Question

Differentiate \[\tan^{- 1} \left( \frac{4x}{1 - 4 x^2} \right), - \frac{1}{2} < x < \frac{1}{2}\] ?

Answer 1

\[\text{ Let, y } = \tan^{- 1} \left\{ \frac{4x}{1 - 4 x^2} \right\}\]

\[\text{ put 2x } = \tan\theta\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{2 \tan\theta}{1 - \tan^2 \theta} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)\]

\[\text{ Here }, - \frac{1}{2} < x < \frac{1}{2}\]

\[ \Rightarrow - 1 < 2x < 1\]

\[ \Rightarrow - 1 < \tan\theta < 1\]

\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}\]

\[ \Rightarrow - \frac{\pi}{2} < 2\theta < \frac{\pi}{2}\]

\[\text{ So, from equation } \left( i \right), \]

\[ y = 2\theta ............\left[ \text{ Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]

\[ \Rightarrow y = 2 \tan^{- 1} \left( 2x \right) ...........\left[ \text{ Since }, 2x = \tan\theta \right]\]

\[\text{ Differentiating it with respect to x }, \]

\[\frac{d y}{d x} = 2\left( \frac{1}{1 + \left( 2x \right)^2} \right)\frac{d}{dx}\left( 2x \right)\]

\[ \Rightarrow \frac{d y}{d x} = 2\left( \frac{1}{1 + 4 x^2} \right) \times 2\]

\[ \therefore \frac{d y}{d x} = \frac{4}{1 + 4 x^2}\]