Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to derive this

$$ f(x)=\frac{6}{1+2e^{-5x}}$$

and getting this

$$ f'(x)=\frac{0(1+2e^{-5x})-6(0-10e^{-5x})}{(1+2e^{-5x})^2}$$


But the answer I get when checking on Wolfram Alpha is


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?

share|cite|improve this question
They both are the same....:) .It's a nice exercise: take your expression, simplify it and show it's the same as in WA. – DonAntonio Oct 29 '12 at 14:26
Oops! The expression must have in the denominator $\,(e^{5x}+2)^2\,$...check this! – DonAntonio Oct 29 '12 at 14:27
up vote 4 down vote accepted

You seemed to have copied the equation from Wolfram wrong (there is no $2$ in front of the $e^{5x}$). To see how the two expressions are the same multiply the top and bottom of the fraction by $(e^{5x})^2$.

share|cite|improve this answer
Oh, the 2 sneaked it's way in there. Thanks to both for verifying and guiding. :) – jiku Oct 29 '12 at 14:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.