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October 1, 2014

October 1, 2014

Posted by **Marissa** on Monday, May 21, 2012 at 12:18pm.

y=ln((1+e^x)/(1-e^x))

I know that the derivative of y=ln(1+e^x) would be (e^x)/(1+e^x), but I'm not sure what to do with the 1-e^x on the bottom here. Can someone help please? Thanks!

- Calculus I -
**Bosnian**, Monday, May 21, 2012 at 12:50pmIn google type:

wolfram alpha

When you see lis of results click on:

Wolfram Alpha:Computational Knowledge Engine

When page be open in rectangle type:

derivartive ln((1+e^x)/(1-e^x))

and click option =

After few secons you will see result.

Then clic option Show steps

- Calculus I -
**Reiny**, Monday, May 21, 2012 at 1:35pmfirst of all recall that

ln (A/B) = ln A - ln B

so

y = ln((1+e^x)/(1-e^x) )

= ln (1+e^x) - ln (1-e^x)

dy/dx = e^x/(1+e^x) - (-e^x)/(1 - e^x)

simplify as needed

- Calculus I -
**Marissa**, Tuesday, May 22, 2012 at 9:39amOkay, thanks guys!

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