Stuck With The Differentiation Of A Inverse Hyperbolic Function

Question

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

Answer 1

You’re working way too hard. Simplify! What is $\cosh^{-1}\big(\cosh x\big)$?

Answer 2

If we are being fussy, the answer is not $2x$.

The function $\cosh x$ is an even function, like $\cos x$. Note that $\cosh^{-1} w$ is the non-negative number whose $\cosh$ is $w$. Now let $x$ be negative. Then $\cosh^{-1}(\cosh x)=-x$. Alternately, $\cosh^{-1}(\cosh x)=|x|$.

So the function we are trying to differentiate is in fact $x^2$ when $x\ge 0$ and $-x^2$ when $x\lt 0$. Its derivative is $2x$ when $x \ge 0$ and $-2x$ when $x\lt 0$, or alternately $2|x|$.