#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 16 maths

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

Given: $\cos ^{-1}\left(4 x^{3}-3 x\right) \text { w.r.t } \tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)$

$\frac{1}{2}

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

\begin{aligned} &v=\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right) \\\\ &\text { Let } x=\cos \theta \end{aligned}

\begin{aligned} &u=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right) \\\\ &u=\cos ^{-1}(\cos 3 \theta) \end{aligned}                        \begin{aligned} &v=\tan ^{-1}\left(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\right) \\ &v=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta) \end{aligned}

\begin{aligned} &\frac{1}{2}

$\begin{array}{ll} u=\cos ^{-1}(\cos 3 \theta)=3 \theta & \theta \in\left(0, \frac{\pi}{3}\right) \\\\ v=\tan ^{-1}(\tan \theta)=\theta & \theta \in\left(0, \frac{\pi}{3}\right) \end{array}$

\begin{aligned} &u=3 \cos ^{-1} x \\\\ &\frac{d u}{d x}=\frac{-3}{\sqrt{1-x^{2}}} \\\\ &v=\cos ^{-1} x \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-3}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=3 \end{aligned}