If $F(x) = \int_0^x e^{\sin(t)} dt$ , so $F'(0)$ is equal to: 
If $F(x) = \int_0^x e^{\sin(t)} dt$ , so $F'(0)$ is equal to:

(A) $-2$  $\quad$  (B) $-1$  $\quad$  (C)  $0$  $\quad$  (D)  $\mathbf1$  $\quad$  (E)  $2$
My attempt:
$$\frac{d}{dt}F(x) = \frac{d}{dt}\int_0^x e^{\sin(t)} dt\Rightarrow F'(x) = \int_0^x \frac{d}{dt}e^{\sin(t)} dt$$
$$F'(x) = e^{\sin(t)}|_{0}^{x} = e^{x} - 1 \implies F'(0) = e^0-1 = 0$$
(wrong answer)
Where am I missing?
 A: When you have $g(x) = \int_0^xf(t)dt$ then $g'(x) = \frac{\delta g}{\delta x} = (\int_0^xf(t)dt)^{'}=  f(x)$ 
So in your function:
$$F'(x) = e^{sin(x)}$$ and for x = 0:
$$F'(0) = e^{sin(0)} = e^0 = 1$$
That's completely different from $g(x)=\int f(x)dx$ where $g'(x) = f(x)$. That's your mistake in your approach of the problem.
A: The variable in the function
$F(x) = \displaystyle \int_0^x e^{\sin t} \; dt \tag 1$
is $x$, not $t$; the variable "$t$" is "integrated out"; we could just as well have written
$F(x) = \displaystyle \int_0^x e^{\sin s} \; ds,  \tag 2$
or used any other variable (besides $x$) under the integral sign.
Therefore,
$F'(x) = \dfrac{d}{dx} \displaystyle \int_0^x e^{\sin t} \; dt = e^{\sin x} \tag 3$
by the Fundamental Theorem of Calculus; evaluating at $x = 0$ yields
$F'(0) = e^{\sin 0} = e^0 = 1; \tag 4$
the correct answer is thus (D).
A: We have $(\int_0^xf)'=f(x)$ by the Fundamental Theorem of Calculus, but you applied the rule for integration 
$$\int_a^bf=F(b) -F(a) $$
where $F'=f$. 
