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Question
Differentiate \[\tan^{- 1} \left( \frac{a + bx}{b - ax} \right)\] ?
Solution
\[\text{ Let, y } = \tan^{- 1} \left[ \frac{a + bx}{b - ax} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a + bx}{b}}{\frac{b - ax}{b}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a}{b} + \frac{bx}{b}}{\frac{b}{b} - \frac{ax}{b}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{\frac{a}{b} + x}{1 - \left( \frac{a}{b} \right)x} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{a}{b} \right) + \tan^{- 1} x \left[ \text{ Since }, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) \right]\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = 0 + \frac{1}{1 + x^2}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^2}\]
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