Confusion on delta function

Question

How should one deal with $\delta(x-y)\partial_x[f(2x+y)\delta(x+y)]$? In other words, what is \begin{equation} \int\phi(x,y)\delta(x-y)\partial_x[f(2x+y)\delta(x+y)] dxdy ? \end{equation}

My attempt was \begin{align} & \int\phi(x,y)\delta(x-y)\partial_x[f(2x+y)\delta(x+y)] dxdy \\ =&\int\phi(x,y)\delta(x-y)[\partial_xf(x)\delta(x+y)+f(x)\partial_x\delta(x+y)] dxdy\\ =&\phi(0,0)f'(0) + \int\phi(x,y)\delta(x-y)f(x)\partial_x\delta(x+y) dxdy \end{align}

I am not 100% sure if the steps above are correct. Also, I am still confused by the term $\int\phi(x,y)\delta(x-y)f(x)\partial_x\delta(x+y) dxdy$.

Please help, thank you very much!

Answer 1

Hint: notice that $$f(2x+y)\delta(x+y)=f(-y)\delta(x+y)$$can you finish now?

Answer 2

The usual product rule still holds. We have $$g(x, y) = \frac \partial {\partial x}(f(2 x + y) \delta(x + y)) = 2 f'(2 x + y) \delta(x + y) + f(2 x + y)\delta'(x + y), \\ \int_{-\infty}^\infty \delta(x - y) g(x, y) \phi(x, y) dy = g(x, x) \phi(x, x), \\ \int_{-\infty}^\infty g(x, x) \phi(x, x) dx = f'(0) \phi(0, 0) - \frac 1 4 \frac d {dx} (f(3 x) \phi(x, x)) \bigg\rvert_{x = 0} = \\ \frac {f'(0) \phi(0, 0) - f(0) \phi^{(1, 0)}(0, 0) - f(0) \phi^{(0, 1)}(0, 0)} 4, \\ \delta(x - y) g(x, y) = \frac {f'(0) \delta(x, y) + f(0) \delta^{(1, 0)}(x, y) + f(0) \delta^{(0, 1)}(x, y)} 4.$$