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Find g'(x) when g(x)= ln{(x-1)/(x+1)}
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\[\ln \frac{x-1}{x+1}=\ln (x-1)-\ln(x+1) \ \ \]
\[\frac{d}{dx}\ln f(x)=\frac{f'(x)}{f(x)}\]
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Oh damn I forgot square root everything inside brackets
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