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प्रश्न
If \[y = \left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)\] , prove that \[\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)\] ?
उत्तर
\[\text{ We have, y } = \left( x - 1 \right) \log\left( x - 1 \right) - \left( x + 1 \right) \log\left( x + 1 \right)\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ \left( x - 1 \right) \log\left( x - 1 \right) - \left( x + 1 \right) \log\left( x + 1 \right) \right]\]
\[ = \left[ \left( x - 1 \right)\frac{d}{dx}\log\left( x - 1 \right) + \log\left( x - 1 \right)\frac{d}{dx}\left( x - 1 \right) \right] - \left[ \left( x + 1 \right)\frac{d}{dx}\log\left( x + 1 \right) + \log\left( x + 1 \right)\frac{d}{dx}\left( x + 1 \right) \right] \]
\[ = \left[ \left( x - 1 \right) \times \frac{1}{\left( x - 1 \right)}\frac{d}{dx}\left( x - 1 \right) + \log\left( x - 1 \right) \times \left( 1 \right) \right] - \left[ \left( x + 1 \right) \times \frac{1}{\left( x + 1 \right)} \times \frac{d}{dx}\left( x + 1 \right) + \log\left( x + 1 \right)\left( 1 \right) \right]\]
\[ = \left[ 1 + \log\left( x - 1 \right) \right] - \left[ 1 + \log\left( x + 1 \right) \right]\]
\[ = \log\left( x - 1 \right) - \log\left( x + 1 \right)\]
\[ = \log\frac{\left( x - 1 \right)}{\left( x + 1 \right)} \]
\[So, \frac{d y}{d x} = \log\frac{\left( x - 1 \right)}{\left( x + 1 \right)}\]
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