Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 45

Because the variable is raised to a variable power, you need to apply the natural logarithm both sides, such that:
ln y = ln (x^(sin x))
Using the property of logarithms, yields:
ln y = sin x*ln x
You need to differentiate both sides, using the chain rule to the left side, since y is a function of x, such that:
(1/y)*y' = cos x*ln x + sin x*(1/x)
y' = y*(cos x*ln x + sin x*(1/x))
Replacing x^(sin x) for y, yields:
y' = (x^(sin x))*(cos x*ln x + sin x*(1/x))
Hence, evaluating the derivative of the function, using logarithmic differentiation, yields y' = (x^(sin x))*(cos x*ln x + sin x*(1/x)).