I'm trying to derive this
$ f(x)=\frac{6}{1+2e^{-5x}}$
and getting this
$ f'(x)=\frac{0(1+2e^{-5x})-6(0-10e^{-5x})}{(1+2e^{-5x})^2}$
$=\frac{60e^{-5x}}{(1+2e^{-5x})^2}$
But the answer I get when checking on Wolfram Alpha is
$f'(x)=\frac{60e^{5x}}{(2e^{5x}+2)^2}$
I don't understand how this works. How do the exponents for e become positive all of a sudden, and where does the +2 in the denominator come from?