For the following we assume that $r = |x-y|$ with $x,y\in\mathbb{R}^3$. We have the following derivative $ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x}\left(\sum_{j=1}^n (x_j - y_j)^2)\right)^{1/2} = \frac{1}{r}(x-y) $ ## Laplace Green's function We have $g(x, y) = \frac{1}{4\pi r}$. For the derivative we obtain $ \begin{align} \frac{\partial g}{\partial x}(x, y) &= \frac{1}{4\pi r^3}(y-x)\nonumber\\ \frac{\partial g}{\partial y}(x, y) &= \frac{1}{4\pi r^3}(x-y)\nonumber. \end{align} $ ## Helmholtz Green's function We have $g(x, y) = \frac{e^{ikr}}{4\pi r}$. For the derivative we obtain $ \frac{\partial}{\partial r}\left(\frac{e^{ikr}}{4\pi r}\right) = \frac{e^{ikr}}{r^2}\left(ikr -1\right). $ Hence, $ \begin{align} \frac{\partial g}{\partial x}(x, y) &= \frac{e^{ikr}}{4\pi r^3}(1-ikr)(y-x)\nonumber\\ \frac{\partial g}{\partial y}(x, y) &= \frac{e^{ikr}}{4\pi r^3}(1-ikr)(x-y)\nonumber. \end{align} $ ## Modified Helmholtz Green's function We have $g(x, y) = \frac{e^{-\omega r}}{4\pi r}$. For the derivative we obtain $ \frac{\partial}{\partial r}\left(\frac{e^{-\omega r}}{4\pi r}\right) = -\frac{e^{-\omega r}}{r^2}\left(\omega r +1\right). $ $ \begin{align} \frac{\partial g}{\partial x}(x, y) &= \frac{e^{-\omega r}}{4\pi r^3}(1+\omega r)(y-x)\nonumber\\ \frac{\partial g}{\partial y}(x, y) &= \frac{e^{-\omega r}}{4\pi r^3}(1+\omega r)(x-y)\nonumber. \end{align} $