Compute the derivative of $S(x) = \int_1^{\arcsin(x)}\frac{\sin(t)}{t}dt$ My book states the following:

FUNDAMENTAL THEOREM OF CALCULUS. Assume the function $f$ is continous in $x\in[a,b].$ Put $$S(x)=\int_a^xf(t)dt, \quad a\leq x\leq b.$$
Then, it follows that the function $S$ is differentiable in $x\in(a,b)$ with the derivative
$$S'(x)=f(x), \quad a<x<b.$$

Using this on my integral, I get
$$S'(x) = \left(\int_1^{\arcsin(x)}\frac{\sin(t)}{t}dt\right)'=\frac{\sin(\arcsin(x))}{\arcsin(x)} =\frac{x}{\arcsin(x)}.$$
This is wrong. I don't see in the theorem what they're doing with $a$.
 A: HINT:
Let $y(x)=\arcsin(x)$ and use the chain rule $$\frac{dS(y(x))}{dx}=\frac{dS(y(x))}{dy(x)}\frac{dy(x)}{dx}$$
A: By the fundamental theorem of calculus you have that $\int^{g(x)}_c f(t)\,\mathrm dt=F(g(x))-F(c)$, but $[F(g(x))-F(c)]'=f(g(x))g'(x)\neq f(g(x))$.

As @Xander pointed out its possible that the OP doesnt knows yet the second part of the FTC. So, using just the stated FTC in the question we have that
$$H(g(x))=\int_c^{g(x)}f(t)\,\mathrm dt\implies [H(g(x))]'=H'(g(x))g'(x)$$
from the chain rule. And after notice that by definition $H(x)=\int_c^x f(t)\,\mathrm dt$, thus $H'(x)=f(x)$, so $H'(g(x))=f(g(x))$.
A: If
$$ F(x) = \int_{1}^{x} \frac{\sin(t)}{t} \,\mathrm{d}t, $$
then you are being asked to find $\frac{\mathrm{d}}{\mathrm{d}x} F(\arcsin(x))$.  First, we note that the fundamental theorem of calculus implies that
$$ F'(x) = \frac{\sin(x)}{x}. $$
We'll use this in the computation below.  Next, by the chain rule, we obtain
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} F(\arcsin(x))
&= F'(\arcsin(x))\cdot  \frac{\mathrm{d}}{\mathrm{d}x} \arcsin(x) \\
&= \frac{\sin(\arcsin(x))}{\arcsin(x)} \cdot  \frac{\mathrm{d}}{\mathrm{d}x} \arcsin(x) && \text{(by FTC)} \\
&= \frac{\sin(\arcsin(x))}{\arcsin(x)} \frac{1}{\sqrt{1-x^2}} && \text{(derivative of arcsin}) \\
&= \frac{x}{\arcsin(x)} \frac{1}{\sqrt{1-x^2}} && \text{($\sin(\arcsin(t)) = 
t\ \forall t$)} \\
&= \frac{x}{\arcsin(x)\sqrt{1-x^2}}.
\end{align}

EDITED to address a different question:
The fundamental theorem of calculus states that if $a$ is any real number, $f$ is "sufficiently nice" (essentially, $f$ must be integrable), and $F$ is the function defined on $\mathbb{R}$ by the integral
$$ F(x) = \int_{a}^{x} f(x) \,\mathrm{d}x, $$
then $F'(x) = f(x)$.  There is no actual dependence on $a$ in the theorem.  To see why this must be true, suppose that $a' \in \mathbb{R}$ is another real number, i.e. suppose that $a'\ne a$.  Then it follows from the linearity of the integral that
$$ \int_{a'}^{x} f(x)\,\mathrm{d}x
= \underbrace{\int_{a'}^{a} f(x)\,\mathrm{d}x}_{=:C} + \int_{a}^{x} f(x)\,\mathrm{d}x
= \int_{a}^{x} f(x)\,\mathrm{d}x + C,
$$
where $C$ is a definite integral, and so equal to some constant (presuming, of course, that all of these integrals exist).  Differentiating, we obtain
$$ \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a'}^{x} f(x)\,\mathrm{d}x \right]
= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a}^{x} f(x)\,\mathrm{d}x + C \right]
= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a}^{x} f(x)\,\mathrm{d}x \right],
$$
since the differentiation is a linear operator, and the derivative of a constant is zero.  In particular,
$$
\frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a'}^{x} f(x)\,\mathrm{d}x \right]
= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a}^{x} f(x)\,\mathrm{d}x \right]
= F'(x),
$$
independent of the choice of $a$ (and/or $a'$).
