## Calculus: Early Transcendentals 8th Edition

$$g'(x) = \sqrt{x + x^3}$$
$$g(x) = \int\limits_0^x{\sqrt{t + t^3}}dt$$ Plug in the upper bound into t and take the derivative of the upper bound. $$g'(x) = \sqrt{(x) + (x)^3} \times (x)'$$ Simplify $$g'(x) = \sqrt{x + x^3}$$