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Math Help - Derivative involving Q-function

  1. #1
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    Derivative involving Q-function

    Hi,

    Please, what is the derivative of

    d/dx{[1-Q(sqrt(x))]^N}

    derivative involving Q-function.pdf

    Thanks
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  2. #2
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    Re: Derivative involving Q-function

    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.
    Last edited by SlipEternal; March 19th 2014 at 11:48 AM.
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  3. #3
    Super Member nycmath's Avatar
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    Re: Derivative involving Q-function

    Interesting function.
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  4. #4
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    Re: Derivative involving Q-function

    Quote Originally Posted by SlipEternal View Post
    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.
    @ SlipEternal, thanks for your help.


    Using the relation between Q-function $Q(x)$ and the cdf $\Phi(x)$ of the normal standard Gaussian distribution, will it be appropriate to do $$\dfrac{dQ(\sqrt{x})}{dx}=\dfrac{d\left[1-\Phi(\sqrt{x})\right]}{dx}$$ $$\dfrac{dQ(\sqrt{x})}{dx}=-\dfrac{d\Phi(\sqrt{x})}{dx}$$. Also, using the relation between the cdf $\Phi(x)$ and pdf $\phi(x)$ of the normal standard Gaussian distribution, then $$-\dfrac{d\Phi(\sqrt{x})}{dx}=-\phi(\sqrt{x})=-\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{1}{2}{\sqrt{x}^2}}$$


    Thanks.
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