Answer
$$F'(x) = -\sqrt{1 + sec(x)}$$
Work Step by Step
$$F(x) = \int\limits_x^0{\sqrt{1+sec(t)}}dt$$
Switch the bounds of the integral (This will make the equation negative).
$$F(x) = - \int\limits_0^x{\sqrt{1+sec(t)}}dt$$
Plug the upper bound into t and multiply by the derivative of the upper bound.
$$F'(x) = - \sqrt{1 + sec(x)} \times (x)'$$
Simplify.
$$F'(x) = - \sqrt{1 + sec(x)}$$
