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