Question

Find the derivative of f(y)= ln(ln(2y^3))

Answer 1

I understand its the chain rule, but i'm having difficulty finding the correct answer supplied by the book.

Answer 2

\((\ln(u))' = \frac1{u}u'\). where \(u(y)=\ln(2y^3)\).
Now compute \(u'\)... \(u'=(\ln(w))' = \frac1ww'\), where \(w=2y^3\).
Now compute \(w'\). and you'll be done.

Answer 3

So, for the first, its 1/(ln(2y^3)) * ln(2y^3)?

Answer 4

I suppose my difficulty lies in determining what f(x) and g(x) are in relation to this problem, given that the chain rule is f '(g(x)) * g '(x)

Answer 5

in f(g(x)), f is the "outer" function. it is the \(\ln\) function. g is the argument of that function. it can still be a compound function. you then just use the formula again.

Answer 6

So, I should end up with 1/(ln(2y^3)) * 1/(2y^3) * 6y^2?

Answer 7

I think so.

Answer 8

and you can make simplifications.

Answer 9

don't forget that \(\ln(x^\alpha) = \alpha\ln(x)\).

Answer 10

thats what I was missing, thank you very much!

Answer 11

yw