Questions regarding The Fundamental Theorem of Calculus FTC1 suggested that
$$
F'(x) = \frac{d}{dx}\int_a^xf(t)dt = f(x)
$$
and the Chain Rule says that 
$$
\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big)=F'\big(g(x)\big)·g'(x).
$$ 
That leads to 
$$
\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big) = g'(x)·\frac{\mathrm{d}}{\mathrm{d}g(x)} \int\limits_a^{g(x)}\!\!f(t)\,\mathrm{d}t = f\big(g(x)\big)·g'(x)
$$
Is it right to combine these two?

If it is, what should I do if I want to find $F'(x)$ instead of $\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big)$ in the combined equation?

I'm actually confused in the following question:

f(t) = $ \int\limits_2^t( {\sqrt {\frac{7}{4}+u^3}}) du  $ 

F(x) = $ \int\limits_1^{\sin x}f(t)dt $

Find: $F''(\pi)$
Leibniz integral rule is what you need 

$$\frac d{dx}F(x)=\frac d{dx}\int_{g(x)}^{h(x)}f(t)dt =f(h(x))h'(x) - f(g(x))g'(x)$$
$$F'(x)=f(\sin x)\cdot\cos(x)=\cos(x)\int_2^{\sin(x)}\sqrt{u^3+7/4}\ \ du$$
Can you take it from here. 

PS: I see you are trying to derive this formula. Look at the wikipedia page for the proofs.
Hint: You know how to differentiate $F(g(x)).$ You want to differentiate $F(x).$ That's just the first one when $g(x)=x.$



I see you've seen that without sufficient context, you may be unclear about what you want, therefore misleading potential answerers about your real problem. Here, instead of being all vague about arbitrary functions, you should have noted right from the get go what it was that was bothering you about a specific question. You only added this later after probably realising that no one seems to understand what you mean. I hope this is a lesson for next time.

In any case, to explain your difficulty, you want to find $F''(π),$ where $$F(x)=\int_1^{\sin x}{f(t)\mathrm d t},$$ and $$f(t)=\int_2^t{\sqrt{\frac74 + u^3}\mathrm d u}.$$

Now, $$F'(x)=(\sin x)'f(\sin x)=\cos x\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}.$$ Thus, we have that $$F''(x)=(\cos x)'\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}+(\cos x)(\sin x)'\sqrt{\frac74 + \sin^3x}=-\sin x\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}+(\cos x)^2\sqrt{\frac74 + \sin^3x}.$$ Now set $x=π.$