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

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1. anonymous

1/(e^x+1)

2. anonymous

how did you get that?

3. anonymous

is it because ln(e^x) = x and the derivative of x = 1??

4. 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.

5. anonymous

oh okay.. thanks!

6. 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

7. anonymous

that's what i've been trying to work with because i realized that the function is f'(x)/f(x)!

8. 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.

9. anonymous

ohh i see what you're saying... so then the derivative of ln(e^x) will be 1

10. anonymous

and the derivative of ln(1+e^x) is e^x/1+e^x?

11. anonymous

Indeed it is

12. anonymous

fantastic! thanks so much.. i'll definitely be posting more up soon.. im studying for a calc 2 final!

13. anonymous

:)

14. 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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