Differentiate
Question:

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

$x \in(-1,0)$

Solution:

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

However, $x=\cos \theta$

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

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

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

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

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

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

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

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

$\Rightarrow \frac{d u}{d x}=0-\left(-\frac{1}{\sqrt{1-x^{2}}}\right)$

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

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

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

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

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

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