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))} ?
Formula for derivative of the inverse function
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derivatives
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0Ok, 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.. – 2012-03-11
2 Answers
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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.
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0now I understand everything, thank you :-) – 2012-03-11
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To find $y=\operatorname{arcsinh} x = \frac{e^x-e^{-x}}{2}$ put $t= e^x$, which leads to an easy equation.