Use the Chain Rule and the Fundamental Theorem of Calculus:

$\dfrac{d}{dx}\left[1-Q(\sqrt{x})\right]^N = -N\left[1-Q(\sqrt{x})\right]^{N-1}\dfrac{dQ(\sqrt{x})}{dx}$

Edit: For the derivative of the $Q$ function, you have:

$\displaystyle Q(\sqrt{x}) = \dfrac{1}{2\pi}\int_{\sqrt{x}}^\infty e^{-u^2/2}du$

So, when you take the derivative, you are just using the fundamental theorem of calculus, and again, the chain rule.