## Calculus 8th Edition

$g'(x) = cos(x^2)$
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $g(x) = \int ^{x}_{1}cos(t^2)dt$ Since $g(x) = \int^{b}_{a}f(t)dt$ $g'(x) = f(x)$ Thus, $g'(x) = cos(x^2)$