Help using Fundemental Theorem of Calculus for the first time So I want to try using the fundamental theorem of calculus:

$F(x) = \int_a^x f(t) dt$ defined in $[a,b]$
$F'(x) = f(x)$

I think that I have a suitable question:

Find $F'(x)$, where $F(x) = \int_0^{x^3} \sin^3t \; dt$

Now I don't understand, is $F'(x)$ actually $\frac{df}{dx}$ here because if it were $\frac{df}{dt}$ I could just take away the integral and it would be solved?
 A: No, $F'(x)$ is $\dfrac{\text{d}F}{\text{d}x}$. Here's how you should approach the question.
By the first part of the fundamental theorem of calculus, you know that 
$$ \int\limits_{0}^{x^3}\sin^{3}(t)\text{ d}t = G(x^3)-G(0)\text{,}$$
where $G$ is an antiderivative of $\sin^{3}$; that is, $G'(x) = \sin^{3}(x)$. Take the derivative of this quantity and you get 
$$\dfrac{\text{d}F}{\text{d}x} = \dfrac{\text{d}}{\text{d}x}\left[\int\limits_{0}^{x^3}\sin^{3}(t)\text{ d}t\right] = \dfrac{\text{d}}{\text{d}x}\left[G(x^3)-G(0)\right] = G'(x^3)(3x^2)-0 = G'(x^3)(3x^2)\text{.} $$
I made use of the chain rule for derivatives and that $G(0)$ is a constant and thus has derivative 0. By definition, $G'(x) = \sin^{3}(x)$ and thus your answer is $3x^2\sin^{3}(x^3)$.
A side note: This is a topic that in my four years of tutoring Calculus I material that students frequently struggle with. My experience is that once one sees how it's done once (using FTC part I and the chain rule), (s)he picks it up very quickly.
A: Hint : 
Observe that $F(x)=H(G(x))$ where $G(x) = x^3$ and $H(x) = \int_0^x \sin^3 t \text{d}t$ and use the chain rule. 
A: I think that you meant $\frac{dF}{dx}$ and $\frac{dF}{dt}$, respectively.
It certainly couldn't be $\frac{dF}{dt}$ because $t$ is internal to the integration operation. But, it's not $\frac{DF}{dx}$ either.
To find $F'(x)$ where $F(x) = \int_0^{x^3}\sin^3tdt$ we use the chain rule
$$\frac{d}{x^3}\int_0^{x^3}\sin^3tdt\cdot \frac{dx^3}{dx} = 3x^2\sin^3x^3$$
If you're just starting out with the the chain rule you may want to re-express it as
$$\frac{d}{dx}\int_0^{u(x)}\sin^3tdt=\frac{d}{du}\int_0^{u(x)}\sin^3tdt\cdot \frac{du}{dx} = u'(x)\sin^3(u(x))=3x^2\sin^3x^3$$
where $u=x^3$
In general, suppose
$$F(x)=\int_a^{u(x)}f(t)dt$$
then
$$F'(x) = f(u(x))\cdot u'(x)$$
