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anonymous
 5 years ago
Find the derivative of the function..
g(x)= ln[(e^x)/(1+e^x)]
anonymous
 5 years ago
Find the derivative of the function.. g(x)= ln[(e^x)/(1+e^x)]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it because ln(e^x) = x and the derivative of x = 1??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wolfram 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
 5 years ago
Best ResponseYou've already chosen the best response.0Wolfram 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
 5 years ago
Best ResponseYou've already chosen the best response.0that's what i've been trying to work with because i realized that the function is f'(x)/f(x)!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.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
 5 years ago
Best ResponseYou've already chosen the best response.0ohh i see what you're saying... so then the derivative of ln(e^x) will be 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the derivative of ln(1+e^x) is e^x/1+e^x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fantastic! thanks so much.. i'll definitely be posting more up soon.. im studying for a calc 2 final!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ouch, 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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