Solve this

Let $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

We need to differentiate $u$ with respect to $v$ that is find $\frac{d u}{d v}$.

We have $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

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

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

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

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

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

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

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

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

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

Given $0

However, $x=\tan \theta$

$\Rightarrow \tan \theta \in(0,1)$

$\Rightarrow \theta \in\left(0, \frac{\pi}{4}\right)$

$\Rightarrow 2 \theta \in\left(0, \frac{\pi}{2}\right)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

But, $\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$

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

However, $\theta \in\left(0, \frac{\pi}{4}\right) \Rightarrow 2 \theta \in\left(0, \frac{\pi}{2}\right)$

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

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

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

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

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

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

We have $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dv}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}$

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

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

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

Thus, $\frac{d u}{d v}=1$