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Question
Differentiate \[\tan^{- 1} \left( \frac{1 + ax}{1 - ax} \right)\] with respect to \[\sqrt{1 + a^2 x^2}\] ?
Solution
\[\text { Let, u }= \tan^{- 1} \left( \frac{1 + ax}{1 - ax} \right)\]
\[\text { Put ax } = \tan\theta\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{1 + \tan\theta}{1 - \tan\theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left( \frac{\tan\frac{\pi}{4} + \tan\theta}{1 - \tan\frac{\pi}{4}\tan\theta} \right)\]
\[ \Rightarrow u = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} + \theta \right) \right]\]
\[ \Rightarrow u = \frac{\pi}{4} + \theta\]
\[ \Rightarrow u = \frac{\pi}{4} + \tan^{- 1} \left( ax \right) \left[ \text { Since}, \tan\theta = ax \right] \]
Differentiating it with respect to x,
\[\frac{du}{dx} = 0 + \frac{1}{1 + \left( ax \right)^2}\frac{d}{dx}\left( ax \right) \]
\[ \Rightarrow \frac{du}{dx} = \frac{a}{1 + a^2 x^2} . . . \left( i \right) \]
\[\text { Now,} \]
\[\text { Let, v } = \sqrt{1 + a^2 x^2}\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 + a^2 x^2}}\frac{d}{dx}\left( 1 + a^2 x^2 \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1 + a^2 x^2}}\left( 2 a^2 x \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{a^2 x}{\sqrt{1 + a^2 x^2}} . . . \left( ii \right) \]
\[\text { Dividing equation } \left( i \right) \text { by }\left( ii \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{a}{1 + a^2 x^2} \times \frac{\sqrt{1 + a^2 x^2}}{a^2 x}\]
\[\frac{du}{dv} = \frac{1}{ax\sqrt{1 + a^2 x^2}}\]
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