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