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Question:

Differentiate $\cos ^{-1}\left(4 x^{3}-3 x\right)$ with respect to $\tan ^{-1}\left(\frac{1-x^{2}}{x}\right)$, if $\frac{1}{2}

Solution:

Let $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$ and $v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$

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

We have $u=\cos ^{-1}\left(4 x^{3}-3 x\right)$

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

$u=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)$

But, $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$

$\Rightarrow u=\cos ^{-1}(\cos 3 \theta)$

Given, $\frac{1}{2}<\mathrm{x}<1 \Rightarrow \mathrm{x} \in\left(\frac{1}{2}, 1\right)$

However, $x=\cos \theta$

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

$\Rightarrow \theta \in\left(0, \frac{\pi}{3}\right)$

$\Rightarrow 3 \theta \in(0, \pi)$

Hence, $u=\cos ^{-1}(\cos 3 \theta)=3 \theta$

$\Rightarrow u=3 \cos ^{-1} x$

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

$\frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \cos ^{-1} \mathrm{x}\right)$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=3 \frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)=-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=3\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=-\frac{3}{\sqrt{1-\mathrm{x}^{2}}}$

Now, we have $\mathrm{v}=\tan ^{-1}\left(\frac{\sqrt{1-\mathrm{x}^{2}}}{\mathrm{x}}\right)$

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

$\mathrm{v}=\tan ^{-1}\left(\frac{\sqrt{1-(\cos \theta)^{2}}}{\cos \theta}\right)$

$\Rightarrow \mathrm{v}=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\right)$

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

$\Rightarrow \mathrm{v}=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)$

$\Rightarrow \mathrm{v}=\tan ^{-1}(\tan \theta)$

However, $\theta \in\left(0, \frac{\pi}{3}\right)$

Hence, $v=\tan ^{-1}(\tan \theta)=\theta$

$\Rightarrow \mathrm{v}=\cos ^{-1} \mathrm{x}$

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

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

We know $\frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^{2}}}$

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

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

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{-\frac{3}{\sqrt{1-\mathrm{x}^{2}}}}{-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=-\frac{3}{\sqrt{1-\mathrm{x}^{2}}} \times \frac{\sqrt{1-\mathrm{x}^{2}}}{-1}$

$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=3$

Thus, $\frac{d u}{d v}=3$

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