Answer
$\frac{cosx}{\sqrt {1-sin^2x}} $
Work Step by Step
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}} $
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