This is an indirect application of the Fundamental Theorem of Calculus which is quickly passed through in many introductory texts. You have an integral function $F(x) = \int^{x}_{1} f(t) dt$; since $t^2 \cos(\pi t )$ is continuous everywhere, we can safely say that $F'(x) = f(x)$ .
When the upper limit is a function of $x$, the FTC will let us write $\int^{u}_{1} f(t) dt = F(u) - F(1)$. When we differentiate this with respect to $x$, the Chain Rule gives us $$\frac{d}{dx}\int^{u}_{1} f(t) dt = \frac{d}{dx} [F(u) - F(1)] = \frac{dF}{du} \cdot \frac{du}{dx} = f(u(x)) \cdot \frac{du}{dx}$$
So for this problem,
$$\frac{d}{dx}\int^{\sqrt{x}}_{1} t^2 \cos(\pi t ) dt = (\sqrt{x})^2 \cos(\pi \sqrt{x} )) \cdot \frac{d}{dx}(\sqrt{x}).$$
To answer your question, you would complete the differentiation and evaluate the result at x = 4 .
(I'll mention, incidentally, that a problem of this type is a favorite final exam question.)