Solve this

Solution:

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 $\frac{d u}{d v}$.

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

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

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

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

$\Rightarrow \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$

$\Rightarrow 2 \theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Now, we have $\mathrm{v}=\sqrt{1-4 \mathrm{x}^{2}}$

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

$\frac{d v}{d x}=\frac{d}{d x}\left(\sqrt{1-4 x^{2}}\right)$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(1-4 \mathrm{x}^{2}\right)^{\frac{1}{2}}$

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

$\Rightarrow \frac{d v}{d x}=\frac{1}{2}\left(1-4 x^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(1-4 x^{2}\right)$

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

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

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

$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}\left[0-4\left(2 x^{2-1}\right)\right]$

$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}[-8 x]$

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

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

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

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

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

Thus, $\frac{d u}{d v}=-\frac{1}{x}$