# Solve this

Question:

Differentiate $\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$ with respect to $\sqrt{1-\mathrm{x}^{2}}$, if $-1 Solution: Let$\mathrm{u}=\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$and$\mathrm{v}=\sqrt{1-\mathrm{x}^{2}}$We need to differentiate$u$with respect to$v$that is find$\frac{d u}{d v}$. We have$u=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$By substituting$x=\tan \theta$, we have$\mathrm{u}=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)\Rightarrow \mathrm{u}=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right)\Rightarrow \mathrm{u}=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)\left[\because \tan (\mathrm{A}-\mathrm{B})=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}}\right]$Given,$-1

However, $x=\tan \theta$

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

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=0-\frac{1}{1+\mathrm{x}^{2}}$

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

Now, we have $\mathrm{v}=\sqrt{1-\mathrm{x}^{2}}$

On differentiating $\vee$ with respect to $x$, we get

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{x}^{2}}}\left[0-2 \mathrm{x}^{2-1}\right]$

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

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

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

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

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

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

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