## Calculus 8th Edition

$F'(x) = -\sqrt {1 + sec(x)}$
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $F(x) = \int ^{0}_{x}\sqrt {1+sec(t)}dt$ Swap upper and lower bounds: (this makes the expression negative) $F(x) = -\int ^{x}_{0}\sqrt {1+sec(t)}dt$ Since $F(x) = \int^{b}_{a}f(t)dt$ $F'(x) = f(x)$ Thus, $F'(x) = -\sqrt {1 + sec(x)}$