Derivative of Dirac delta distribution

Question

How is the derivative of $\delta(x-y)$ with respect to $x$ related to the derivative with respect to $y$? I suspect they differ by a minus sign but I'm not sure.

Both $x$ and $y$ are real variables.

Answer 1

Recall that $$\delta(x\!-\!y), \quad\frac{\partial}{\partial x}\delta(x\!-\!y)\quad\text{and}\quad \frac{\partial}{\partial y}\delta(x\!-\!y)\tag{A}$$ are informal kernel notations for the distributions $$u,u_x,u_y~\in~ D^{\prime}(\mathbb{R}^2)\tag{B}$$ defined as

$$u[f]~:=\int_{\mathbb{R}}\!\mathrm{d}z~f(z,z),\quad u_x[f]~:=-\int_{\mathbb{R}}\!\mathrm{d}z~\frac{\partial f(z,z)}{\partial x},\quad u_y[f]~:=-\int_{\mathbb{R}}\!\mathrm{d}z~\frac{\partial f(z,z)}{\partial y},\tag{C}$$

respectively, for testfunctions $f\in D(\mathbb{R}^2)$. Here $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ mean the partial derivative of $f$ wrt. 1st and 2nd entry, respectively.

In this notation OP wants to prove that $$u_x+u_y~=~0.\tag{D}$$ Proof of eq. (D):

$$u_x[f]+u_y[f]~=~-\int_{\mathbb{R}}\!\mathrm{d}z~\frac{df(z,z)}{d z} ~=~f(-\infty,-\infty)-f(\infty,\infty)~=~0, \tag{E}$$ because the testfunction $f$ vanishes at infinity.