Density of Maximum brownian motion

Question

Why is it that for $M(t):=\max\limits_{0\leq s\leq t}B(s)$ $$P(M(t)\geq a) = 2P(B(t)\geq a)=\frac{2}{\sqrt{2\pi}}\int\limits_{a/\sqrt{\sigma^2t}}^{\infty}e^{-y^2/2}dy, \ a \geq 0$$ Doesn't result in: $$f_{M(t)}(a)=\frac{-2}{\sqrt{2\pi\sigma^2t}}e^{-a^2/2\sigma^2t}$$ when differentiating with respect to $a$?