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What is the derivative of $\ln\left(\left(\frac{1-x}{1+x}\right)^2\right)$ with respect to $x$? I used chain rule and my answer was $-4\frac{1-x}{1+x}$. But the answer should be $\frac{-4}{1-x^2}$. Is that right?

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  • $\begingroup$ Use the chain-rule. en.wikipedia.org/wiki/Chain_rule $\endgroup$
    – Cornman
    Jul 28 '18 at 0:28
  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Find", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$
    – Clement C.
    Jul 28 '18 at 0:29
  • $\begingroup$ You accepted my answer, but I calculated the derivative of a "wrong" function, because I interpreted your notation like that. But your edit makes clear it is an other function. See user534382. $\endgroup$
    – Cornman
    Jul 28 '18 at 0:42
  • $\begingroup$ @Cornman, I apologize for that. I typed it wrongly the first time. $\endgroup$
    – user139289
    Jul 28 '18 at 0:44
  • $\begingroup$ You are welcome. I should have asked before editing. $\endgroup$
    – Cornman
    Jul 28 '18 at 0:46
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I assume you meant $y=\ln\left(\left(\frac{1-x}{1+x}\right)^2\right)$?

Then $y = 2 \left[ \ln(1-x) - \ln(1+x) \right]$.

With chain-rule:

$$\frac{\mathrm dy}{\mathrm dx} = 2 \left[ \frac{-1}{1-x} - \frac{1}{1+x} \right] = -2\cdot\frac{2}{1-x^2} = \frac{-4}{1-x^2}$$

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