If $g(x) = \text{arctanh}\ (\log x)$, find $g'(x)$.
I tried to separate the terms first and I got $\dfrac12 (\log(1+\log x) - \log(1-\log x))$.
The answer is $\dfrac1{x(1-\log x)^2}$.
2026-04-08 00:22:59.1775607779
If $g(x) = \text{arctanh}\ (\log x)$, find $g'(x)$.
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HINT :
Let $y=\log x$, then you have $g=\text{arctanh}\ y$. Apply chain rule $$ \frac{dg}{dx}=\frac{dg}{dy}\cdot\frac{dy}{dx}. $$