## Calculus: Early Transcendentals 8th Edition

$$h'(u) = \frac{\sqrt{u}}{u+1}$$
$$h(u) = \int\limits_0^u {\frac{\sqrt{t}}{t+1}}dt$$ Plug in the upper bound into t and multiply by the derivative of the upper bound due to chain rule. $$h'(u) = \frac{\sqrt{u}}{u+1} \times (u)'$$ Simplify. $$h'(u) = \frac{\sqrt{u}}{u+1}$$