# Solve this

Question:

Differentiate $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ with respect to $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, if $-1 Solution: Let$\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)$and$\mathrm{v}=\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$We need to differentiate$u$with respect to$v$that is find$\frac{\text { du }}{\text { dv }}$. We have$\mathrm{u}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right)$By substituting$x=\tan \theta$, we have$u=\tan ^{-1}\left(\frac{\sqrt{1+(\tan \theta)^{2}}-1}{\tan \theta}\right)\Rightarrow u=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)\Rightarrow u=\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right)\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{1-\cos \left(2 \times \frac{\theta}{2}\right)}{\sin \left(2 \times \frac{\theta}{2}\right)}\right)$But,$\cos 2 \theta=1-2 \sin ^{2} \theta$and$\sin 2 \theta=2 \sin \theta \cos \theta\Rightarrow u=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)\Rightarrow u=\tan ^{-1}\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right)\Rightarrow u=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$Given$-1

However, $x=\tan \theta$

$\Rightarrow \tan \theta \in(-1,1)$

$\Rightarrow \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$

$\Rightarrow \frac{\theta}{2} \in\left(-\frac{\pi}{8}, \frac{\pi}{8}\right)$

Hence, $u=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}$

$\Rightarrow \mathrm{u}=\frac{1}{2} \tan ^{-1} \mathrm{x}$

On differentiating $u$ with respect to $x$, we get

$\frac{d u}{d x}=\frac{d}{d x}\left(\frac{1}{2} \tan ^{-1} x\right)$

$\Rightarrow \frac{d u}{d x}=\frac{1}{2} \frac{d}{d x}\left(\tan ^{-1} x\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^{2}}$

$\Rightarrow \frac{d u}{d x}=\frac{1}{2} \times \frac{1}{1+x^{2}}$

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

Now, we have $v=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

By substituting $x=\tan \theta$, we have

$\mathrm{v}=\sin ^{-1}\left(\frac{2 \tan \theta}{1+(\tan \theta)^{2}}\right)$

$\Rightarrow \mathrm{v}=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$

$\Rightarrow \mathrm{v}=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right)\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$\Rightarrow \mathrm{v}=\sin ^{-1}\left(\frac{2 \times \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}}\right)$

$\Rightarrow \mathrm{v}=\sin ^{-1}\left(2 \times \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right)$

$\Rightarrow v=\sin ^{-1}(2 \sin \theta \cos \theta)$

But, $\sin 2 \theta=2 \sin \theta \cos \theta$

$\Rightarrow v=\sin ^{-1}(\sin 2 \theta)$

However, $\theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \Rightarrow 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence, $v=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\Rightarrow v=2 \tan ^{-1} x$

On differentiating $v$ with respect to $x$, we get

$\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \tan ^{-1} \mathrm{x}\right)$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^{2}}$

$\Rightarrow \frac{d v}{d x}=2 \times \frac{1}{1+x^{2}}$

$\therefore \frac{d v}{d x}=\frac{2}{1+x^{2}}$

We have $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{1}{2\left(1+\mathrm{x}^{2}\right)}}{\frac{2}{1+\mathrm{x}^{2}}}$

$\Rightarrow \frac{d u}{d v}=\frac{1}{2\left(1+x^{2}\right)} \times \frac{1+x^{2}}{2}$

$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{1}{4}$

Thus, $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{1}{4}$

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