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प्रश्न
Differentiate \[\tan^{- 1} \left( \frac{2 a^x}{1 - a^{2x}} \right), a > 1, - \infty < x < 0\] ?
उत्तर
\[\text{Let, y } = \tan^{- 1} \left\{ \frac{2 a^x}{1 - a^{2x}} \right\}\]
\[\text{ put }a^x = \tan\theta\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{2 \times a^x}{1 - \left( a^x \right)^2} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{2 \tan\theta}{1 - \tan^2 \theta} \right) \]
\[ \Rightarrow y = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)\]
\[\text{ Here }, - \infty < x < 0\]
\[ \Rightarrow a^{- \infty} < a^x < 2^ \circ\]
\[ \Rightarrow 0 < \tan\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]
\[ \Rightarrow 0 < 2\theta < \frac{\pi}{2}\]
\[\text{ So, from equation } \left( i \right), \]
\[ y = 2\theta ............\left[ Since, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow y = 2 \tan^{- 1} \left( a^x \right) \]
\[\text{ Differentiating it with respect to x }, \]
\[\frac{d y}{d x} = \frac{2}{1 + \left( a^x \right)^2}\frac{d}{dx}\left( a^x \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{2 \times a^x \log_e a}{1 + a^{2x}}\]
\[ \therefore \frac{d y}{d x} = \frac{2 a^x \log_e a}{1 + a^{2x}}\]
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