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Question
If \[y = \log \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] __________ .
Options
\[\frac{4 x^3}{1 - x^4}\]
\[- \frac{4x}{1 - x^4}\]
\[\frac{1}{4 - x^4}\]
\[- \frac{4 x^3}{1 - x^4}\]
Solution
\[- \frac{4x}{1 - x^4}\]
\[\text { We have, y } = \log\left( \frac{1 - x^2}{1 + x^2} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\frac{1 - x^2}{1 + x^2}}\frac{d}{dx}\left( \frac{1 - x^2}{1 + x^2} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1 + x^2}{1 - x^2}\left[ \frac{\left( 1 + x^2 \right)\left( - 2x \right) - \left( 1 - x^2 \right)\left( 2x \right)}{\left( 1 + x^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{1 - x^2}\left[ \frac{- 2x - 2 x^3 - 2x + 2 x^3}{\left( 1 + x^2 \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 4x}{1 - x^4}\]
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