Question

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

Answer 1

1/(e^x+1)

Answer 2

how did you get that?

Answer 3

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

Answer 4

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.

Answer 5

oh okay.. thanks!

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

Indeed it is

Answer 12

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

Answer 13

:)

Answer 14

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*)