Question

Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) w . r . t . \sin^{- 1} \frac{2x}{1 + x^2},\]tan-11+x2-1x w.r.t. sin-12x1+x2, if x ∈ (–1, 1) .

Answer 1

\[\text { Let }u = \tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\]

\[\text { Put } x = \tan\theta\]

\[ \therefore u = \tan^{- 1} \left( \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan\theta} \right)\]

\[ \Rightarrow u = \tan^{- 1} \left( \frac{\sec\theta - 1}{\tan\theta} \right) \]

\[ \Rightarrow u = \tan^{- 1} \left( \frac{1 - \cos\theta}{\sin\theta} \right) \]

\[ \Rightarrow u = \tan^{- 1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \right) \]

\[ \Rightarrow u = \tan^{- 1} \left( \tan\frac{\theta}{2} \right)\]

\[\text { Here }, - 1 < x < 1\]

\[ \Rightarrow - 1 < \tan\theta < 1 \]

\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4} \]

\[ \therefore u = \frac{\theta}{2} \left[ \because \tan^{- 1} \left( \tan\theta \right) = \theta, if \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right] \]

\[ \Rightarrow u = \frac{1}{2} \tan^{- 1} x \left[ \because x = \tan\theta \right]\]

Differentiating both sides with respect to x, we get

\[\frac{du}{dx} = \frac{1}{2\left( 1 + x^2 \right)}\]

\[v = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\]

\[\Rightarrow v = 2 \tan^{- 1} x\]

Differentiating both sides with respect to x, we get

\[\frac{dv}{dx} = \frac{2}{1 + x^2}\]

\[\therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{2\left( 1 + x^2 \right)}}{\frac{2}{1 + x^2}}\]

\[ \Rightarrow \frac{du}{dv} = \frac{1}{4}\]