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Math Help - Derivative of Q function

  1. #1
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    Derivative of Q function

    Q(x) = \frac{1}{\sqrt{2\cdot \pi}} \int_x^\infty e^{- \frac{t^2}{2}}dt

    what is Q'(x) ???

    thanks
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  2. #2
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    Quote Originally Posted by graticcio View Post
    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:
    \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 Q(x) = 1 - \Phi(x) where \Phi(x) is the cdf of the standard gaussian distribution. Now by differentiating this we get:

    Q'(x) = - \Phi '(x)

    But the differential of the cdf is the pdf ! Hence:

    Q'(x) = -\frac{1}{\sqrt{2\pi}}e^{- \frac{x^2}{2}}
    Thanks from Phummy
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