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