# Solve this

Question:

Differentiate $\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)$ with respect to $\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)$, if $0<\mathrm{x}<1$.

Solution:

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

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

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

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

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

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

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