# Derivatives of ln(x)

1. Jan 21, 2009

### jumbogala

1. The problem statement, all variables and given/known data
I know how to take the derivative of ln(x), it's just 1/x. But what if you had something more complicated than just x?

For example, ln(x4(2x+5)5)?

2. Relevant equations

3. The attempt at a solution

I guess you would still do 1/(x4(2x+5)5), then multiply it by the derivative of the denominator.

Which would be 4x3(2x+5)5 + x4(5(2x+5)4)(2).

Is that correct?

The problem I'm supposed to be doing is actually more complicated, it's ln[x5(x+4)3(x3+4)6]. Would the procedure be similar? I guess I'm not sure about taking the derivative of something with 3 terms, I've only ever seen it done with two.

2. Jan 21, 2009

### prime-factor

Yes. You are correct:

For example:

y= ln(x2)

y' = (1 / x2) .dx2dx

y' = 2x/x2

Also with three terms:

y= u.v.w

y' = u'.v.w + u.v'.w + u.v.w'

or you can deal with u and v and then w.
I think the u'.v.w + u.v'.w + u.v.w' will be a bit quicker

let u = x5 , v= (x+4)3 , w = (x3+4)6

(Note that any of these could be u, v w) But remember not to forget the natural log function when differentiating :)

Last edited: Jan 21, 2009