Find $ F''(1)$ if $F(x)=\int_1^x f(t)\,\mathrm dt$ and $f(t)=\int_1^{t^2}\frac{\sqrt{5+u^4}}{u}\,\mathrm du$ Find $ F''(1)$ if: $$F(x)=\int_1^x f(t) \,\mathrm dt$$ and: $$f(t)=\int_1^{t^2}\dfrac{\sqrt{5+u^4}}{u}\,
\mathrm du$$ I tried using substitution and integration by parts but none of the methods work.
 A: Using the Fundamental Theorem of Calculus which states if:
$$F(x) = \int_a^xf(t)\,\mathrm dt$$
Then:
$$F'(x)=f(x)$$
Thus:
$$F'(x) = \dfrac{\mathrm d}{\mathrm dx}\int_1^xf(t)\,\mathrm dt=f(x)$$
$$F''(x) = f'(x) = \dfrac{\mathrm d}{\mathrm dx}\int_1^{x^2}\dfrac{\sqrt{5+u^4}}{u}\,\mathrm du=\dfrac{2x\sqrt{5+x^8}}{x^2} \Rightarrow \dfrac{2(1)\sqrt{5+1^8}}{1^2}=2\sqrt{6}$$
Note that you must apply the chain rule. This is the reason $2x$ is multipled, as it is the derivative of $x^2$. You can see why you need to apply the chain rule by applying the Second Part of the Fundamental Theorem of Calculus. Given $F'(x) = f(x)$ then:
$$\int_{h(x)}^{g(x)} f(t)\,\mathrm dt = F(g(x)) - F(h(x))$$
$$\dfrac{\mathrm d}{\mathrm dx} \int_{h(x)}^{g(x)} f(t)\,\mathrm dt = \dfrac{\mathrm d}{\mathrm dx} (F(g(x)) - F(h(x)))$$
$$\dfrac{\mathrm d}{\mathrm dx} (F(g(x)) - F(h(x))) = F'(g(x))g'(x) - F'(h(x))h'(x)$$
$$F'(g(x))g'(x) = f(x^2)\cdot 2x = \dfrac{\sqrt{5+x^8}}{x^2} \cdot 2x$$
In this case, $h(x)$ is constant so the latter term becomes $0$, and you're left with the chain rule's consequences.
