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Question
If \[y = \tan^{- 1} \left( \frac{1 - x}{1 + x} \right), \text{ find} \frac{dy}{dx}\] ?
Solution
\[\text{ We have, y } = \tan^{- 1} \left( \frac{1 - x}{1 + x} \right)\]
\[\Rightarrow \frac{dy}{dx} = \frac{1}{1 + \left( \frac{1 - x}{1 + x} \right)^2}\frac{d}{dx}\left( \frac{1 - x}{1 + x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( 1 + x \right)^2}{1 + x^2 + 2x + 1 + x^2 - 2x}\left[ \frac{\left( 1 + x \right)\frac{d}{dx}\left( 1 - x \right) - \left( 1 - x \right)\frac{d}{dx}\left( 1 + x \right)}{\left( 1 + x \right)^2} \right] \left[ \text{ using quotient rule } \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( 1 + x \right)^2}{2 x^2 + 2}\left[ \frac{\left( 1 + x \right)\left( - 1 \right) - \left( 1 - x \right)\left( 1 \right)}{\left( 1 + x \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( 1 + x \right)^2}{2\left( x^2 + 1 \right)}\left( \frac{- x - 1 - 1 + x}{\left( 1 + x \right)^2} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( 1 + x \right)^2}{2\left( x^2 + 1 \right)} \times \frac{- 2}{\left( 1 + x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{1}{x^2 + 1}\]
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