Determine the derivative $\frac{dy}{dx}$ of the integral 
Determine the derivative of the integral $$ \,\int_{\sqrt x}^{0}\sin (t^2)dt $$ 

What does this question mean.
I do not understand it and I think you can't integrate $\sin t^2\,$.

Think about the problem in the following way, you have a function $G(z)=-\int_0^z\sin(t^2)dt$.  Now, the function you're interested in is $F(x)=G(\sqrt{x})$.  You want to take the derivative of this function.
Although the formula looks scary (with all the integrals), you can think of it as a black box and use your typical difference quotient to define the derivative.  After breaking up the function as above, the derivative is fairly easy.  Namely, we use the chain rule:
$$F'(x)=G'(\sqrt{x})\cdot(\sqrt{x})'.$$
Using the fundamental theorem of calculus, $G'(z)=-\sin(z^2)$.  Therefore, 
$$
F'(x)=-\sin((\sqrt{x})^2)\cdot\frac{1}{2\sqrt{x}}=\frac{-\sin(|x|)}{2\sqrt{x}}=\frac{-\sin(x)}{2\sqrt{x}}.
$$
