anonymous
  • anonymous
Find the derivative of the function.. g(x)= ln[(e^x)/(1+e^x)]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1/(e^x+1)
anonymous
  • anonymous
how did you get that?
anonymous
  • anonymous
is it because ln(e^x) = x and the derivative of x = 1??

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anonymous
  • anonymous
Wolfram told me so. I really don't feel like doing that derivative. If you want to see the steps type it in yourself, it shows you the steps.
anonymous
  • anonymous
oh okay.. thanks!
anonymous
  • anonymous
Wolfram do it in an ugly way. In general: \[\frac{\mathbb{d}}{\mathbb{d}x} \ln f(x) = \frac{f'(x)}{f(x)} \] Make it a bit slicker by splitting it up using Ln(a/b) = Ln a - Ln b before differentiating
anonymous
  • anonymous
that's what i've been trying to work with because i realized that the function is f'(x)/f(x)!
anonymous
  • anonymous
i.e \[ \frac{\mathbb{d}}{\mathbb{d}x} \ln \frac{e^x}{1+e^x} = \frac{\mathbb{d}}{\mathbb{d}x} \ln e^x - \frac{\mathbb{d}}{\mathbb{d}x}\ln 1+ e^x = \text{blah...} \] Unfortunately, I think that is nicer than something involving using the fact it is on the form ln[f'(x)/f(x)] , but I could be missing something.
anonymous
  • anonymous
ohh i see what you're saying... so then the derivative of ln(e^x) will be 1
anonymous
  • anonymous
and the derivative of ln(1+e^x) is e^x/1+e^x?
anonymous
  • anonymous
Indeed it is
anonymous
  • anonymous
fantastic! thanks so much.. i'll definitely be posting more up soon.. im studying for a calc 2 final!
anonymous
  • anonymous
:)
anonymous
  • anonymous
Ouch, sorry. Calc 2 was easy, but I absolutely hated series haha. Easy while you are doing them, but even easier to forget :). Vector calculus is so much more fun.. (*sarcasm*)

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