`y = x^x` Use logarithmic differentiation to find the derivative of the function.

Textbook Question

Chapter 3, 3.6 - Problem 43 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Since variable is raised to a variable power in this function, the common rules of differentiation do not work. First, you need to apply the natural logarithm both sides, such that:

`ln y = ln(x^x)`

Using properties of logarithm, yields:

`ln y = x*ln x`

Differentiating both sides, yields:

`(1/y)*y' = x'*ln x + x*(ln x)'`

`(1/y)*y' = ln x + x*(1/x)`

Reducing like terms, yields:

`(1/y)*y' = ln x + 1`

`y' = y*( ln x + 1)`

Replacing `x^x` for y, yields:

`y' =(x^x)*( ln x + 1)`

Hence, evaluating the derivative of the function, using logarithmic differentiation, yields `y' =(x^x)*( ln x + 1).`

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