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 $e^{-x}$ $$y = \frac{e^x.e^{-x}}{e^{-x} + e^x.e^{-x}} = \frac{1}{e^{-x} + 1}$$

  • $\begingroup$ 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. $\endgroup$ – user106860 May 15 at 4:56
  • $\begingroup$ You didn't show your inversion method while posting the question and edited after my answer. So I didn't know that you've inverted the fraction. $\endgroup$ – Ak19 May 15 at 5:11
  • $\begingroup$ Sorry about that. $\endgroup$ – user106860 May 15 at 5:53
  • $\begingroup$ No problem!! :) $\endgroup$ – Ak19 May 15 at 5:58

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