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Question
Differentiate \[\tan^{- 1} \left( \frac{x - 1}{x + 1} \right)\] with respect to \[\sin^{- 1} \left( 3x - 4 x^3 \right), \text { if }- \frac{1}{2} < x < \frac{1}{2}\] ?
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Solution
\[\text { Let, u } = \tan^{- 1} \left( \frac{x - 1}{x + 1} \right)\]
\[\text { Put x }= \tan\theta\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\tan\theta - 1}{\tan\theta + 1} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\tan\theta - \tan\frac{\pi}{4}}{1 + \tan\theta \tan\frac{\pi}{4}} \right) \]
\[ \Rightarrow u = \tan^{- 1} \left[ \tan\left( \theta - \frac{\pi}{4} \right) \right] . . . \left( i \right) \]
\[\text { Here }, - \frac{1}{2} < x < \frac{1}{2}\]
\[ \Rightarrow - \frac{1}{2} < \tan\theta < \frac{1}{2}\]
\[ \Rightarrow - \tan^{- 1} \left( \frac{1}{2} \right) < \theta < \tan^{- 1} \left( \frac{1}{2} \right)\]
\[\text { So, from equation } \left( i \right), \]
\[u = \theta - \frac{\pi}{4} .......\left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow u = \tan^{- 1} x - \frac{\pi}{4} .........\left[ \text { Since, x } = \tan\theta \right]\]
differentiating it with respect to x,
\[\frac{du}{dx} = \frac{1}{1 + x^2} - 0 \]
\[ \Rightarrow \frac{du}{dx} = \frac{1}{1 + x^2} . . . \left( ii \right) \]
\[\text{ And }, \]
\[\text { Let, v } = \sin^{- 1} \left( 3x - 4 x^3 \right)\]
\[\text { Put x } = \sin\theta\]
\[ \Rightarrow v = \sin^{- 1} \left( 3\sin\theta - 4 \sin^3 \theta \right)\]
\[ \Rightarrow v = \sin^{- 1} \left( \sin3\theta \right) . . . \left( iii \right)\]
\[\text { Now }, - \frac{1}{2} < x < \frac{1}{2}\]
\[ \Rightarrow - \frac{1}{2} < \sin\theta < \frac{1}{2}\]
\[ \Rightarrow - \frac{1}{6} < \theta < \frac{\pi}{6}\]
\[\text { So, from equation } \left( iii \right), \]
\[v = 3\theta .........\left[ \text { Since,} \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow v = 3 \sin^{- 1} x .......\left[ \text { Since,} x = \sin\theta \right]\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{3}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) \text { by } \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{1}{1 + x^2} \times \frac{\sqrt{1 - x^2}}{3}\]
\[ \therefore \frac{du}{dv} = \frac{\sqrt{1 - x^2}}{3\left( 1 + x^2 \right)}\]
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