x(2 y-1)}$

Question 9

$\frac{d y}{d x}$ निकालें यदि 

[Find $\frac{d y}{d x}$ if]

(i) $y=x^{\frac{1}{x}}$
Sol :
Taking log both sides

$\log y=\log x^{\frac{1}{x}}$

$\log y=\frac{1}{x} \cdot \log x$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=\frac{-1}{x^{2}} \log {x}+\frac{1}{x} \times \frac{1}{x}$

$\frac{d y}{d x}=y\left[\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right]$

$=\frac{x^{\frac{1}{x}}}{x^{2}}\left[1-\log x\right]$

(ii) y=x^y
Sol :
Taking log both sides

log y=log x^y

log y=y.log x

Differentiating w.r.t.x

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d y}{d x} \cdot \log x+y \cdot \frac{1}{x}$

$\frac{1}{y} \cdot \frac{d y}{d x}-\log x \frac{d y}{d x}=\frac{y}{x}$

$\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}$

$\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}$

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

(iii) y=x^{log x}
Sol :
Taking log both sides

log y=log x^{log x}

log y=log x.log x

log y=(log x)²

Differentiating w.r.t.x

$\frac{d y}{d x}=y \cdot 2 \frac{(\log x)}{x}$

$=x^{\log x} \cdot \frac{2\left(\log x\right)}{x}$

(iv) $y=x^{x+\frac{1}{x}}$
Sol :
Taking log both sides

$\log {y}=\log {x}\left(x+\frac{1}{x}\right)$

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

Differentiating w.r.t.x

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

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

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

(v) $y=x^{\cos ^{-1} x}$
Sol :
$y=x^{\cos ^{-1} x}$

Taking log both sides

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

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

Differentiating w.r.t.x

$\frac{1}{y}. \frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \log x+\cos^{-1} x\times\frac{1}{x}$

$\frac{d y}{dx}=y\left[\frac{1}{x} \cos ^{-1} x-\frac{1}{\sqrt{1-x^{2}}} \log x\right]$

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

Question 10

निम्नलिखित फलनों को x के सापेक्ष अवलिकत करें।

[Differentiate the following functions w.r.t. x]

(i) $(\log x)^{\log x}, x>1$
Sol :
Let y=$(\log x)^{\log x}$

Taking log both sides

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

log y=log x. log(log x)

Differentiating w.r.t.x

$\frac{1}{y} \cdot \frac{d y}{dx}=\frac{1}{x} \cdot \log (\log x)+\log x \times \frac{1}{\log x} \times \frac{1}{x}$

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

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

(ii) $(\sin x)^{\sin x}, 0<x<\pi$
Sol :
Let y=$(\sin x)^{\sin x}$

Taking log both sides

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

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

Differentiating w.r.t.x

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

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

$=(\sin x)^{\sin x} \cos x[ \log(\sin x)+1]$

(iii) $x^{\sin x}, x>0$
Sol :
Let y=x^{sin x}

Taking log both sides

log y=log x^{sin x}

log y=sin x.log x

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}=\cos x \log x+\sin x \times \frac{1}{x}$

$\frac{dy}{dx}=y\left[\cos x \log x+\frac{\sin x}{x}\right]$

$=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]$

(iv) $(\log x)^{\cos x}$

Sol :
Let y=(log x)^{cos x}

Taking log both sides

log y=log(log x)^{cos x}

log y=cos x.log(log x)

Differentiating w.r.t.x

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

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

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

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

Taking log both sides

log y=log(sin x-cos x)^(sinx-cosx)

log y=(sinx-cosx).log(sinx-cosx)

Differentiating w.r.t.x

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

$\frac{dy}{dx}=y\left[\left(\cos x+\sin x+(\cos x+\sin x) \cdot \log \left(\sin x-\cos x\right.\right)\right]$

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

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

Taking log both sides

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

log y=3 cos2x.log(5x)

Differentiating w.r.t.x

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

$\frac{d{y}}{d{x}}=y\left[-6 \sin 2 x \log (5 x)+\frac{3\cos x}{x}\right]$

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

Question 11

Find $\frac{d y}{d x}$ if

(i) x^{tan y}=y^{tan x}
Sol :
Taking log both sides

log x^{tan y}=log y^{tan x}

tan y.log x=tan x.log y

Differentiating w.r.t.x

$\sec ^{2} y \cdot \frac{d y}{d x} \cdot \log x+\tan y \times \frac{1}{x}=\sec ^{2} x \cdot \log y+\tan x \cdot \frac{1}{y}\times \frac{ dy}{d x}$

$\sec ^{2} y \log x \cdot \frac{d y}{dx}-\frac{\tan x}{y} \cdot \frac{d y}{d x}=\sec ^{2} x \log y-\frac{\tan y}{x}$

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

$\frac{d y}{d x}=\frac{y}{x} \cdot \frac{x \sec ^{2} x \log y-\tan y}{y \sec ^{2} y \log x-\tan x}$

(ii) (sin y)^x=(cos x)^y
Sol :
Taking log both sides

log(sin y)^x=log(cos x)^y

x.log(sin y)=y.log(cos x)

Differentiating w.r.t.x

$\text { 1. } \log (\sin y)+x \frac{1}{\operatorname{sin} y} \times \cos y \times \frac{d y}{d x}=\frac{d y}{d x} \log (\cos x)+y \cdot \frac{1}{\cos x} (-\sin x)$

$\log (\sin y)+y \tan x=\log (\cos x) \frac{d y}{d x}-x \cot y \frac{d y}{d x}$

$\frac{\log (\sin y)+y \tan x}{\log (\cos x)-x \cot y}=\frac{d y}{dx}$

(iii) (sec x)^y=(tan y)^x
Sol :
Taking log both sides

log(sec x)^y=log(tan y)^x

y.log(sec x)=x.log(tan y)

Differentiating w.r.t.x

$\frac{d y}{d x} \cdot \log (\sec x)+y \cdot \frac{1}{\sec x} \times \sec x \tan x=1 \cdot \log (\tan y)+x.\frac{1}{\tan y}\times \sec^2 y.\frac{dy}{dx}$

$\log (\sec x).\frac{d y}{d x}-x \sec y .\operatorname{cosec}y \frac{d y}{d x}=\log (\tan y)-y \tan x$

$[\log(\sec x)-x \sec y \operatorname{cosec} y] \frac{d y}{d x}=\log (\tan y)-y \tan x$

$\frac{dy}{dx}=\frac{\log (\tan y)-y \tan x}{\log (\operatorname{sec} x)-x \sec x . cosec~y}$

(iv) x^yy^x=1
Sol :
Taking log both sides

log x^y.y^x=log 1

log x^y+log y^x=0

y log x+x log y=0

Differentiating w.r.t.x

$\frac{dy}{d x} \cdot \log x+y \times \frac{1}{x}+1 \cdot \log y+x \cdot \frac{1}{y} \frac{dy}{d x}=0$

$\left(\log {x}+\frac{x}{y}\right) \frac{d y}{dx}=-\left(\frac{y}{x}+\log y\right)$

$\left(\frac{y \log x+x}{y}\right) \frac{d y}{d x}=-\left(\frac{y+x \log y}{x}\right)$

$\frac{d y}{d x}=\frac{-y}{x}\times\frac{y+x \log y}{x+y \log x}$

(v) y^x=x^y
Sol :
Taking log both sides

log y^x=log x^y

x log y=y log x

Differentiating w.r.t.x

$1 \cdot \log y+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d y}{d x} \cdot \log x+y \cdot \frac{1}{x}$

$\log y -\frac{y}{x}=\log x \frac{d y}{d x}-\frac{x}{y} \frac{d y}{d x}$

$\frac{x \log y -y}{x}=\left(\frac{y \log x-x}{y}\right) \frac{d y}{d x}$

$\frac{y(x \log y-y)}{x(y \log x-x)}=\frac{d y}{d x}$

(vi) (cos x)^y=(cos y)^x
Sol :
Taking log both sides

log(cos x)^y=log(cos y)^x

y.log(cos x)=x.log(cos y)

Differentiating w.r.t.x

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

$\log (\cos x) \frac{dy}{dx}+x \tan y \frac{d y}{d x}=\log (\cos x)+y \tan x$

$\frac{d y}{d x}=\frac{\log (\cos y)+y \tan x}{\log (\cos x)+x \tan y}$

Question 12 

यदि(If) $y=e^{x^{x}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=e^{x^{x}}$

Taking log both sides

$\log y=\log e^{x^{x}}$

log y=x^x log e

log y=x^x

Taking log both sides

log(log y)=log x^x

log(log y)=x.log.x

Differentiating w.r.t.x

$\frac{1}{\log {y}}+\frac{1}{y} \times \frac{d y}{dx}=1 \cdot \log x+x \cdot \frac{1}{x}$

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

$\frac{d y}{d x}=e^{x} \cdot x^{x}(1+\log x)$

Question 13

यदि (If) (x-y)^m+n=x^m.yⁿ  दिखाएँ कि (show that) $\frac{d y}{d x}=\frac{y}{x}$
Sol :
$(x-y)^{m+n}=x^{m} y^{n}$

Taking log both sides

$\log (x-y)^{m+n}=\log x^{m} \cdot y^{n}$

$(m+n) \cdot \log (x-y)=\log x^{m}+\log y^{n}$

$(m+n) \log (x-y)=m \cdot \log x+n \cdot \log y$

Differentiating w.r.t.x

$(m+n) \cdot \frac{1}{x-y} \times\left[1-\frac{dy}{dx}\right]=m \times \frac{1}{x}+n \times \frac{1}{y} \times \frac{dy}{d x}$

$\frac{m+n}{x-y}-\frac{m+n}{x-y} \frac{dy}{dx}=\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}$

$\frac{m+n}{x-y}-\frac{m}{x}=\left(\frac{m+n}{x-y}+\frac{n}{y}\right) \frac{dy}{d x}$

$\frac{m x+n x-m x+m y}{(x-y) x}=\left(\frac{m y+n y+n x-ny}{(x-y) \cdot y}\right) \frac{dy}{dx}$

$\frac{n x+n y}{x}=\left(\frac{n x+m y}{y}\right) \frac{d y}{d x}$

Question 14

यदि (If) $x=e^{\frac{x}{y}}$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{x-y}{x \log x}$
Sol :
$x=e^{\frac{x}{y}}$

Taking log both sides

$\log x=\log e^{\frac{x}{y}}$

$\log x=\frac{x}{y} \cdot \log e$

$\log x=\frac{x}{y}$

y.log x= x

Differentiating w.r.t.x

$\frac{d y}{d x} \cdot \log x+y \cdot \frac{1}{x}=1$

$\log {x} \frac{d y}{d x}=1-\frac{y}{x}$

$\log x \frac{dy}{dx}=\frac{x-y}{x}$

$\frac{d y}{dx}=\frac{x-y}{x \log x}$

(ii) यदि (If) $x y=e^{x-y}$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{y}{x}\left(\frac{x-1}{y+1}\right)$
Sol :
$x y=e^{x \cdot y}$

Taking log both sides

$\log x \cdot y=\log e^{x-y}$

Differentiating w.r.t.x

$\frac{1}{x}+\frac{1}{y} \times \frac{d y}{d x}=1-\frac{dy}{d x}$

$\frac{1}{y} \frac{d y}{d x}+\frac{d y}{dx}=1-\frac{1}{x}$

$\left(\frac{1}{y}+1\right) \frac{d y}{d x}=1-\frac{1}{x}$

$\left(\frac{1+y}{y}\right) \frac{d y}{dx}=\frac{x-1}{x}$

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

Question 15

यदि (If) $y=(\cos x)^{(\cos x)^{(\cos x)}\dots \text{to }\infty }$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$
Sol :
$y=(\cos x)^{(\cos x)^{(\cos x)}\dots \text{to }\infty }$

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

Taking log both sides

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

log y=y.log(cos x)

Differentiating w.r.t.x

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

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

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

$\frac{d y}{dx}=\frac{-y^{2} \tan x}{1-y \log (\cos x)}$

(i) $(\log x)^{x}+x^{\log x}$
Sol :
Let $y=u+v=(\log x)^{x}+x^{\log _{x}}$

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

Taking log both sides

$\log u=\log \left(\log {x}\right)^{x}$

log u=x log(log x)

Differentiating w.r.t.x

$\frac{1}{u} \cdot \frac{d u}{d x}=1.\log (\log x)+x \times \frac{1}{\log x} \times \frac{1}{2}$

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

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

$\frac{du}{dx}=\left(\log x\right)^{x}\left[\frac{\left[\log x \log (\log x)+1\right.}{\log x}\right]$

$\frac{du}{dx}=\left(\log x\right)^{x-1}\left[1+\log {x} \cdot \log (\log x)\right]$

Now ,

$v=x^{\log x}$

Taking log both sides

$\log v=\log {x} ^{\log x}$

log v=log x.log x

$\log v=(\log x)^{2}$

Differentiating w.r.t.x

$\frac{1}{v} \cdot \frac{d v}{d x}=2 \log x \times \frac{1}{x}$

$\frac{d v}{d x}=v \cdot \frac{2 \log x}{x}$

$\frac{d v}{d x}=x^{\log x} \cdot \frac{2 \log x}{x}$

$\frac{d v}{dx}=2 x^{\log x-1} \cdot \log x$

∵ y=u+v

Differentiating w.r.t.x

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

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

(ii) $x^{\sin x}+(\sin x)^{\cos x}$
Sol :
Let y=u+v=$x^{\sin x}+(\sin x)^{\cos x}$

$u=x^{\sin x}$

Taking log both sides

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

log u=sin x.log x

Differentiating w.r.t.x

$\frac{1}{u}\times \frac{d u}{d x}=\cos x \cdot \log x+\sin x \times \frac{1}{x}$

$\frac{du}{d x}=4\left[\frac{\sin x}{x}+\cos x \cdot \log x\right]$

$\frac{du}{dx}=x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]$

Now , $v=(\sin x)^{\cos x}$

Taking log both sides

$\log v=\log (\sin x)^{\cos x}$

$\log v=\cos x \cdot \log (\sin x)$

Differentiating w.r.t.x

 $\frac{1}{v} \times \frac{d v}{d x}=-\sin x \log (\sin x)+\cos x\times \frac{1}{\sin x} \times \cos x$

$\frac{d{v}}{dx}=v[\cos x \cdot \cot x-\sin x \log (\sin x)]$

$\frac{d v}{d x}=(\sin x)^{\cos x}[\cos x \cot x-\sin x \log (\sin x)]$

∵ y=u+v

Differentiating w.r.t.x

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

(iii) $x^{x}-2 \sin x$
Sol :
Let y=u-v=$x^{x}-2 \sin x$

$u=x^{x}$

Taking log both sides

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

log u=x.log x

Differentiating w.r.t.x

$\frac{1}{u} \cdot \frac{d y}{dx}=1 \cdot \log x+x \times \frac{1}{x}$

$\frac{d u}{d x}=u[1+\log x]$

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

Now ,  $v=2^{\sin x}$

Taking log both sides

$\log v=\log 2^{\sin x}$

log v=sin x .log 2

Differentiating w.r.t.x

$\frac{1}{v} \cdot \frac{d v}{d x}=\cos x \cdot \log {2}$

$\frac{d v}{dx}=2^{\sin x} \cdot \cos x \log 2$

∵y=u-v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d u}{dx}-\frac{d v}{d x}$

$\frac{d y}{d x}=x^{2}(1+\log x)-2^{\sin x} \cos x\log 2$

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

$u=(\sin x)^{x}$

Taking log both sides

$\log u=\log (\sin x)^{x}$

log u=x.log(sin x)

Differentiating w.r.t.x

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

$\frac{du}{dx}=u[x \cot x+\log (\sin x)]$

$\frac{du}{dx}=(\sin x)^{x}\left[x \cot x+\log \left(\sin x\right)\right]$

Now , $u=\sin ^{-1} \sqrt{2}$

Differentiating w.r.t.x

$\frac{d v}{d u}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \times \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{x} \sqrt{1-2}}$

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

∵ y=u+v

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{d y}{d x}+\frac{d v}{d x}=(\sin x)^{x}[x \cot x+\log (\sin x)]+\frac{1}{2 \sqrt{x-x^{2}}}$

(v) $x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Sol :
Let y=u+v=$x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$

$u=x^{x \cos x}$

Taking both sides log

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

$\log u=x \cos x \cdot \log {x}$

Differentiating w.r.t.x

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

$\frac{du}{dx}=u\left[\cos x+\cos x \log x-x \sin x \log x\right]$

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

Now , $v=\frac{x^{2}+1}{x^{2}-1}$

Differentiating w.r.t.x

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

$=\frac{2 x^{3}-2 x-2 x^{3}-2 x}{\left(x^{2}-1\right)^{2}}$

$=\frac{-4 x}{\left(x^{2}-1\right)^{2}}$

∵y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$

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

(vi) $(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
Sol :
Let y=$(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

$u=(x \cos x)^{x}$

Taking log both sides

$\log u=\log (x \cos x)^{x}$

$\log u=x \cdot \log (x \cos x)$

Differentiating w.r.t.x

$\frac{1}{u} \times \frac{d y}{d x}=1 . \log \left(x \cos x\right)+x \cdot \frac{1}{x \cos x}[1.\cos x+x(-\sin x)]$

$\frac{du}{dx}=u[\log (x \cos x)+1-x \tan x]$

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

Now , $v=(x \sin x)^{\frac{1}{x}}$

Taking log both sides

$\log v=\log (x \sin x)^{\frac{1}{x}}$

$\log v=\frac{1}{x} \log (x \sin x)$

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d x}=\frac{\frac{1}{x \sin x}[1 . \sin x+x \cos x] x-\log {y}(x \sin 3 x).1]}{x^2}$

$\frac{1}{v} \times \frac{du}{dx}=\frac{1+x \cot x-\log (x\sin x)}{x^{2}}$

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

y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$

$=(x \cos x)^x[1-x \tan x+\log(x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{1+x\cot x-\log (x\sin x)}{x^{2}}\right]$

(vii) $x^{x^{2}-3}+(x-3)^{x^{2}}, x>3$
Sol :
Let y=u+v=$x^{x^{2}-3}+(x-3)^{x^{2}}$

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

Taking log both sides

$\log u=\log x^{x^2}-3$

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

Differentiating w.r.t.x

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

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

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

Now , $v=(x-3)^{x^{2}}$

Taking log both sides

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

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

Differentiating w.r.t.x

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

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

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

∵ y=u+v

Differentiating w.r.t.x

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

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

Question 17

(i) यदि (If) $y=x^{x}+x^{\frac{1}{x}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=x^{x}, v=x^{\frac{1}{x}}$ and

y=u+v

$u=x^{x}$

Taking log both sides

$\log u=\log x^x$

$\log u=x \cdot \log {x}$

Differentiating w.r.t.x

$\frac{1}{u} \times \frac{d y}{d x}=1 \cdot \log x+x \times \frac{1}{x}$

$\frac{d u}{d x}=x^{x}[\log x+1]$

$v=x^{\frac{1}{x}}$

Taking log both sides

$\log v=\log {x} \frac{1}{x}$

$\log v=\frac{1}{x} \cdot \log x$

$\log v=\frac{\log {x}}{x}$

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d u}=\frac{\frac{1}{x} \times x-\lg x \cdot 1}{x^{2}}$

$\frac{d v}{dx}=x^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}\right]$

y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d u}{dx}+\frac{d v}{dx}$

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

(ii) यदि (If) $y=x+x^{\frac{1}{x}}$ (find) $\frac{dy}{dx}$ निकालें।
Sol :

(iii) यदि (If) $y=(\sin x)^{\cos x}+(\cos x)^{\sin x}$ निकाले (Find) $\frac{d y}{d x}$
Sol :
Let $u=(\sin x)^{\cos x}, \quad v=(\cos x)^{\sin x}$

y=u+v

$u=(\sin x)^{\cos x}$

Taking log both sides

$\log u=\log (\sin x)^{\cos x}$

log u=cos x. log (sin x)

Differentiating w.r.t.x

$\frac{1}{u} \times \frac{du}{dx}=-\sin x \log (\sin x)+\cos x \times \frac{1}{x}\times \cos x$

$\frac{du}{dx}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]$

$\frac{d u}{dx}=(\sin x)^{\cos x}\left[\cot x-\log (\tan x)^{\sin x}\right]$

Now , $v=(\cos x)^{\sin x}$

Taking log both sides

$\log v=\log (\cos x)^{\sin x}$

$\log v=\sin x \cdot \log (\cos x)$

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d x}=\cos x\log(\sin x)+\sin x \frac{1}{\cos x} \times(-\sin x)$

$\frac{d v}{d x}=(\cos )^{\sin x}\left[\log (\cos x)^{\cos x}-\tan x \sin x\right]$

y=u+v

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{dx}$

$\frac{d y}{dx}=(\sin x)^{\cos x}\left[\cot x \cos x-\log (\sin x)^{\sin x}\right]+\left(\cos\right)^{\sin x }\left[\log \left(\cos x\right)^{\cos x}-\tan x \sin x\right]$

(iv) यदि (If) $y=x^{\tan x}+(\tan x)^{\cot x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=x^{\tan x}$ , $v=(\tan x)^{\cot x}$ , $y=u+v$

$u=x^{\tan x}$

Taking log both sides

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

log u=tan x.log x

Differentiating w.r.t.x

$\frac{1}{u} \cdot \frac{du}{dx}=\sec ^{2} x \cdot \log {x}+\tan x \times\frac{1}{x}$

$\frac{d y}{d x}=x^{\tan x}\left[\frac{\tan x}{2}+\sec^{2} x \cos x\right]$

Now , $v=(\tan x)^{\operatorname{cot} x}$

Taking log both sides

$\log {v}=\log (\tan x)^{\cot x}$

log v=cot x.log (tan x)

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d x}=-\operatorname{cosec}^{2} x \cdot \log (\tan x)+\cot x \times\frac{1}{\tan x} \times \operatorname{sec}^{2} x$

$\frac{1}{v} \times \frac{d v}{d x}=\frac{\cos x}{\sin x} \times \frac{1}{\frac{\sin x}{\cos x}} \times \frac{1}{\cos ^{2} x}-\operatorname{cosec}^{2} x \log \left(\tan x\right)$

$\frac{d v}{dx}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x-\operatorname{cosec}^{2} x \log (\tan x)\right]$

$\frac{dv}{dx}=(\tan x)^{\cot x} \operatorname{cosec}^{2} x[1-\log (\tan x)]$

y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d y}{dx}+\frac{d y}{d x}$

$\frac{d y}{dx}=x^{\tan x}\left[\frac{\tan x}{x}+\sec ^{2} x \log x\right]+(\tan x)^{\cot x} \cdot \text{cosec x}^{2} x(1-\log (\tan x)]$

(v) यदि (If) $y=(\tan x)^{\cot x}+(\cot x)^{\tan x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=(\tan x)^{\cot x}$ $ v=(\cot x)^{\tan x}$

y=u+v

$u=(\tan x)^{\cot x}$

Taking log both sides

$\log u=\log (\tan x)^{\cot x}$

log u=cot x.log(tan x)

Differentiating w.r.t.x

$\frac{1}{u} \times \frac{d y}{dx}=-\operatorname{cosec}^{2} x \cdot \log (\tan x)+\cot x \times \frac{1}{\tan x} \times \operatorname{sec}^{2} x$

$\frac{d u}{dx}=u\left[\frac{\cos x}{\sin x} \times \frac{1}{\frac{\sin x}{\cos x}} \times \frac{1}{\cos ^{2} x}-\operatorname{cosec}^{2} x \log (\tan x)\right]$

$\frac{d u}{d x}=(\tan x)^{\cot x}.\text{cosec}^{2}x[1-\log(\tan x)]$

Now , $v=(\cot x)^{\tan x}$

Taking log both sides

$\log v=\log \left(\cot {x}\right)^{\tan x}$

$\log v=\tan x \cdot \log (\cot x)$

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{dx}=\sec ^{2} x \log (\operatorname{cot x})+\tan x \cdot \frac{1}{\operatorname{cot} x} \times\left(-\text{cosec} ^{2} x\right)$

$\frac{d v}{d x}=v\left[\sec ^{2} x \log (\cot x)-\frac{\sin x}{\cos x} \times \frac{1}{\frac{\cos x}{\sin x}} \times \frac{1}{\sin^{2}x}\right.$

$\frac{d v}{d x}=(\cot x)^{\tan x} \sec ^{2} x[\log (\cot x)-1]$

y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d u}{d x}-\frac{d v}{d x}$

$\frac{d y}{d x}=(\tan x)^{\cot x} \operatorname{cosec}^{2} x(1-\log (\tan x)]+(\cot x)^{\tan x}.\sec^2x[\log(\cot x)-1]$

(vi) यदि (If) $y=x^{x}+(\sin x)^{\cot x}$ निकाले (Find) $\frac{d y}{d x}$
Sol :
Let $u=x^{x}$ $v=(\sin x)^{\cos x}$

y=u+v

$u=x^{x}$

Taking log both sides

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

log u=x.logx

Differentiating w.r.t.x

$\frac{1}{u} \cdot \frac{d y}{dx}=1 \cdot \log x+x \cdot \frac{1}{x}$

$\frac{du}{dx}=x^{2}\left[1+\log {x}\right]$

Now ,  $v=(\sin x)^{\operatorname{cot x}}$

log v=cot x.log(sin x)

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d x}=-\text{cosec} ^{2} x \cdot \log (\sin x)+\cot x \frac{1}{\sin x} \times \cos x$

$\frac{d v}{dx}=v\left[\cot^{2} x-\text{cosec} ^{2} x \log (\sin x)\right]$

$\frac{d v}{dx}=(\sin x)^{\cot x}\left[\cot ^{2} x-\text{cosec} ^{2} x \log (\sin x)\right]$

∵ y=u+v

Differentiating w.r.t.x

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

$\frac{d y}{dx}=x^{x}\left(1+\log x\right)+(\sin x)^{\cot x}[\cot^2 x-\text{cosec}^2x.\log(\sin x)]$

(vii) यदि(If) $y=(x)^{\cos x}+(\cos x)^{\sin x}$ निकाले (find) $\frac{d y}{d x}$
Sol :
Let $u=x^{\cos x}$ $ v=(\cos x)^{\sin x}$

y=u+v

$u=x^{\cos x}$

Taking log both sides

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

log u=cos x.log x

Differentiating w.r.t.x

$\frac{1}{u} \times \frac{du}{dx}=-\sin x \cdot \log x+\cos x \cdot \frac{1}{x}$

$\frac{d u}{dx}=u\left[\frac{\cos x}{x}-\sin x \log x\right]$

$\frac{d y}{d x}=x^{\cos x}\left[\frac{\cos x}{x}-\operatorname{sin} x \log x\right]$

Now , $v=(\cos x)^{\sin x}$

Taking log both sides

$\log v=\log(\cos x)^{\sin x}$

log v=sin x.log(cos x)

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{du}{dx}=\cos x \cdot \log (\cos x)+\sin x \times \frac{1}{\cos x} \times(-\sin x)$

$\frac{d v}{dx}=v[\cos x \log (\cos x)-\sin x \tan x]$

$\frac{d v}{d x}=(\cos )^{\sin} [(\cos x \log x(\cos x)-\sin x \tan x]$

∵ y=u+v

Differentiating w.r.t.x

$\frac{d y}{d x}=\frac{d u}{dx}+\frac{d v}{d x}$

$\frac{d y}{dx}=x^{\cos x}\left(\frac{\cos x}{x}-\sin x \log x )+(\cos x)^{\sin x}.[\cos x \log \cos x-\sin x \tan x]\right.$

(viii) $e^{\sin x}+(\tan x)^x$ को x के सापेज्ञ अवकलित करें ।
Sol :
Let y=u+v$=e^{\sin x}+(\tan x)^{x}$

$u=e^{\sin x}$

Differentiating w.r.t.x

$\frac{d u}{d x}=e^{\sin x} \cdot \cos x$

$v=(\tan x)^{x}$

Taking log both sides

$\log u=\log (\tan x)^{2}$

log u=x.log(tan x)

Differentiating w.r.t.x

$\frac{1}{v} \times \frac{d v}{d x}=1 . \log (\tan x)+x \frac{1}{\tan x} \times \sec ^{2} x$

$\frac{d v}{d x}=(\tan x)^{x}[\log (\tan x)+x \operatorname{cosec} x+\sec x]$

∵ y=u+v

Differentiating w.r.t.x

$\frac{d y}{dx}=\frac{d y}{d x}+\frac{d v}{d x}$

$\frac{d y}{dx}=e^{\sin x} \cdot \cos x+(\tan x)^{x}\left[\log {\tan(x-2)+x \text{cosec x} \sec x}\right.$

Question 18

(i) यदि (If) $2^{x}+x^{y}=1$ निकाले (find) $\frac{d y}{d x}$
Sol :
$2^{x}+x^{y}=1$

$2^{x}+e^{\log x^{y}}=1$

$2^{x}+e^{y \cdot \log x}=1$

Differentiating w.r.t.x

$2^{x} \cdot \log 2+e^{y.\log x} \cdot\left[\frac{d y}{d x} \log x+y \cdot \frac{1}{x}\right]=0$

$2^{x} \log {2}+x^{y}\left[\log {x} \frac{d y}{d x}+\frac{y}{x}\right]=0$

$2^{x} \log 2+x^{y} \log _{x} \frac{d y}{d x}+x^{y} \frac{y}{x}=0$

$x^{y} \cdot \log x {\frac{dy}{dx}}=-\left(2^{x} \log x+x^{y} \frac{y}{x}\right)$

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

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

(ii) यदि (If) $x^{y}+y^{x}=1$ निकाले ( find ), $\frac{d y}{d x}$
Sol :
$x^{y}+y^{x}=1$

$e^{\log {x^y}}+e^{\log y^{x}}=1$

$e^{y \log x}+e^{x \log y}=1$

Differentiating w.r.t.x

$e^{y \cdot \log x\left[\frac{d y}{dx} \cdot \log x+y \cdot \frac{1}{x}\right]+e^{x} \log y}\left[1 \cdot \log y+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right]=0$

$x^{y}\left[\log x \frac{d y}{dx}+\frac{y}{x}\right]+y^{x}\left[\log y+\frac{x}{y} \frac{d y}{dx}\right]=0$

$x^{y} \cdot \log x \frac{d y}{dx} \frac{y x^{y}}{x}+y^{x} \log{ y}+\frac{x y^{x}}{y} \cdot \frac{d y}{d x}=0$

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

$\left[x^{y} \log x+x y^{x-1}\right] \frac{d y}{dx}=-\left[y \cdot x^{y-1}+y^{x} \log y\right]$

$\frac{d y}{d x}=-\frac{y \cdot x^{y-1}+y^{x} \log {y}}{x^{y} \log {x}+x y^{x-1}}$

Question 19

(i) यदि (If) $y=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}}$

Taking log both sides

$\log y=\log \frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}}$

$\log {y}=\log {x^{2}} \sqrt{4 x+3}-\log (3 x+1)^{2}$

$\log y=\log x^{2}+\log \sqrt{4 x+3}-\log (3 x+1)^{2}$

$\log y=2 \log x+\frac{1}{2} \log (4 x+3)-2 \log (3 x+1)$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d{y}}{d{x}}=2 \times \frac{1}{x}+\frac{1}{2} \times \frac{1}{4 x+3} \times 4+2\times \frac{1}{3 x+1} \times 3$

$\frac{d y}{d x}=y^{2}\left[\frac{1}{x}+\frac{1}{4 x+3}-\frac{3}{3 x+1}\right]$

$\frac{d y}{d x}=\frac{x^{2} \sqrt{4 x+3}}{(3 x+1)^{2}} \cdot 2\left[\frac{1}{x}+\frac{1}{4 x+3}-\frac{3}{3 x+1}\right]$

(ii) यदि (If) $y=\frac{2(x-\sin x)^{\frac{3}{2}}}{\sqrt{x}}$ निकाले (find) $\frac{d y}{d x}$
Sol :
$y=\frac{2\left(x-\sin x\right)^{\frac{3}{2}}}{\sqrt{x}}$

Taking log both sides

$\log y=\log \frac{2[x-\sin x]^{\frac{3}{2}}}{x^{1 / 2}}$

$\log y=\log {2}(x-\sin x)^{\frac{3}{2}}-\log x^{\frac{1}{2}}$

$\log y=\log 2+\frac{3}{2} \log (x-\sin x)-\frac{1}{2} \log x$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=0+\frac{3}{2} \times \frac{1}{x-\sin x} \times[1-\cos x]-\frac{1}{2} \times \frac{1}{x}$

$\frac{1}{y} \frac{d y}{d x}=\frac{3\left(1-\cos x\right)}{2(x-\sin x)}-\frac{1}{2 x}$

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

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

Question 20

निम्नलिखित फलनों को x के सापेक्ष अवकलित करें ।

[Differentiate the following functions w.rt.x]

(i) cosx cos2x cos3x
Sol :
Let y=cosx cos2x cos3x

Taking log both sides

log y=log cosx cos2x cos3x

log y=log cos x+log cos2x+log cos3x

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}=\frac{1}{\cos x} \times(-\sin x)+\frac{1}{\cos 2 x} \times(-\sin 2 x) \cdot 2+\frac{1}{\cos 3 x} \times(-\sin 3x).3$

$\frac{d y}{d x}=y[-\tan x-2 \tan 2 x-3 \tan 3 x]$

$\frac{d y}{d x}=-\cos \cos 2 x \cos 3x[\tan x+2 \tan 2 x+3 \tan 3 x]$

(ii) sinx sin2x sin3x
Sol :
Let y=sinx sin2x sin3x

Taking log both sides

log y=log sinx sin2x sin3x

log y=log sin x+log sin 2x+log sin 3x

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=\frac{1}{\sin x} \times \cos x+\frac{1}{\sin 2 x} \times \cos 2 x\times 2+\frac{1}{\sin 3 x} \times \cos 3 x \times 3$

$\frac{d y}{d x}=y\left[\text{cot x}+2 \cot 2 x+3 \cot 3 x\right]$

$\frac{d y}{d x}=\sin 2x \sin 2 x \sin 3 x[\cot x+2 \cot 2 x+3 \cot 3 x]$

(iii) $\frac{(x-1)^{2}(2 x+1)(x+3)}{e^{x} \sin x}$
Sol :
Let y=$\frac{(x-1)^{2}(2 x+1)(x+3)}{e^{x} \sin x}$

Taking log both sides

$\log y=\log \frac{(x-1)^{2}(2 x+1)(x+3)}{e^{x} \sin x}$

$\log y=\log (x-1)^{2}(2 x+1)(x+3)-\log e^{x} \cdot \sin x$

$\log y=\log (x-1)^{2}+\log (2 x+1)+\log y(x+3)-\log e^{x}-\log \sin x$

log y=2log(x-1)+log(2x+1)+log(x+3)-x-logsinx

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{dy}{dx}=2 \times \frac{1}{x-1}+\frac{1}{2 x+1} \times 2+\frac{1}{x+3}-1-\frac{1}{\sin x}\times \cos x$

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

(iv) $\frac{(x+1)^{3 / 2} \cdot \log x}{x^{3} \sin x}$
Sol :
Let y=$\frac{(x+1)^{3 / 2} \cdot \log x}{x^{3} \sin x}$

Taking log both sides

$\log {y}=\log \frac{(x+1)^{3 / 2} \log x}{x^{3} \sin x}$

$\log y=\log (x+1)^{\frac{3}{2}} \cdot \log x-\log x^{3}\sin x$

$\log y=\log (x+1)^{3 / 2}+\log \log x-\left[\log x^{3}+\log \sin x]\right.$

$\log y=\frac{3}{2} \log (x+1)+\log (\log x)-3 \log x-\log \sin x$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}=\frac{3}{2} \times \frac{1}{x+1}+\frac{1}{\log x}\times\frac{1}{x}-3\times \frac{1}{x}-\frac{1}{\sin x} \times \operatorname{cos} x$

$\frac{d y}{d x}=\frac{(x+1)^{3 / 2} \log x}{x^{3}-\sin x}\left[\frac{3}{2(x+1)}+\frac{1}{x \log x}-\frac{3}{x}-\cot x\right]$

(v) $(x+1)^{2}(x-2)^{3}(x+4) \log x$
Sol :
Let y=$(x+1)^{2}(x-2)^{3}(x+4) \log x$

Taking log both sides

$\log y=\log y(x+1)^{2}(x-2)^{3}(x+4) \log x$

$\log y=\log (x+1)^{2}+\log (x-2)^{3}+\log (x+4)+\log (\log x)$

$\log y=2 \log (x+1)+3 \log (x-2)+\log (x+4)+\log (\log 4)$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}=2 \times\frac{x}{x+1}+3 \times\frac{x }{x-2}+\frac{1}{x+4}+\frac{1}{\log x} \times \frac{1}{x}$

$\frac{d y}{d x}=(x+1)^{2}(x-2)^{3}(x+4) \cdot \log x\left[\frac{2}{x+1}+\frac{3}{x-2}+\frac{1}{x+4}+\frac{1}{x \log x}\right]$

(vi) $(x+3)^{2}(x+4)^{3}(x+5)^{4}$
Sol :
Let y=$(x+3)^{2}(x+4)^{3}(x+5)^{4}$

Taking log both sides

$\log y=\log (x+3)^{2}(x+4)^{3}(x+5)^{4}$

$\log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}$

$\log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)$

Differentiating w.r.t.x

$\frac{1}{y} \times \frac{d y}{d x}-2 \times \frac{x }{x+3}+3 \times \frac{1}{x+4}+4 \times \frac{1}{x+5}$

$\frac{d y}{dx}=(x+3)^{2}(x+4)^{3}(x+5)^{4}\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$=(x+3)^{2}(x+4)^{3}(x+5)^{4}\left[\frac{2\left(x^{2}+9 x+20\right)+3\left(x^{2}+8 x+15\right)+4(x^2+7x+12)}{(x+3)(x+4)\left(x+5\right)}\right]$

$=(x+3)(x+4)^{2}(x+5)^{3}\left[\begin{array}{c}2 x^{2}+18 x+40+3 x^{2}+24x+45+4 x^{2}+28 x+48\end{array}\right.$

$\frac{dy}{dx}(x+3)(x+4)^{2}(x+5)^{3}\left(5 x^{2}+70 x+133\right)$