Suppose that $F(x)=\int_{1}^{x} f(t) d t$, where \[ f(t)=\int_{1}^{t^{4}} \frac{\sqrt{6+u^{6}}}{u} d u \] Find $F^{\prime \prime}(2)$. \[ F^{\prime \prime}(2)= \]

Step 1 ：Suppose that \(F(x)=\int_{1}^{x} f(t) d t\), where \(f(t)=\int_{1}^{t^{4}} \frac{\sqrt{6+u^{6}}}{u} d u\)

Step 2 ：We need to find \(F^{\prime \prime}(2)\), which is the second derivative of the function \(F(x)\) at \(x=2\)

Step 3 ：The first derivative of \(F(x)\) is \(f(x)\), and the derivative of \(f(x)\) is \(f'(x)\)

Step 4 ：The function \(f(t)\) is an integral of a function of \(u\), with the upper limit of the integral being \(t^4\)

Step 5 ：By the Fundamental Theorem of Calculus, the derivative of this integral with respect to \(t\) is just the integrand evaluated at \(t^4\), multiplied by the derivative of \(t^4\) with respect to \(t\)

Step 6 ：This gives us \(f'(t) = 4t^3 \frac{\sqrt{6+(t^4)^6}}{t^4}\)

Step 7 ：We can then find the second derivative of \(F(x)\), which is \(f'(x)\), by substituting \(x=2\) into the expression for \(f'(t)\)

Step 8 ：\(f'(2) = 4*2^3 \frac{\sqrt{6+(2^4)^6}}{2^4} = 2\sqrt{16777222}\)

Step 9 ：The second derivative of the function \(F(x)\) at \(x=2\) is \(2\sqrt{16777222}\)

Step 10 ：So, \(F^{\prime \prime}(2) = \boxed{2\sqrt{16777222}}\)