# Solve this

Question:

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

However, $x=\tan \theta$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

But, $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=2 \times \frac{1}{1+\mathrm{x}^{2}}$

$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2}{1+\mathrm{x}^{2}}$

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

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

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

$\therefore \frac{d u}{d v}=1$

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