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

giorgiana1976 | Student

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

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