# How do you find the derivative of g(x) = int 9*sqrt(1+t^8)dt from 7 to x^2?

I assume you want to find $g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\setminus {\int}_{7}^{{x}^{2}} 9 \setminus \sqrt{1 + {t}^{8}} \setminus \mathrm{dt}\right)$, where $g \left(x\right) = \setminus {\int}_{7}^{{x}^{2}} 9 \setminus \sqrt{1 + {t}^{8}} \setminus \mathrm{dt}$.
Let $F \left(x\right) = \setminus {\int}_{7}^{x} 9 \setminus \sqrt{1 + {t}^{8}} \setminus \mathrm{dt}$ and $h \left(x\right) = {x}^{2}$. Note that $g \left(x\right) = F \left(h \left(x\right)\right)$. By the Chain Rule, $g ' \left(x\right) = F ' \left(h \left(x\right)\right) \setminus \cdot h ' \left(x\right)$
By the Fundamental Theorem of Calculus, $F ' \left(x\right) = 9 \setminus \sqrt{1 + {x}^{8}}$. Since $h ' \left(x\right) = 2 x$, the final answer is:
$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\setminus {\int}_{7}^{{x}^{2}} 9 \setminus \sqrt{1 + {t}^{8}} \setminus \mathrm{dt}\right) = 9 \setminus \sqrt{1 + {x}^{16}} \setminus \cdot 2 x = 18 x \setminus \sqrt{1 + {x}^{16}}$.