Question
If $F(x)={\int }_2^{4x^5}\frac{2}{\ln \left(t^2\right)}dt$, what is $F^{\prime }(x)$ ? (Do not include " $F^{\prime }(x)=$ " in your answer.)
Provide your answer below:

Step 1 ：The problem is asking for the derivative of the function $F(x)={\int }_2^{4x^5}\frac{2}{\ln (t^2)}dt$.

Step 2 ：This is a problem of the Fundamental Theorem of Calculus part 1, which states that if a function $f$ is continuous over the interval $[a,b]$ and $F$ is an antiderivative of $f$ on $[a,b]$, then ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

Step 3 ：To find $F^{\prime }(x)$, we need to differentiate the integral with respect to $x$.

Step 4 ：The derivative of the integral of a function is just the function itself, evaluated at the upper limit of the integral, times the derivative of the upper limit with respect to $x$.

Step 5 ：So, $F^{\prime }(x)=\frac{2}{\ln ((4x^5{)}^2)}\ast \frac{d}{dx}(4x^5)$.

Step 6 ：Let's calculate this.

Step 7 ：The derivative of the function $F(x)$ is $\boxed{{\displaystyle \frac{40x^4}{\ln (16x^{10})}}}$.