If $\displaystyle F(x) = \int_{1}^{\,x}{f(t)}\,dt$, where $\displaystyle f(t) = \int_{1}^{\,t^2}{\frac{\sqrt{9+u^4}}{u}}\,du$, find $F''(2).$
I used FTC to get If $\displaystyle F(x) =\int_{1}^{\,x}{\frac{\sqrt{9+x^8}}{x}}\,dx $
Then I tried to use FTC again to find $F'(x)$ but then I got lost cause it's just the same thing over again. So then I decided that i'll just plug in 2 to the function and got 8.14 but I know this is incorrect. Any ideas?
The FTC says that $F'(x)=f(x)=\int_{1}^{x^2}{\frac{\sqrt{9+u^4}}{u}}\,du$. Now use the FTC again along with the chain rule. To do that note that $f(x)=g(h(x))$ where $g(x):=\int_{1}^{x}{\frac{\sqrt{9+u^4}}{u}}\,du$ and $h(x):=x^2$. Hence $F''(x)=f'(x)=g'(h(x))h'(x)=\frac{\sqrt{9+(x^2)^4}}{x^2}\cdot2x$. Evaluating at $2$ gives $\sqrt{265}$.
We want to know:
$$\text{F}\space''\left(x\right)=\frac{\partial^2}{\partial x^2}\left\{\int_1^x\int_1^{t^2}\frac{\sqrt{9+\text{u}^4}}{\text{u}}\space\text{d}\text{u}\space\text{d}t\right\}=2\cdot \frac{\sqrt{9+x^8}}{x}\tag1$$
Because, when $\text{f}\left(\text{u}\right)$ is a continuous function:
$$\frac{\partial^2}{\partial x^2}\left\{\int_1^x\int_1^{t^2}\text{f}\left(\text{u}\right)\space\text{d}\text{u}\space\text{d}t\right\}=2\cdot x\cdot\text{f}\left(x^2\right)\tag2$$
So:
$$\text{F}\space''\left(2\right)=2\cdot \frac{\sqrt{9+2^8}}{2}=\sqrt{265}\tag3$$