I'am suppose to show that $\frac{\mathrm{d} }{\mathrm{d} x}[x \operatorname{cosh}^{-1}(\cosh x)] = 2x$
And this is what i've tried.Upon differentiating the above function wrt $x$ using the product rule and applying the formula $\cosh^{-1}f(x) = \frac{f\prime(x)}{\sqrt{[f(x)]^2 - 1}}$, I end up getting $x\frac{\sinh x}{\sqrt{(\cosh )^2 - 1}} + \cosh^{-1}(\cosh x)$
How to proceed from here?
Thank You