Thread: 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.

Interesting function.

Originally Posted by SlipEternal

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.