Answer:
[tex]\displaystyle g'(s) = (s-s^4)^6[/tex]
Step-by-step explanation:
The Fundamental Theorem of Calculus states that:
[tex]\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)[/tex]
Where a is some constant.
We can let:
[tex]\displaystyle g(t) = (t-t^4)^6[/tex]
By substitution:
[tex]\displaystyle g(s) = \int_6^s g(t)\, dt[/tex]
Taking the derivative of both sides results in:
[tex]\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right][/tex]
Hence, by the Fundamental Theorem:
[tex]\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\ & = (s-s^4)^6\end{aligned}[/tex]
