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Question
Differentiate \[\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] with respect to \[\tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text{ if } - 1 < x < 1\] ?
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Solution
\[\text { Let, u }= \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\]
\[\text { Put x } = \tan\theta \Rightarrow \theta = \tan^{- 1} x, \]
\[ \Rightarrow u = \sin^{- 1} \left( \frac{2\tan\theta}{1 + \tan^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)\]
\[\text { Let, v } = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \tan2\theta \right) . . . \left( ii \right)\]
\[\text { Here, }- 1 < x < 1\]
\[ \Rightarrow - 1 < \tan\theta < 1\]
\[ \Rightarrow - \frac{\pi}{4} < \tan\theta < \frac{\pi}{4}\]
\[\text{ So, from equation } \left( i \right), \]
\[u = 2\theta \left[ \text {Since,} \sin^{- 1} \left( \sin\theta \right) = \theta, \text { if } \theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow u = 2 \tan^{- 1} x\]
Differentiating it with respect to x,
\[\frac{du}{dx} = \frac{2}{1 + x^2} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right), \]
\[v = 2\theta \left[ \text {Since}, \tan^{- 1} \left( \tan\theta \right) = \theta , \text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow v = 2 \tan^{- 1} x\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{2}{1 + x^2} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) \text { by } \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2}\]
\[ \therefore \frac{du}{dv} = 1\]
