Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 327: 10

$h'(u) =\frac{\sqrt{u}}{u+1}$

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $h(u) = \int ^{u}_{0}\frac{\sqrt{t}}{t+1} dt$ Since $h(u) = \int^{b}_{a}f(t)dt$ $h'(u) = f(u)$ Thus, $h'(u) =\frac{\sqrt{u}}{u+1}$