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

$\frac{cosx}{\sqrt {1-sin^2x}}$
Let's apply the fundamental theorem of Calculus ($t=sinx$, and multiply by$(sin x)'=cosx)$. $\frac{dy}{dx}=(\frac{1}{\sqrt {1-sin^2x}})\cdot cosx=\frac{cosx}{\sqrt {1-sin^2x}}$