Exercise 11.4
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।
(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
$\left(\frac{d y}{d x}\right)_{x=0}=a^{0} \cdot \log _{e} a+\sqrt{\frac{1-0}{1+0}} \cdot \frac{1}{(1-0)^{2}}$
$=\log _{e} a+1$
No comments:
Post a Comment