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\[{d \over dx} \ln {e^x \over 1+e^x}\]
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d/dx(e^x/(1+e^x)/ [e^x/(1+e^x)] = [e^x(1+e^x)-e^x.(e^x) ] / [e^x/(1+e^x)]= 1+e^x
the quotient rule and the derivative of ln
\[\ln \frac{e^x}{1+e^x}=\ln(e^x)-\ln(1+e^x)\]
\[=x-\ln(1+e^x)\]
now its easy to differentiate
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so the answer would be 1- (e^x)/(1+e^x)?
Yes, the answer is: \[1-\frac{e^{x}}{1+e^{x}} \]
hmm, that's what I had at the beginning but i didn't get it on my calculator
\[x-\ln(1+e^x) =>1-\frac{1}{1+e^{x}} *e^x = 1-\frac{e^{x}}{1+e^{x}} = \frac{1+e^{x}-e^{x}}{1+e^{x}} = \frac{1}{1+e^{x}} \]
...oh. could've done that I guess :P Thanks, all!
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