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I am trying to find the 2-D derivative of an elongated Gaussian density. The Gaussian has standard deviations $\sigma_x$ and $\sigma_y$. How can I get the scale-normalized 2-D Gaussian derivative in this case? Normally, I'd multiply by $\sigma^2$, but what to do when the standard deviation in different in each dimension? Thank you.

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  • $\begingroup$ What is the use of the 2-D derivative (I assume you mean $\frac{\partial^2 f_{X,Y}(x,y)}{\partial x\partial y}$ where $f_{X,Y}(x,y)$ is the bivariate Gaussian density) and why do you think that the result is a Gaussian derivative, whatever that means? Since you don't mention the correlation between $X$ and $Y$, are they independent? If so, $f_{X,Y}(x,y) = f_X(x)f_Y(y)$ which makes the double partial derivative just the product of the individual Gaussian "derivatives" which presumably you know how to find. $\endgroup$ – Dilip Sarwate Dec 8 '15 at 17:05
  • $\begingroup$ I am actually looking for scale-normalized 2D derivative of a non-uniform Gaussian function. Since it is non-uniform so it is defined by two standard deviations like sigmax and sigmay. Since I will generate a stake of these derivatives for my image, i need to find the max response. However, I have to first make them scale-normalized.. So how to do that. $\endgroup$ – Mohammad Asmat Ullah Khan Jan 4 '16 at 23:15
  • $\begingroup$ Could you write down the function and what you are after? $\endgroup$ – Royi May 4 '16 at 17:08
  • $\begingroup$ Votes and best answer validation are required $\endgroup$ – Laurent Duval Jul 28 at 12:04
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If I understand the question, you have three options:

  • You can take the partial derivative with respect to either variable.
  • You can take the total derivative with respect to another variable on which they both depend.
  • You can take the gradient, which is a vector whose components are the partial derivatives of the components.

The problem should determine the appropriate choice.

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If you differentiate in the $x$ direction, due to the separability, you simply replace the scale factor $\sigma$ that goes out of the exponential by $\sigma_x$ (potentially with its square), the same in the $y$-direction. If you differentiate two times, you will get some $\sigma_x^2\sigma_y^2$, $\sigma_x^4$ or $\sigma_y^4$ instead of $\sigma^4$.

Your formula should be consistent with the former ones when you set $\sigma_x = \sigma_y = \sigma$.

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