Answer
$g'(x) = 4x^3~cos(x^8)$
Work Step by Step
The function $f(t) = cos(t^2)$ is continuous for all $t$
Let $f(x) = x^4$
Then $f'(x) = 4x^3$
$f'(x)$ is continuous for all values of $x$
According to the Substitution Rule:
$g(x) = \int_{0}^{x^4}cos(t^2)~dt = \int_{0}^{x}cos[(u^4)^2]\cdot 4u^3~du$
The function $~~cos[(u^4)^2]\cdot 4u^3~~$ is continuous for all $u$
According to the Fundamental Theorem of Calculus (Part 1):
$g'(x) = cos[(x^4)^2]\cdot 4x^3 = 4x^3~cos(x^8)$
