## Calculus: Early Transcendentals 8th Edition

$$F'(x) = -\sqrt{1 + sec(x)}$$
$$F(x) = \int\limits_x^0{\sqrt{1+sec(t)}}dt$$ 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. $$F'(x) = - \sqrt{1 + sec(x)}$$