# How to derive $\frac{d}{d\lambda}=-\mathbb{y}\cdot\nabla$ for $\frac{1}{\lvert\mathbb{x}-\lambda\mathbb{y}\rvert}$?

When talking about the multipole expansion of an electromagnetic potential, my professor noted that for the function

$$\tag{1}\frac{1}{\lvert\mathbb{x}-\lambda\mathbb{y}\rvert},$$

the two operators

$$\tag{2}\frac{d}{d\lambda}$$

and

$$\tag{3}-\mathbb{y}\cdot\nabla_\mathbb{x}$$

are the same, i.e. $\frac{d^n}{d\lambda^n}=(-\mathbb{y}\cdot\nabla_\mathbb{x})^n$ for all $n$.

I checked that explicitly for $n=1$ and $n=2$.

Is there a way to derive it for an arbitrary $n\in\mathbb N$?

EDIT: I have to clarify: I did not prove that $\frac{d}{d\lambda}$ and $-\mathbb{y}\cdot\nabla$ are the same operators. I just checked that their action on the function $(1)$ yields the same result if they are applied one or two times. However, this does not prove they are the same operators.

Use the fact that $$\frac{\partial^2}{\partial\lambda\partial x_i}=\frac{\partial^2}{\partial x_i \partial \lambda}$$ we have $$\left[\frac{d}{d\lambda},-\textbf{y}\cdot\nabla_{\textbf{x}}\right]=0$$
Given you said that you've proved $$\frac{d}{d\lambda}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}=-\textbf{y}\cdot\nabla_{\textbf{x}}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ then use induction.
Suppose $$\frac{d^n}{d\lambda^n}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}=(-\textbf{y}\cdot\nabla_{\textbf{x}})^n\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ $$\frac{d^{n+1}}{d\lambda^{n+1}}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}= \frac{d}{d\lambda}\frac{d^n}{d\lambda^n}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ $$=\frac{d}{d\lambda}(-\textbf{y}\cdot\nabla_{\textbf{x}})^n\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ $$=(-\textbf{y}\cdot\nabla_{\textbf{x}})^n\frac{d}{d\lambda}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ $$=(-\textbf{y}\cdot\nabla_{\textbf{x}})^n(-\textbf{y}\cdot\nabla_{\textbf{x}})\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$ $$=(-\textbf{y}\cdot\nabla_{\textbf{x}})^{n+1}\frac{1}{|\textbf{x}-\lambda\textbf{y}|}$$