# Thread: Differentiate a mixture of e's

1. ## Differentiate a mixture of e's

Hi i'm just looking to see if i've got this right:

differentiate:

$\displaystyle \frac{e^x}{e^x+e^{-x}}$

I got this:

$\displaystyle \frac{2}{e^{2x}+e^{-2x}}$

Something at the back of my head says "that looks wrong" so i'm seeking clarifcation and/or guidance

Thank you

2. Originally Posted by aceband
Hi i'm just looking to see if i've got this right:

differentiate:

$\displaystyle \frac{e^x}{e^x+e^{-x}}$

I got this:

$\displaystyle \frac{2}{e^{2x}+e^{-2x}}$

Something at the back of my head says "that looks wrong" so i'm seeking clarifcation and/or guidance

Thank you
Not sure how you got the denominator. I got

$\displaystyle \frac{2}{\left(e^x + e^{-x}\right)^2}$

3. $\displaystyle \frac{d}{dx}\frac{e^x}{e^x+e^{-x}}= \frac{e^x(e^x+e^{-x})-e^x(e^x-e^{-x})}{(e^x+e^{-x})^2}=\frac{2}{e^{2x}+e^{-2x}+2}$

4. Thanks guys,

How did you get that +2 in the bottom though?

5. $\displaystyle 2e^xe^{-x} = 2e^{x-x} = 2e^0 = 2$

6. Originally Posted by aceband
Thanks guys,

How did you get that +2 in the bottom though?
$\displaystyle \left(e^x + e^{-x}\right)^2 = \left(e^{x}\right)^2 + 2e^x \cdot e^{-x} + \left(e^{-x}\right)^2$ and the middle term when simplified give 2.

7. AHH!!! Fantastic now i get it - i stupidly thought 2 * e^0 was 0

Thank you so much for the help.