Formula for derivative of the inverse function

How to deduce formula for $\operatorname{arcsinh}(x)$ knowing that $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$, $\sinh'(x)=\cosh x$ and $(f^{-1}(x))'=\dfrac{1}{f'(f^{-1}(x))}$ ?

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Did you even try anything? All you need is given, so if you had just plugged in everything in that last formula you would have obtained the result yourself. – TMM Mar 11 '12 at 15:19
I do not understand your indignation. I've tried, but I couldn't. I've explained in my comment under the answer which fragment I don't understand.. – xan Mar 11 '12 at 15:24
If Alex had answered simply $\mathrm{arcsinh}'(x) = \frac{1}{\sqrt{x^2 + 1}}$, would it have helped you? You want others to explain what they did, so the least we can ask is you to do the same. – TMM Mar 11 '12 at 15:32
Ok, in future I will think twice before I ask, but nobody yelled at me because of my questions before, so I didn't realise that.. – xan Mar 11 '12 at 15:50

Yes. Letting $f(x)=\sinh(x)$ we have $f'(x)=\frac{e^x+e^{-x}}{2}=\cosh(x)$ thus $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}=\frac{1}{\cosh(\mathrm{arcsinh}(x))}=\frac{1}{\sqrt{x^2+1}}.$$
It is worth noting that the formula $(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$ comes from the chain rule, as $$1=\frac{d}{dx} x=\frac{d}{dx}f(f^{-1}(x))=f'(f^{-1}(x))(f^{-1}(x))'$$ so rearranging gives the desired result.
I don't understand this: $\frac{1}{\cosh(\mathrm{arcsinh}(x))}=\frac{1}{\sqrt{x^2+1}}$ Where it came from? – xan Mar 11 '12 at 15:21
@xan: It comes from $\cosh^2 w-\sinh^2 w=1$, which can be verified directly from the definition of $\cosh$ and $\sinh$. – André Nicolas Mar 11 '12 at 15:32
To find $$y=\operatorname{arcsinh} x = \frac{e^x-e^{-x}}{2}$$ put $t= e^x$, which leads to an easy equation.