Solve this


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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{4}{\sqrt{1-4 \mathrm{x}^{2}}} \times 1$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=\frac{4}{\sqrt{1-4 \mathrm{x}^{2}}}$

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

We have $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

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

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

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

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

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