Find the indicated derivatives or value and simplify

Question

√x

g(x) = ∫ √(t³-1) dt, g'(x)

1

Answer 1

g(x) = ∫_{(1 to √x)} [t³ - 1]dt

By the Fundamental Theorem of Calculus, since the lower limit of integration is a constant and the upper limit is a nonconstant function of x, g'(x) is found by multiplying the derivative of the upper limit by the result of replacing t in the integrand by the upper limit of integration.

So, g'(x) = [(√x)³ - 1](√x)' = [x^3/2 - 1](1/2)x^-1/2) = [x^3/2 - 1] / (2x^1/2) = (1/2)x - 1 / (2√x)