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 $a^2-2ay-1= 0$
 A: Let $y=\ln(x+\sqrt{x^2+1})$, then:
$$e^{y}=x+\sqrt{x^2+1}$$
$$e^{y}-x=\sqrt{x^2+1}$$
$$(e^{y}-x)^2=(\sqrt{x^2+1})^2$$
$$e^{2y}-2xe^{y}+x^2=x^2+1$$
$$e^{2y}-2xe^{y}=1$$
$$e^{2y}-1=2xe^{y}$$
$$x=\frac{e^{y}-e^{-y}}{2}$$
So $\displaystyle f^{-1}(y)=\frac{e^{y}-e^{-y}}{2}$.
A: here is a trick in this particular situation. that is to recognize 
$$(\sqrt{x^2 + 1} +x)(\sqrt{x^2 + 1} - x) = 1 $$ so that 
$$\ln(\sqrt{x^2 + 1} + x) = -\ln(\sqrt{x^2 + 1} - x)$$
now we can find the inverse function. suppose $$\ln(\sqrt{x^2 + 1} + x) = y,$$
then $$\ln(\sqrt{x^2 + 1} - x) = -y$$ exponentiating these two equations and subtracting gives you $$x = {e^y - e^{-y} \over 2} = \sinh y $$
A: $$y=\log(x+\sqrt{x^2+1})\implies x+\sqrt{x^2+1}=e^y\implies x^2+1=e^{2y}-2xe^y+x^2\implies$$
$$2e^yx=e^{2y}-1\implies x=\frac{e^{2y}-1}{2e^y}=\frac{e^y-e^{-y}}2=\sinh y (=\text{hyperbolic sine})$$
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 $\;f^{-1}\;$ .
