Prove $\frac{d}{dx}{\rm arctanh}(\ln \cosh x) = \frac{\tanh x}{1-(\ln \cosh x)^2}$

Question

In the book "Lehrbuch der Analysis Teil I" of Heuser page 303, there was a task: Prove $$\frac{d}{dx}{\rm arctanh}(\ln \cosh x) = \frac{\tanh x}{1-(\ln \cosh x)^2}.$$ When I tried, I ended up with $$\frac{d}{dx}{\rm arctanh}(\ln \cosh x) = \tanh x \frac 1 2 \frac{1+\ln \cosh x}{1-\ln \cosh x},$$ from which it should be able to say $$\frac 1{1-(\ln \cosh x)^2} = \frac 1 2 \frac{1+\ln \cosh x}{1-\ln \cosh x}.$$ How should it lead to the proof? I made already mistake at the earlier step?

Answer 1

$$\ \frac{d}{dx}{\rm arctanh[f(x)]}=\frac{f'(x)}{1-[f(x)]^2}$$

So you have: $$\ \frac{d}{dx}{\rm arctanh}(\ln(\cosh(x)))=(\ln(\cosh(x)))'\cdot\frac{1}{1-[\ln(\cosh(x))]^2}=$$ $$\ =\frac{\tanh(x)}{1-[\ln(\cosh(x))]^2}$$

I do not understand how you get your result