What is the result when differentiate the function y=[ln(3x)]^1/6?

This is a composed function and we'll differentiate from
the last function to the first, using the chain rule:

We'll
re-write the function:

y = f(x) =
[ln(3x)]^1/6

The last function is the power function and
the first function is the logarithmic function.

The
function f(x) is the result of composition between u(x) = u^1/6 and v(x) = ln
3x

u(v(x)) =
[ln(3x)]^1/6

u'(x) = (u^1/6)' = (1/6)*u^(1/6 -
1)

u'(x)
=  (1/6)*u^(1-6)/6

u'(x) =
1/6*u^-5/6

u'(x) =
1/(6u^5/6)

v'(x) = (ln 3x)' =
(1/3x)*(3x)'

v'(x) =
3/3x

v'(x) =
1/x

[u(v(x))]' = 1/6x*(ln
3x)^5/6

We'll use the power property of
logarithms:

a*ln b = ln
(b^a)

f'(x) = 1/(ln
3x)^5*6x/6

We'll simplify and we'll
get:

f'(x) = 1/(ln
3x)^5x