Suppose that $F(x)={\int }_1^{x}f(t)dt$, where
$$f(t)={\int }_1^{t^4}\frac{\sqrt{6+u^6}}{u}du$$
Find $F^{\prime \prime }(2)$.
$$F^{\prime \prime }(2)=$$

Step 1 ：Suppose that $F(x)={\int }_1^{x}f(t)dt$, where $f(t)={\int }_1^{t^4}\frac{\sqrt{6+u^6}}{u}du$

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 $f^{\prime }(x)$

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 $f^{\prime }(t)=4t^3\frac{\sqrt{6+(t^4{)}^6}}{t^4}$

Step 7 ：We can then find the second derivative of $F(x)$, which is $f^{\prime }(x)$, by substituting $x=2$ into the expression for $f^{\prime }(t)$

Step 8 ：$f^{\prime }(2)=4\ast 2^3\frac{\sqrt{6+(2^4{)}^6}}{2^4}=2\sqrt{16777222}$

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

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