Derivatives of ln(x): Taking on Complex Problems

[jumbogala]

Homework Statement

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)?

Homework Equations

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.

 

[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

your question: ln[x⁵(x+4)³(x³+4)⁶]

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 :)