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I'm trying to derive this

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

and getting this

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

$$=\frac{60e^{-5x}}{(1+2e^{-5x})^2}$$

But the answer I get when checking on Wolfram Alpha is

$$f'(x)=\frac{60e^{5x}}{(2e^{5x}+2)^2}$$

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?

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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

1 Answer 1

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 $(e^{5x})^2$.

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Oh, the 2 sneaked it's way in there. Thanks to both for verifying and guiding. :) –  eee3 Oct 29 '12 at 14:34

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