# Differentiate the following questions

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Question:

Differentiate $\sin ^{-1} \sqrt{1-\mathrm{x}^{2}}$ with respect to $\cos ^{-1} \mathrm{x}$, if

$x \in(0,1)$

Solution:

Let $u=\sin ^{-1} \sqrt{1-x^{2}}$ and $v=\cos ^{-1} x$

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

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

$\Rightarrow \mathrm{u}=\sin ^{-1} \sqrt{1-\cos ^{2} \theta}$

$\Rightarrow \mathrm{u}=\sin ^{-1} \sqrt{\sin ^{2} \theta}\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

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

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

(i) Given $x \in(0,1)$

However, $x=\cos \theta$.

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

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

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

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

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

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

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

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

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

Now, on differentiating $v$ with respect to $x$, we get

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

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

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

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

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

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

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