KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.4 (Q1-Q3)

Exercise 11.4










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 ecos 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 ex)
Sol :
Let y=sin(tan-1 ex)

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

Differentiating w.r.t.x

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

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


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

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

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


2 comments:

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