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Log Function

If y=x-1 log ...

Question

If $y = (x - 1) \log (x - 1) - (x + 1) \log (x + 1)$ , prove that $\frac{d y}{d c} = \log (\frac{x - 1}{1 + x})$

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Solution

$We have, y = (x - 1) \log (x - 1) - (x + 1) \log (x + 1)$
Differentiating with respect to x,
$\frac{d y}{d x} = \frac{d}{d x} [(x - 1) \log (x - 1) - (x + 1) \log (x + 1)] = [(x - 1) \frac{d}{d x} \log (x - 1) + \log (x - 1) \frac{d}{d x} (x - 1)] - [(x + 1) \frac{d}{d x} \log (x + 1) + \log (x + 1) \frac{d}{d x} (x + 1)] = [(x - 1) \times \frac{1}{(x - 1)} \frac{d}{d x} (x - 1) + \log (x - 1) \times (1)] - [(x + 1) \times \frac{1}{(x + 1)} \times \frac{d}{d x} (x + 1) + \log (x + 1) (1)] = [1 + \log (x - 1)] - [1 + \log (x + 1)] = \log (x - 1) - \log (x + 1) = \log \frac{(x - 1)}{(x + 1)} So, \frac{d y}{d x} = \log \frac{(x - 1)}{(x + 1)}$

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