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Differentiate $\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right)$ with respect to $\sqrt{1-4 x^{2}}$, if

$x \in\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2}\right)$


Let $u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right)$ and $v=\sqrt{1-4 x^{2}}$

We need to differentiate $u$ with respect to $v$ that is find $\frac{\text { du }}{\text { dv }}$.

We have $u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right)$

$\Rightarrow u=\sin ^{-1}\left(4 x \sqrt{1-(2 x)^{2}}\right)$

By substituting $2 x=\cos \theta$, we have

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

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

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

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

$\Rightarrow u=\sin ^{-1}(\sin 2 \theta)$

Given $x \in\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2}\right)$

However, $2 x=\cos \theta \Rightarrow x=\frac{\cos \theta}{2}$

$\Rightarrow \frac{\cos \theta}{2} \in\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2}\right)$

$\Rightarrow \cos \theta \in\left(\frac{1}{\sqrt{2}}, 1\right)$

$\Rightarrow \theta \in\left(0, \frac{\pi}{4}\right)$

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

Hence, $u=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\Rightarrow u=2 \cos ^{-1}(2 x)$

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

$\frac{d u}{d x}=\frac{d}{d x}\left[2 \cos ^{-1}(2 x)\right]$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=2 \frac{\mathrm{d}}{\mathrm{dx}}\left[\cos ^{-1}(2 \mathrm{x})\right]$

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

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

$\Rightarrow \frac{d u}{d x}=-\frac{2}{\sqrt{1-4 x^{2}}}\left[\frac{d}{d x}(2 x)\right]$

$\Rightarrow \frac{d u}{d x}=-\frac{2}{\sqrt{1-4 x^{2}}}\left[2 \frac{d}{d x}(x)\right]$

$\Rightarrow \frac{d u}{d x}=-\frac{4}{\sqrt{1-4 x^{2}}} \frac{d}{d x}(x)$

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$

$\Rightarrow \frac{d u}{d x}=-\frac{4}{\sqrt{1-4 x^{2}}} \times 1$

$\therefore \frac{d u}{d x}=-\frac{4}{\sqrt{1-4 x^{2}}}$

In part (i), we found $\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{4 \mathrm{x}}{\sqrt{1-4 \mathrm{x}^{2}}}$

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

$\Rightarrow \frac{d u}{d v}=\frac{-\frac{4}{\sqrt{1-4 x^{2}}}}{-\frac{4 x}{\sqrt{1-4 x^{2}}}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=-\frac{4}{\sqrt{1-4 \mathrm{x}^{2}}} \times\left(-\frac{\sqrt{1-\mathrm{x}^{2}}}{4 \mathrm{x}}\right)$

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

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

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