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Question
Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text { if }0 < x < 1\] ?
Solution
\[\text { Let, u } = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[\text { Put x } = \tan\theta\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)\]
\[\text { let, v} = \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)\]
\[ \Rightarrow v = \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow v = \cos^{- 1} \left( \cos2\theta \right) . . . \left( ii \right)\]
\[\text { Here, } 0 < x < 1\]
\[ \Rightarrow 0 < \tan\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]
\[\text { So, from equation } \left( i \right), \]
\[u = 2\theta \left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta,\text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow u = 2 \tan^{- 1} x \left[ \text { Since,} x = \tan\theta \right]\]
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,
\[\frac{dv}{dx} = \frac{2}{1 + x^2} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) \text { by }\left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2}\]
\[ \therefore \frac{du}{dv} = 1\]
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