Solve this

Solution:

Let $u=\tan ^{-1}\left(\frac{x-1}{x+1}\right)$ and $v=\sin ^{-1}\left(3 x-4 x^{3}\right)$

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

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

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

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

$\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{\tan \theta-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{4} \tan \theta}\right)$

$\Rightarrow \mathrm{u}=\tan ^{-1}\left(\tan \left(\theta-\frac{\pi}{4}\right)\right)\left[\because \tan (\mathrm{A}-\mathrm{B})=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}}\right]$

Given, $-\frac{1}{2}

However, $x=\tan \theta$

$\Rightarrow \tan \theta \in\left(-\frac{1}{2}, \frac{1}{2}\right)$

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

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

As $\tan 0=0$ and $\tan \frac{\pi}{4}=1$, we have $\tan ^{-1}\left(\frac{1}{2}\right) \in\left(0, \frac{\pi}{4}\right)$

Thus, $\theta-\frac{\pi}{4}$ lies in the range of $\tan ^{-1} x$.

Hence, $u=\tan ^{-1}\left(\tan \left(\theta-\frac{\pi}{4}\right)\right)=\theta-\frac{\pi}{4}$

$\Rightarrow \mathrm{u}=\tan ^{-1} \mathrm{x}-\frac{\pi}{4}$

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

$\frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}-\frac{\pi}{4}\right)$

$\Rightarrow \frac{d u}{d x}=\frac{d}{d x}\left(\tan ^{-1} x\right)-\frac{d}{d x}\left(\frac{\pi}{4}\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^{2}}$ and derivative of a constant is 0 .

$\Rightarrow \frac{d u}{d x}=\frac{1}{1+x^{2}}+0$

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

Now, we have $v=\sin ^{-1}\left(3 x-4 x^{3}\right)$

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

$v=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)$

But, $\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta$

$\Rightarrow v=\sin ^{-1}(\sin 3 \theta)$

Given, $-\frac{1}{2}

However, $x=\sin \theta$

$\Rightarrow \sin \theta \in\left(-\frac{1}{2}, \frac{1}{2}\right)$

$\Rightarrow \theta \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$

$\Rightarrow 3 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence, $v=\sin ^{-1}(\sin 3 \theta)=3 \theta$

$\Rightarrow v=3 \sin ^{-1} x$

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

$\frac{d v}{d x}=\frac{d}{d x}\left(3 \sin ^{-1} x\right)$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=3 \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)$

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

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=3 \times \frac{1}{\sqrt{1-\mathrm{x}^{2}}}$

$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{3}{\sqrt{1-\mathrm{x}^{2}}}$

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

$\Rightarrow \frac{d u}{d v}=\frac{\frac{1}{1+x^{2}}}{\frac{3}{\sqrt{1-x^{2}}}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{1}{1+\mathrm{x}^{2}} \times \frac{\sqrt{1-\mathrm{x}^{2}}}{3}$

$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\sqrt{1-\mathrm{x}^{2}}}{3\left(1+\mathrm{x}^{2}\right)}$

Thus, $\frac{d u}{d v}=\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)}$