# How to find derivative of 2-D elliptical Gaussian function with different standard deviations along $x$ and $y$ directions?

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.

• 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. – Dilip Sarwate Dec 8 '15 at 17:05
• 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. – Mohammad Asmat Ullah Khan Jan 4 '16 at 23:15
• Could you write down the function and what you are after? – Royi May 4 '16 at 17:08
• Votes and best answer validation are required – Laurent Duval Jul 28 '19 at 12:04

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 $\sigma_x = \sigma_y = \sigma$.