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Question
Differentiate\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\] with respect to \[\sin^{-1} \left( \frac{2x}{1 + x^2} \right)\], If \[- 1 < x < 1, x \neq 0 .\] ?
Solution
\[\text { Let, u }= \tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\]
\[\text { put x }= \tan\theta\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan\theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{sec\theta - 1}{\tan\theta} \right) \]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{1 - \cos\theta}{\sin\theta} \right) \]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \right) \]
\[ \Rightarrow u = \tan^{- 1} \left( \tan\frac{\theta}{2} \right) . . . \left( i \right)\]
\[\text { And,} \]
\[ v = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\]
\[ \Rightarrow v = \sin^{- 1} \left( \frac{2\tan\theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow v = \sin^{- 1} \left( \sin2\theta \right) . . . \left( ii \right)\]
\[\text { Here }, \]
\[ - 1 < x < 1\]
\[ \Rightarrow - 1 < \tan\theta < 1 \]
\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4} . . . \left( A \right) \]
\[\text { So, from equation } \left( i \right), \]
\[u = \frac{\theta}{2} .........\left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right] \]
\[ \Rightarrow u = \frac{1}{2} \tan^{- 1} x ..........\left[ \text { since, } x = \tan\theta \right]\]
Differentiating it with respect to x,
\[\frac{du}{dx} = \frac{1}{2}\left( \frac{1}{1 + x^2} \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{1}{2\left( 1 + x^2 \right)} . . . \left( i \right)\]
\[\text { Now, from equation } \left( ii \right) \text { and } \left( A \right), \]
\[v = 2\theta .........\left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow v = 2 \tan^{- 1} x .........\left[ \text { Since, } x = \tan\theta \right]\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = 2\left( \frac{1}{1 + x^2} \right) . . . \left( iv \right)\]
\[\text { dividing equation } \left( iii \right) \text { by } \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{1}{2\left( 1 + x^2 \right)} \times \frac{1 + x^2}{2}\]
\[ \therefore \frac{du}{dv} = \frac{1}{4}\]
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