# Solve this

Question:

Differentiate $\sin ^{-1} \sqrt{1-x^{2}}$ with respect to $\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$, if $0 Solution: Let$u=\sin ^{-1} \sqrt{1-x^{2}}$and$v=\cot ^{-1}\left(\frac{x}{\sqrt{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} \sqrt{1-x^{2}}$By substituting$x=\cos \theta$, we have$\mathrm{u}=\sin ^{-1} \sqrt{1-(\cos \theta)^{2}}\Rightarrow u=\sin ^{-1} \sqrt{1-\cos ^{2} \theta}\Rightarrow u=\sin ^{-1} \sqrt{\sin ^{2} \theta}\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\Rightarrow u=\sin ^{-1}(\sin \theta)$Given,$0

However, $x=\cos \theta$

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

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

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

$\Rightarrow u=\cos ^{-1} x$

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

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

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

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

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

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

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

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

$\Rightarrow \mathrm{v}=\cot ^{-1}\left(\frac{\cos \theta}{\sqrt{\sin ^{2} \theta}}\right)\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

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

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

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

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

$\Rightarrow v=\cos ^{-1} x$

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

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

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

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

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

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

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

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

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