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प्रश्न
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?
उत्तर
\[\text{ Let, y } = \tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right)\]
\[\text{ put ax } = \tan\theta\]
\[ \therefore y = \tan^{- 1} \left( \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan\theta} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{sec\theta - 1}{\tan\theta} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{1 - \cos\theta}{\sin\theta} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2\frac{\theta}{2}\frac{\theta}{2}} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \tan\frac{\theta}{2} \right)\]
\[ \Rightarrow y = \frac{\theta}{2}\]
\[ \Rightarrow y = \frac{1}{2} \tan^{- 1} \left( ax \right)\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = \frac{1}{2} \times \left( \frac{1}{1 + \left( ax \right)^2} \right)\frac{d}{dx}\left( ax \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2\left( 1 + a^2 x^2 \right)}\left( a \right)\]
\[ \therefore \frac{d y}{d x} = \frac{a}{2\left( 1 + a^2 x^2 \right)}\]
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