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 $a$.

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 $H'(g(x))=f(g(x))$.
