Differentiate $\sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right)$ with respect to $\sec ^{-1}\left(\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)$, if
$x \in(0,1 / \sqrt{2})$
Let $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ and $v=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)$
We need to differentiate $u$ with respect to $v$ that is find $\frac{\mathrm{du}}{\mathrm{dv}}$.
We have $u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$
By substituting $x=\sin \theta$, we have
$u=\sin ^{-1}\left(2 \sin \theta \sqrt{1-(\sin \theta)^{2}}\right)$
$\Rightarrow u=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right)$
$\Rightarrow u=\sin ^{-1}\left(2 \sin \theta \sqrt{\cos ^{2} \theta}\right)\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\Rightarrow u=\sin ^{-1}(2 \sin \theta \cos \theta)$
$\Rightarrow u=\sin ^{-1}(\sin 2 \theta)$
Now, we have $v=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)$
By substituting $x=\sin \theta$, we have
$\mathrm{v}=\sec ^{-1}\left(\frac{1}{\sqrt{1-(\sin \theta)^{2}}}\right)$
$\Rightarrow \mathrm{v}=\sec ^{-1}\left(\frac{1}{\sqrt{1-\sin ^{2} \theta}}\right)$
$\Rightarrow \mathrm{v}=\sec ^{-1}\left(\frac{1}{\sqrt{\cos ^{2} \theta}}\right)\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\Rightarrow \mathrm{v}=\sec ^{-1}\left(\frac{1}{\cos \theta}\right)$
$\Rightarrow \mathrm{v}=\sec ^{-1}(\sec \theta)$
Given $x \in\left(0, \frac{1}{\sqrt{2}}\right)$
However, $x=\sin \theta$
$\Rightarrow \sin \theta \in\left(0, \frac{1}{\sqrt{2}}\right)$
$\Rightarrow \theta \in\left(0, \frac{\pi}{4}\right)$
$\Rightarrow 2 \theta \in\left(0, \frac{\pi}{2}\right)$
Hence, $u=\sin ^{-1}(\sin 2 \theta)=2 \theta$.
$\Rightarrow u=2 \sin ^{-1}(x)$
On differentiating $u$ with respect to $x$, we get
$\frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \sin ^{-1} \mathrm{x}\right)$
$\Rightarrow \frac{d u}{d x}=2 \frac{d}{d x}\left(\sin ^{-1} x\right)$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=2 \times \frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}}$
We have $\theta \in\left(0, \frac{\pi}{4}\right)$'
Hence, $v=\sec ^{-1}(\sec \theta)=\theta$
$\Rightarrow v=\sin ^{-1} x$
On differentiating $v$ with respect to $x$, we get
$\frac{d v}{d x}=\frac{d}{d x}\left(\sin ^{-1} x\right)$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
We have $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{2}{\sqrt{1-\mathrm{x}^{2}}}}{\frac{1}{\sqrt{1-\mathrm{x}^{2}}}}$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}} \times \sqrt{1-\mathrm{x}^{2}}$
$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=2$
Thus, $\frac{\mathrm{du}}{\mathrm{dv}}=2$
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