Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 43

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).