Differentiate $\sin ^{-1} \sqrt{1-\mathrm{x}^{2}}$ with respect to $\cos ^{-1} \mathrm{x}$, if
$x \in(0,1)$
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, $\frac{\mathrm{du}}{\mathrm{dv}}=1$
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