TITLE: The derivative of $S(x) = \int_1^{x^2}\frac{\sin{t}}{t} dt$?
QUESTION [5 upvotes]: I need to find $S'(x)$ where:
$$S(x) = \int_1^{x^2}\frac{\sin{t}}{t} dt$$
My first thought was to solve the definite integral first getting a function that only depends on $x$ and then derivate that.
However, trying to find the primitive of $\frac{\sin t}{t}$ seems impossible. Wolfram Alpha gives me $\int\frac{\sin{t}}{t} dt=\mathrm{Si}(t)+C$ where $\mathrm{Si}(x)$ is the sine integral. So I don't get an expression that I can derivate but a definition.
On the other side this problem should be easier than that. Let me explain:
If $$F(x) = \int{x^2} dx$$ then $$F'(x) = x^2 +C$$
I should be able to apply that to $S(t)$ but how?

REPLY [8 votes]: Use the Fundamental Theorem of Calculus: $${d\over dx}\int_a^x f(t)\, dt=f(x)$$ and
the chain rule. 
So if $F(x)=\int_1^{x } {\sin t\over t}\,dt$, then 
$$S(x)=F(x^2)$$ and $${d\over dx}S (x)=F'(x^2)\cdot (x^2)'.$$
You have:
$${d\over dx}\underbrace{\int_1^{x^2} {\sin t\over t}\,dt}_{F(x^2)}= 
\underbrace{\sin(x^2)\over x^2}_{F'(x^2)}\cdot (x^2)'=2x{\sin(x^2)\over x^2}.$$