## University Calculus: Early Transcendentals (3rd Edition)

$-1$
Let's take the derivative of both sides, and let's apply the fundamental theorem of Calculus (substitute the limits, and multiply by their derivatives) to both integrals on the right! $\frac{dy}{dx}=(\frac{1}{1+0^2}\cdot 0-\frac{1}{1+tan^2x}\cdot sec^2x)=-\frac{sec^2x}{1+tan^2x}=-1$