The two are equivalent, as a check with wolfram alpha shows.
I can also solve $\frac{e^x}{1+e^{x}} = A+ \frac{1}{1+e^{-x}}$? and I get that $A=0$.
But is there a way that I can directly simplify $\frac{e^x}{1+e^{x}$}$ to get $\frac{1}{1+e^{-x}}$?
I think one way that works is to write $$\frac{e^x}{1+e^{x}$} = \frac{1}{\frac{1+e^{x}}{e^x}} = \frac{1}{e^{-x}+1}$$ which is the desired result.
Is there another way? I ask because inverting the fraction in order to simplify it seems a bit round-about to me, granted iv'e found a few cases where it seems useful
$$y = \frac{e^x}{1+e^x}$$
Dividing Nr. and Dr. by $e^x$
$$y = \frac{1}{\frac{1}{e^x} +1} = \frac{1}{1+\frac{1}{e^x}}$$ $$y = \frac{1}{1+e^{-x}}$$
or simply multiply Nr. and Dr. by $e^{-x}$ $$y = \frac{e^x.e^{-x}}{e^{-x} + e^x.e^{-x}} = \frac{1}{e^{-x} + 1}$$