Power rule for fundamental theorem of calculus Find $\frac{dy}{dx}$, where $y$ is given by
$$y=\left( \int_{0}^{x} (t^3+1)^{10}dt\right)^3$$
The solution says
$$\frac{dy}{dx}=3\left( \int_0^x (t^3+1)^{10}dt\right) \frac{d}{dx}\left( \int_0^x (t^3+1)^{10}dt\right) = 3(x^3+1)^{10}\left( \int_0^x (t^3+1)^{10}dt\right)$$
I'm confused because I thought that the power rule needs to be applied. So that the correct answer should be:
$$\frac{dy}{dx}=3\left( \int_0^x (t^3+1)^{10}dt\right)^2 \frac{d}{dx}\left( \int_0^x (t^3+1)^{10}dt\right) = 3((x^3+1)^{10})^{20}\left( \int_0^x (t^3+1)^{10}dt\right)$$

It is very likely that the "solution" just has a typo: missing power of $2$ in the factor $(\int_0^x(t^3+1)^{10}\;dt)$.
Your first step is correct. But your final answer takes the power of $2$ in the wrong place:
$$\frac{dy}{dx}=3\left( \int_0^x (t^3+1)^{10}dt\right)^2 \frac{d}{dx}\left( \int_0^x (t^3+1)^{10}dt\right) 
= 3\left( \int_0^x (t^3+1)^{10}dt\right)^2(x^3+1)^{10}$$
