# How to simplify $\frac{e^x}{1+e^{x}}$ to $\frac{1}{1+e^{-x}}$?

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 $$e^{-x}$$ $$y = \frac{e^x.e^{-x}}{e^{-x} + e^x.e^{-x}} = \frac{1}{e^{-x} + 1}$$
• Dividing numerator and denominator both by $e^x$ seems nicer to me than inverting the fraction (although I think they are the same thing). Thanks. – user106860 May 15 at 4:56