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 $a$.

HINT:
Let $y(x)=\arcsin(x)$ and use the chain rule $$\frac{dS(y(x))}{dx}=\frac{dS(y(x))}{dy(x)}\frac{dy(x)}{dx}$$
