FTC Double Derivative of Two Integrals I need help working through this FTC problem I've encountered. It looks like such:
$$\frac{d^{2}}{d x^{2}}\left(\int_1^{\sin (x)}\left(\int_1^t\sqrt{1+u^2} d u\right)dt\right)$$
I know that in such cases, you can just substitute the bounds of the leftmost integral for the bounds of the right integral, then just multiply the derivative of the upper bound times $f(\text{upper bound})$. This can be seen here.
$$\frac{d^{2}}{d x^{2}}\left(\int_1^{x}\left(\int_1^{sin(t)}\sqrt{1+u^4} d u\right)dt\right)$$
However, I'm not sure how to handle it when the $\sin(x)$ is in the outer integral. My first thought is to do $\arctan()$ of something to get it in the form of $x$ for substitution, but I really don't know. Any help would be appreciated.
 A: By the fundamental theorem of calculus, if $g(u)$ is a "nice" function and you define the function $F(x)$ by
$$
F(x) = \int_0^x g(u) du,
$$
then
$$
\frac{d}{dx} F(x) = g(x).
$$
In your problem, you are differentiating a function that looks like
$$
(F \circ \sin)(x) = F(\sin x) = \int_0^{\sin x} g(u) du.
$$
To differentiate functions like this, you would use the chain rule. And in particular,
$$
\frac{d}{dx} F(\sin x) = F'(\sin x) \cos x.
$$
Fortunately, as we noted above, you know the derivative of $F$. In your case, you happen to have an annoying function $g(u)$ and you will be wanting to compute a second derivative --- you'll need to use the fundamental theorem of calculus (twice total), the chain rule (twice total), and the product rule in the second derivative.
A: Consider the functions $F(x) = \int_1^{\sin x} t \, dt$ and $G(x) = \int_1^x \sqrt{1 + u^2} \, du.$ Observe that we have $$F(G(x)) = \int_1^{\sin x} \int_1^t \sqrt{1 + u^2} \, du \, dt,$$ hence by the Chain Rule, it follows that $$\frac{d^2}{d^2 x} \int_1^{\sin x} \int_1^t \sqrt{1 + u^2} \, du \, dt = \frac{d^2}{d^2 x} F(G(x)) = \frac d {dx} \biggr[ F'(G(x)) G'(x) \biggl] \phantom{blah blah blah bl}$$ $$\phantom{\frac{d^2}{d^2 x} \int_1^{\sin x} \int_1^t \sqrt{1 + u^2} \, du \, dt = \frac{d^2}{d^2 x} F(G(x))}= F'(G(x)) G''(x) + F''(G(x)) [G'(x)]^2.$$ Further, by the Fundamental Theorem of Calculus, we have that $F'(x) = \sin x \cos x$ and $G'(x) = \sqrt{1 + x^2}.$ Can you finish the solution from here? (Hint: you will need to implement the Product Rule to compute $F''(x)$ and the Chain Rule to compute $G''(x).$)
