TITLE: Derivative of $\sec^{-1} e^{2x}$, my answer differs from wolfram alpha.
QUESTION [6 upvotes]: The problem: Find the derivative of $\sec^{-1} e^{2x}$ (\arcsec doesn't seem to work)
My work: 
$u= e^{2x}$
$\mathrm{d}u = 2e^{2x}\,\mathrm{d}x$ 
The formula I know for the derivative of arcsec(u) is $\frac{u^{\prime}}{\vert u \vert \sqrt{u^2-1}}$
Which gives me:
$$\frac{2e^{2x}}{\vert e^{2x} \vert \sqrt{\left(e^{2x}\right)^2 -1 }}$$
simply! 
$$\frac{2}{\sqrt{e^{4x}-1}} $$
always check your work:
Wolframalpha gives: $$  \frac{2}{e^{2x}\sqrt{1-e^{-4x}}}$$ 
I have tried to several times to make sure they are algebraically the same, but atlas, I have failed on that attempt.   I know that wolfram alpha is more trustworthy than my mathematical skills. 
THus, I wonder where my mistake lies.

REPLY [9 votes]: You haven't made a mistake.
$$e^{2x}\sqrt{1-e^{-4x}}=\sqrt{e^{4x}}\sqrt{1-e^{-4x}}=\sqrt{e^{4x}(1-e^{-4x})}=\sqrt{e^{4x}-1}$$