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 $\sin t^2\,$.
 A: With the fundamental theorem of calculus, since $\;\sin t^2\;$ is continuous everywhere, you have
$$\int_{\sqrt x}^0\sin t^2\;dt=F(0)-F(\sqrt x)\;,\;\;\text{with}\;\;F'(t)=\sin t^2$$
Thus, to differentiate the above, just apply the chain rule and use the fact that $\;F\;$ is a primitive of the integrand:
$$\left(\int_{\sqrt x}^0\sin t^2\;dt\right)'=\left(F(0)-F(\sqrt x)\right)'=-F'(\sqrt x)\cdot\frac1{2\sqrt x}=-\frac{\sin x}{2\sqrt x}$$
A: 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,
$$R = 0 - \sin x \cdot \frac{1}{2 \sqrt x} = \frac{- \sin x}{2 \sqrt x}$$
A: 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, 
$$
F'(x)=-\sin((\sqrt{x})^2)\cdot\frac{1}{2\sqrt{x}}=\frac{-\sin(|x|)}{2\sqrt{x}}=\frac{-\sin(x)}{2\sqrt{x}}.
$$
