The problem is very simple, but I'm not sure of the result. I have the function $f(x,y)$ in the Euclidian plan and two constants $(a,b)$:
$ f(x,y) = \delta(a x + b y) $
Now I need the partial derivative of this Dirac delta's. Are these following results corrects (using Differentiation of dirac delta function: I think yes)?:
$ \partial_{x} f(x,y) \stackrel{?}{=} a \delta(a x + b y) \partial_{x} \\ \partial_{x} f(x,y) \stackrel{?}{=} b \delta(a x + b y) \partial_{y}$
Thank you for reading.
So lets come back to the basis ... a distribution $T$ is rigorously defined as a linear form on smooth test functions with compact support, i.e. to any test function $\varphi$, it associates a number $T(\varphi) =\langle T,\varphi\rangle$. For instance if $T=f$ is a "classical" function then its distribution is characterized by the fact that for any smooth function $\varphi$, $$\langle T,\varphi\rangle = ∫_{\mathbb R} f\,\varphi.$$ The main point is that it allows to generalize functions to more general distributions, such as the Dirac delta (which I denote by $\delta_0$) verifying $\langle\delta_0,\varphi\rangle = \varphi(0)$. One can then generalize a lot of formulas doing as if the braket was just the integral of a product of functions. For instance, the integration by parts formula allows to define the derivative of a distribution, and the change of variable in integrals allows to change variable in the distribution, so $T'$ and $T(ax)$ are defined by $$ \langle T',\varphi\rangle = -\langle T,\varphi'\rangle \\ \langle T(ax),\varphi\rangle = \tfrac{1}{|a|}\,\langle T,\varphi(\tfrac{x}{a})\rangle $$ In the case of a two variables distribution, one should use two variables test functions (so $\varphi = \varphi(x,y)$). In particular, for $a≠0$ one can define $T(ax+by)$ by $$\tag{1} \langle T(ax+by),\varphi\rangle = \tfrac{1}{|a|} \langle T(x),\varphi(\tfrac{x-by}{a},y)\rangle $$
In particular we deduce that $\langle\delta_0(ax+by),\varphi\rangle = \tfrac{1}{|a|} \int_{\mathbb R} \varphi(\tfrac{-by}{a},y) \,\mathrm d y$, and so $$\langle\delta_0(ax+by),\varphi\rangle = \int_{\mathbb R} \varphi(-by,ay) \,\mathrm d y$$
From $(1)$, we also obtain that since $\langle \partial_x T,\varphi\rangle = -\langle T,\partial_x\varphi\rangle$, $$\tag{2} \langle \partial_x(T(ax+by)),\varphi\rangle = -\tfrac{1}{|a|} \langle T(x),(\partial_x\varphi)(\tfrac{x-by}{a},y)\rangle \\= -\tfrac{1}{|a|} \langle T(x),a\,\partial_x(\varphi(\tfrac{x-by}{a},y))\rangle = \tfrac{1}{|a|} \langle a\,T'(x),\varphi(\tfrac{x-by}{a},y)\rangle $$ which proves that the formula $\partial_x(T(ax+by)) = a \,T'(ax+by)$ remains valid for distributions. Hence $$ \boxed{\partial_x(\delta_0(ax+by)) = a \,\delta_0'(ax+by) \\ \partial_y(\delta_0(ax+by)) = b \,\delta_0'(ax+by)} $$
If you want the action on test functions, then from $(2)$ we see that $$ \langle \partial_x(\delta_0(ax+by)),\varphi\rangle = - \int_{\mathbb R} (\partial_x\varphi)(-by,ay)\, \mathrm d y = - \tfrac{a}{b}\int_{\mathbb R} (\partial_y\varphi)(-by,ay))\, \mathrm d y $$