Let $g(x) = \int_{0}^{2^x} \sin(t^2)\,dt $. What is the $g'(0)$? I'm kinda stuck in this exercise : 

Let $\displaystyle g(x) = \int_{0}^{2^x}\sin{(t^2)\,dt}$. What is the $g'(0)$?

How do I approach this kind of thing? I was thinking about Riemann sums but does this have any application here?
 A: $g'(x)= 2^x \sin ((2^x)^2)\ln 2$
For more information you can refer to this Differentiation of integral.
A: We will use the Fundamental Theorem of Calculus. Let $F(t)$ be an antiderivative of $\sin(t^2)$. We will not find an explicit formula for $F(t)$, but we won't need to.
We have
$$g(x)=F(2^x)-F(0).$$
Differentiate, using the Chain Rule. Note that since $2^x=e^{x\ln 2}$, the derivative of $2^x$ is $(\ln 2)e^{x\ln 2}=(\ln 2)2^x$.
It follows that
$$g'(x)=(\ln 2)2^x F'(2^x).$$
But $F'(2^x)=\sin((2^x)^2)=\sin(2^{2x})$. Putting things together, we find that
$$g'(x)=(\ln 2)2^x\sin(2^{2x}).$$
One can write the above argument more compactly. By the Fundamental Theorem of Calculus, we have for any continuous function $f(t)$, that the derivative of $\int_a^u f(t)\,dt$ with respect to $u$ is $f(u)$. Letting $u=2^x$, and $a=0$, we find that 
$$g(x)=\int_a^u \sin(t^2)\,dt.$$
Then
$$\frac{dg}{dx}=\frac{dg}{du}\frac{dg}{dx}.$$
But $\frac{dg}{du}=\sin(u^2)$ and $\frac{du}{dx}=(\ln 2)2^x$, and we are essentially finished. 
