The question derives from page 240 of the Essential Calculus book.
$$\int_5^x (t-t^2)^8\,dt.$$
Finding the derivative is done by using fundamental theorem of calculus. Yes if we can just use that then it should be $(x-x^2)^8$. However, I still don't get this.
According to the wikipedia, it explains the fundamental theorem of calculus as:
Let $f$ be a continuous real-valued function defined on a closed interval $[a, b].$ Let $F$ be the function defined, for all $x$ in $[a, b],$ by
$$F(x) = \int_a^x f(t)\,dt.$$
Then, $F$ is uniformly continuous on $[a, b],$ differentiable on the open interval $(a, b),$ and
$$F'(x) = f(x) \text{ for all } x \in (a,b).$$
In the question above, the $a$ value is specified but we don't have $b$ value. In other words, the interval has not been defined. How can we still apply it? Furthermore, x is not even defined it can be any real number hence $(-\infty,\infty)$ since $a$ is 5 in this case, it can be $4,$ $8$ or $1000000.$ Given the absence of $b$ how can we apply the fundamental theorem of calculus? what if $x$ is less than $5$? say $-1$ or $-2$?