- #1

jumbogala

- 423

- 4

## 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(x

^{4}(2x+5)

^{5})?

## Homework Equations

## The Attempt at a Solution

I guess you would still do 1/(x

^{4}(2x+5)

^{5}), then multiply it by the derivative of the denominator.

Which would be 4x

^{3}(2x+5)

^{5}+ x

^{4}(5(2x+5)

^{4})(2).

Is that correct?

The problem I'm supposed to be doing is actually more complicated, it's ln[x

^{5}(x+4)

^{3}(x

^{3}+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.