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प्रश्न
Differentiate the following with respect to x:
\[\cot^{- 1} \left( \frac{1 - x}{1 + x} \right)\]
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उत्तर
\[\cot^{- 1} \left( \frac{1 - x}{1 + x} \right) = \frac{d}{dx}\left[ \tan^{- 1} \left( \frac{1 + x}{1 - x} \right) \right]\]
\[ = \frac{1}{1 + \left( \frac{1 + x}{1 - x} \right)^2} \times \frac{1 - x + 1 + x}{\left( 1 - x \right)^2}\]
\[ = \frac{\left( 1 - x \right)^2}{\left( 1 - x \right)^2 + \left( 1 + x \right)^2} \times \frac{2}{\left( 1 - x \right)^2}\]
\[ = \frac{\left( 1 - x \right)^2}{1 - 2x + x^2 + 1 + 2x + x^2}\]
\[ = \frac{\left( 1 - x \right)^2}{2\left( 1 + x^2 \right)} \times \frac{2}{\left( 1 - x \right)^2}\]
\[ = \frac{1}{1 + x^2}\]
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