Exercise 11.4
Question 10
निम्नलिखित फलनों को x के सापेक्ष अवलिकत करें।
[Differentiate the following functions w.r.t. x]KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q10-Q12)
(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=xsin x
Taking log both sides
log y=log xsin 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) $(\log x)^{\cos x}$
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) xtan y=ytan x
Sol :
Taking log both sides
log xtan y=log ytan 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) xyyx=1
Sol :
Taking log both sides
log xy.yx=log 1
log xy+log yx=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) yx=xy
Sol :
Taking log both sides
log yx=log xy
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
Sol :
$y=e^{x^{x}}$
Taking log both sides
$\log y=\log e^{x^{x}}$
log y=xx log e
log y=xx
Taking log both sides
log(log y)=log xx
log(log y)=x.log.x
Differentiating w.r.t.x
$\frac{1}{\log {y}}+\frac{1}{y} \times \frac{d y}{dx}=1 \cdot \log x+x \cdot \frac{1}{x}$
$\frac{d y}{dx}=y \cdot \log y(1+\log x)$
$\frac{d y}{d x}=e^{x} \cdot x^{x}(1+\log x)$
Thank you so much sir
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