Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 47

$-\frac{sinx}{2\sqrt x}$

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)! $\frac{dy}{dx}=sin(0^2)\cdot 0-sin(\sqrt x)^2\cdot(\frac{1}{2}x^{-\frac{1}{2}})=-\frac{sinx}{2\sqrt x}$