Show function $F: [0,1] \rightarrow \int_x^{x^3}f(t)dt$ is differentiable given continous function f
Define $G(x)=\int_0^x f(t)dt$, then $G$ is differentiable on $(0,1)$. Note that $F(x)=G(x^3)-G(x)$. Differentiability of $F$ follows from differentiability of $G$ and chain rule.