$\displaystyle Q(x) = \frac{1}{\sqrt{2\cdot \pi}} \int_x^\infty e^{- \frac{t^2}{2}}dt$
what is Q'(x) ???
thanks
General note:
Use fundamental theorem of calculus. The general statement is:
$\displaystyle \left(\int_{b(x)}^{a(x)} f(x) \, dx \right)' = f(a(x)) a'(x) - f(b(x)) b'(x)$
Here a(x) and b(x) represent functions of x.
However for Q function, you need not do all that.
Recall that $\displaystyle Q(x) = 1 - \Phi(x)$ where $\displaystyle \Phi(x)$ is the cdf of the standard gaussian distribution. Now by differentiating this we get:
$\displaystyle Q'(x) = - \Phi '(x)$
But the differential of the cdf is the pdf ! Hence:
$\displaystyle Q'(x) = -\frac{1}{\sqrt{2\pi}}e^{- \frac{x^2}{2}} $