<s>[INST] Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$ 
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.

Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$, 
then  $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$  y^2+ e^{2x}-2(e^x)y= 1$$ 
So if   $e^x= a$, then $a^2-2ay-1= 0$
 [/INST] $$y=\log(x+\sqrt{x^2+1})\implies x+\sqrt{x^2+1}=e^y\implies x^2+1=e^{2y}-2xe^y+x^2\implies$$
$$2e^yx=e^{2y}-1\implies x=\frac{e^{2y}-1}{2e^y}=\frac{e^y-e^{-y}}2=\sinh y (=\text{hyperbolic sine})$$
Also:
$$f'(x)=\left(1+\frac x{\sqrt{x^2+1}}\right)\frac1{x+\sqrt{x^2+1}}=\frac1{\sqrt{x^2+1}}$$
Thus, by the theorem of the derivative of the inverse function:
$$(f^{-1})'(x)=\left.\frac1{f'(x)}\right|_{x\leftrightarrow y}=\left.\sqrt{x^2+1}\right|_{x\leftrightarrow y} =\sqrt{\left(\frac{e^y-e^{-y}}2\right)^2+1}=\frac{e^y+e^{-y}}2=\cosh y$$
You can, of course, also differentiate directly the explicit formula for $\;f^{-1}\;$ .
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