Maths Puzzles

What is the first derivative of x^x and hence of f(x)^g(x)?

Hint: try differentiating the logarithm of the function.

Solution:

For brevity let y' be the first derivative of y, and similarly f'(x) for f(x) and g'(x) for g(x). This also assumes that f(x) and g(x) are differentiable, of course.

Let y = x^x
log y = x log x
y' y⁻¹ = 1 + log x
y' = (1 + logx) x^x

Now let y = f(x)^g(x)
log y = g(x) log f(x)
y' y⁻¹ = g(x) f'(x) f(x)⁻¹ + g'(x) log f(x)
y' = [ g(x) f'(x) f(x)⁻¹ + g'(x) log f(x) ] f(x)^g(x)

It looks ugly, but it's true.