Chapter 1 - Differentiation
Exercise 1.4
Q.4
i)
$\begin{aligned}
&Differentiate \quad x sinx \quad w. r . t \quad tanx\\
sol^n\\
Let,
u&=x \sin x, \quad v=\tan x\\
\frac{d u}{d v}&=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}\\
u&=x \sin x\\
&Diff \quad w.r.t \quad x\\
\frac{d u}{d x}&=x \cdot \frac{d}{d x}(\sin x)+\sin x \cdot \frac{d}{d x}(x)\\
&=x \cos x+\sin x\\
And,\\
v&=\tan x\\
&Diff \quad w.r.t \quad x\\
\frac{d v}{d x}&=\frac{d}{d x} \tan x \\
&=\sec ^{2} x \\
Now,\\
\frac{d u}{d v}&=\frac{x \cos x+\sin x}{\sec ^{2} x}\end{aligned}$
ii)
$\begin{aligned}
&Differentiate \quad \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad w.r.t \quad \tan ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\\
Let \quad u&=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\\
u&=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\\
Put,\\
x&=\tan \theta\\
u&=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\\
u&=\sin ^{-1}(\sin 2 \theta)\\
u&=2\theta\\
u&=2 \tan^{-1}x\\
&Diff \quad w.r.t \quad x\\
\frac{d u}{d x}&=\frac{2}{1+x^2}\\
And,\\
\quad v&=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\\
put,\\
x&=\tan \theta\\
v&=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\\
v&=\cos ^{-1}(\cos 2 \theta)\\
v&=2\theta\\
v&=2 \tan^{-1}x\\
&Diff \quad w.r.t \quad x\\
\frac{d v}{d x}&=\frac{2}{1+x^2}\\
Now,\\
\frac{d u}{d v}&=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}\\\\
&=\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}\\\\
&=1\end{aligned}$
iii)
$\begin{aligned}
&Differentiate \quad \tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \quad w.r.t \quad \sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\\
Let,\\
u&=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\\
&Put,\quad x=\sin \theta\\
u&=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right)\\
u&=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}\right)\\
u&=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)\\
u&=\tan ^{-1}(\tan \theta)\\
u&=\theta\\
u&=\sin^{-1}x\\
&Diff \quad w.r.t \quad x \\
\frac{d u}{d x}&=\frac{1}{\sqrt{1-x^2}}\\\\
And,\\
v&=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right) \\
&Put,\quad x=\cos \theta)\\
v&=\sec ^{-1}\left(\frac{1}{2 \cos ^{2} \theta-1}\right)\\
v&=\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)\\
v&=\sec ^{-1}(\sec 2 \theta)\\
v&=2\theta\\
v&=2\cos^{-1}x\\
&Diff \quad w.r.t \quad x \\
\frac{d v}{d x}&=\frac{-2}{\sqrt{1-x^2}}\\\\
Now,\\
\frac{d v}{d x}&=\frac{\frac{d u}{d x}}{\frac{d v }{d x}}\\
\frac{d u}{d v}&=\frac{\frac{1}{\sqrt{1-x^2}}}{\frac{-2}{{\sqrt{1-x^2}}}}\\\\
\frac{d u}{d v}&=\frac{-1}{2}
\end{aligned}$
iv)
$\begin{aligned}
&Differentiate \quad \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \quad w.r.t \quad \tan ^{-1}(x)\\
Let,\\
u&=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), \quad v=\tan ^{-1}(x)\\
Put, x&=\tan \theta\\
u&=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\\
u&=\cos ^{-1}(\cos 2 \theta)\\
u&=2\theta\\
u&=2\tan^{-1}x\\
&Diff \quad w.r.t \quad x\\
\frac{d u}{d x}&=\frac{2}{1+x^2}\\\\
And, \\
\quad v&=\tan ^{-1}(x)\\
v&=\tan ^{-1}(\tan \theta)\\
v&=\theta\\
u&=\tan^{-1}x\\
&Diff \quad w.r.t \quad x\\
\frac{d v}{d x}&=\frac{1}{1+x^2}\\
Now,\\
\frac{d u}{d v}&=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}\\\\
&=\frac{\frac{2}{1+x^2}}{\frac{1}{1+x^2}}\\\\
&=2\end{aligned}$
v)
$\begin{aligned}
&Differentiate \quad u=3^{x}\quad w.r.t \quad v=\log _{x} 3\\
Let,\\
u&=3^{x}\\
&Diff \quad w.r.t \quad x \\
\frac{d u}{d x}&=3^{x} \log 3\\\\
And,\\
\quad v&=\log _{x} 3\\
v&=\frac{\log 3}{\log x}\\
&Diff \quad w.r.t \quad x \\
\frac{d v}{d x} &=\log 3 \left(\frac{1}{\log x}\right) \\
&=\log 3 \frac{d}{d x}(\log x)^{-1} \\
&=\log 3 \cdot(-1)(\log x)^{-2} \frac{d}{d x} \log x \\
&=\frac{\log 3}{x(\log x)^{2}} \\
Now,\\
\frac{d u}{d v}&=\frac{\frac{d u}{dx}}{\frac{d v}{d x}}\\\\
&=\frac{3^{x} \log 3}{\frac{\log 3}{x(\log x)^{2}}}\\\\
&=3^{x}(\log x)^{2} x \end{aligned}$
vi)
$\begin{aligned}
&Differentiate \quad \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right) \quad w.r.t \quad \sec^{-1}x\\
Let,\\
u&=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\\
u&=\tan ^{-1}\left[\frac{\cos ^{2}\left(\frac{x}{2}\right)-\sin ^{2}\left(\frac{x}{2}\right)}{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^{2}}\right]\\
\\
u&=\tan ^{-1}\left[\frac{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left(\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right)}{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^{2}}\right]\\
\\
u&=\tan ^{-1}\left[\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}\right]\\
&Divide \quad by \cos \left(\frac{x}{2}\right)\\
\\
u&=\tan ^{-1}\left[\frac{1-\tan (x / 2)}{1+\tan \left(\frac{x}{2}\right)}\right]\\
u&=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]\\
u&=\frac{\pi}{4}-\frac{x}{2}\\
&Diff \quad w.r.t \quad x\\
\frac{d u}{d x}&=\frac{-1}{2}\\\\
And,\\
\quad v&=\sec ^{-1} x\\
&Diff \quad w.r.t \quad x\\
\frac{d v}{d x}&=\frac{1}{x \sqrt{x^{2}-1}}\\\\
Now,\\
\frac{d u}{d v} &=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \\ \\
\frac{d u}{d v} &=\frac{\frac{-1}{2}}{\frac{1}{x \sqrt{x^{2}-1}}} \\ \\
\frac{d u}{d v} &=\frac{-x \sqrt{x^{2}-1}}{2} \end{aligned}$
vii)
$\begin{aligned}
&Differentiate \quad x^{x} \quad w.r.t \quad x^{\sin x}\\
Let,\\
u&=x^{x}\\
&Taking \quad log\\
\log u&=\log x^{x}\\
\log u&=x \log x\\
&Diff \quad w.r.t \quad x\\
\frac{1}{u} \cdot \frac{d u}{d x}&=x \cdot \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x} x\\
&=\frac{x}{x}+\log x\\
\frac{d u}{d x}&=u(1+\log x)\\
\frac{d u}{d x}&=x^{x}(1+\log x)\\\\
And, \\
\quad v&=x^{\sin x}\\
\log v&=\log x^{\sin x}\\
\log v&=\sin x \cdot \log x\\
&Diff \quad w.r.t \quad x\\
\frac{1}{v} \frac{d v}{d x}&=\sin x \cdot \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x} \sin x\\
\\
&=\frac{\sin x}{x}+\log x \cdot \cos x\\
\frac{d v}{d x}&=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cdot \cos x\right)\\
\\
Now,\\
\frac{d u}{d v}&=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}\\
\\
&=\frac{x^{x}(1+\log x)}{x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)}\\
&=\frac{x^{x} \cdot (1+\log x)}{x^{\sin x} \cdot \frac{1}{x}(\sin x+x \log x \cos x)}\\
&=\frac{x^{x} \cdot x(1+\log x)}{x^{\sin x} \cdot (\sin x+x \log x \cos x)}\\
&=\frac{x^{+1-\sin x}(1+\log x)}{(\sin x+x \log x \cos x)}\end{aligned}$
viii)
$\begin{aligned}
&Differentiate \quad u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right),
v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\\
Let,\\
u&=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\\
&Put,x=\tan \theta\\
u&=\tan ^{1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)\\
u&=\tan ^{-1}\left(\frac{\sqrt{\sec^2\theta}-1}{\tan \theta}\right)\\
u&=\tan ^{-1}\left(\frac{{\sec\theta}-1}{\tan \theta}\right)\\
u&=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\\
u&=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\\
u&=\tan ^{-1}\left(\frac{2 \cdot \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cdot \sin (\theta/2) \cos \left(\frac{\theta}{2}\right)}\right)\\
u&=\tan ^{-1}\left(\frac{\sin \left(\frac{\theta}{2}\right)}{\cos ( \theta/ 2)}\right)\\
u&=\tan ^{-1}\left(\tan \left(\frac{\theta}{2}\right)\right)\\
u&=\frac{\theta}{2}\\
u&= \frac{\tan^{-1 }x}{2}\\
&Diff \quad w.r.t \quad x \\
\frac{d u}{d x}&=\frac{1}{2\left(1+x^{2}\right)}\\\\
And,\\
v&=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\\
&put,x=\sin x\\
v&=\tan ^{1}\left(\frac{2 \sin \theta \sqrt{1-\sin ^{2} \theta}}{1-2 \sin^2\theta}\right)\\
v&=\tan ^{1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)\\
v&=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)\\
v&=\tan ^{-1}(\tan 2\theta)\\
v&=2 \theta\\
v&=2 \cdot \sin ^{-1}(x)\\
&Diff \quad w.r.t \quad x \\
\frac{d v}{d x}&=\frac{2}{\sqrt{1-x^{2}}}\\\\
Now,\\
\frac{d u}{d v}&=\frac{\frac{d y}{d x}}{\frac{d v}{d x}}\\\\
\frac{d u}{d v}&=\frac{\frac{1}{2(1+x)}}{\frac{2}{\sqrt{1-x^{2}}}}\\\\
\frac{d u}{d v}&=\frac{1}{4} \frac{\sqrt{1-x^{2}}}{\left(1+x^{2}\right)}\end{aligned}$
(4) (i) Differentiate x sin x w. r. t. tan x.
(ii) Differentiate sin^-1(2x/(1 + x2)) w. r. t. cos−1(1 − x2/1 + x2 )
(iii) Differentiate tan−1(x/sqrt( 1 − x^2) w. r. t. sec−1(1/2x^2 − 1)
(iv) Differentiate cos−1(1 − x^2)/(1 + x2) w. r. t. tan−1 x.
(v) Differentiate 3^x w. r. t. logx 3.
(vi) Differentiate tan−1(cos x/1 + sin x)w. r. t. sec−1 x.
(vii) Differentiate xx w. r. t. x^sin x.
(viii) Differentiate tan−1 (sqrt( 1 + x^2) − 1)/x w. r. t. tan−1 (2xsqrt1 − x21 − 2x2
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