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Question:

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

Solution:

Given $y=(x-1) \log (x-1)-(x+1) \log (x+1)$

On differentiating $y$ with respect to $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[(\mathrm{x}-1) \log (\mathrm{x}-1)-(\mathrm{x}+1) \log (\mathrm{x}+1)]$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}[(x-1) \log (x-1)]-\frac{d}{d x}[(x+1) \log (x+1)]$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}[(x-1) \times \log (x-1)]-\frac{d}{d x}[(x+1) \times \log (x+1)]$

Recall that (uv)' $=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{d y}{d x}=\log (x-1) \frac{d}{d x}(x-1)+(x-1) \frac{d}{d x}[\log (x-1)]$

$-\left(\log (x+1) \frac{d}{d x}(x+1)+(x+1) \frac{d}{d x}[\log (x+1)]\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\log (\mathrm{x}-1)\left[\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}(1)\right]+(\mathrm{x}-1) \frac{\mathrm{d}}{\mathrm{dx}}[\log (\mathrm{x}-1)]$

$-\left(\log (x+1)\left[\frac{d}{d x}(x)+\frac{d}{d x}(1)\right]+(x+1) \frac{d}{d x}[\log (x+1)]\right)$

We know $\frac{d}{d x}(\log x)=\frac{1}{x}$ and $\frac{d}{d x}(x)=1$.

Also, the derivative of a constant is 0 .

$\begin{aligned} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=& \log (\mathrm{x}-1)[1-0]+(\mathrm{x}-1) \times \frac{1}{\mathrm{x}-1} \\ &-\left(\log (\mathrm{x}+1)[1+0]+(\mathrm{x}+1) \times \frac{1}{\mathrm{x}+1}\right) \end{aligned}$

$\Rightarrow \frac{d y}{d x}=\log (x-1)+1-(\log (x+1)+1)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\log (\mathrm{x}-1)-\log (\mathrm{x}+1)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$

Thus, $\frac{\mathrm{dy}}{\mathrm{dx}}=\log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$

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