Question

Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text {  if }0 < x < 1\] ?

Answer 1

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

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

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

\[ \Rightarrow u = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)\]

\[\text { let, v} = \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)\]

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

\[ \Rightarrow v = \cos^{- 1} \left( \cos2\theta \right) . . . \left( ii \right)\]

\[\text { Here, } 0 < x < 1\]

\[ \Rightarrow 0 < \tan\theta < 1\]

\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]

\[\text { So, from equation } \left( i \right), \]

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

\[ \Rightarrow u = 2 \tan^{- 1} x \left[ \text { Since,} x = \tan\theta \right]\]

differentiating it with respect to x,

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

\[\text { From equation } \left( ii \right), \]

\[v = \theta .........\left[ \text { Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]

\[ \Rightarrow v = 2 \tan^{- 1} x ......\left[ \text { Since, } x = \tan\theta \right]\]

Differentiating it with respect to x,

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

\[\text { Dividing equation } \left( iii \right) \text { by }\left( iv \right), \]

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

\[ \therefore \frac{du}{dv} = 1\]