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 $\sin t^2\,$.

You can take the derivative of an integral, even though you can't directly integrate it.
let
$I = \int \sin t^2 dt$
The value that you care about is $$R = I(0) - I(\sqrt x)$$
since we are asked to evaluate the integral from $\sqrt x$ to $0$.
Now, differentiating $R$ by $dx$, we get the expression
$$\frac{dR}{dx} = \frac{d I(0)}{dx} - \frac{dI(\sqrt x)}{dx}$$
Since $I(0)$ is a value (constant), it simply evaluates to $0$
using chain rule on the second term,
$$R = 0 - \frac{dI(\sqrt x)}{dx} = 0 - I'(\sqrt x) \cdot \frac{1}{2 \sqrt x}$$
Now, we need to find $I'$. Since $$I = \int \sin t^2 dt\\I' = dI = \sin t^2$$
Hence, $$I'(\sqrt x) = \sin ({\sqrt x}^2) = \sin x$$
Therefore,
$$R = 0 - \sin x \cdot \frac{1}{2 \sqrt x} = \frac{- \sin x}{2 \sqrt x}$$
