Solve this

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\$

Leave a comment

Click here to get exam-ready with eSaral