Question

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

Answer 1

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

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

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

\[ \Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)\]

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

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

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

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

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

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

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

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

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

Differentiating it with respect to x,

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

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

\[v = 2\theta ........\left[ \text { Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, if \theta \in \left[ 0, \pi \right] \right]\]

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

Differentiating it with respect to x,

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

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

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

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