# Differential operator - what is $\frac{\partial f}{\partial x}x$?

given $f$ - a smooth function, $f\colon\mathbb{R}^2\to \mathbb{R}$.

I have a differential operator that takes $f$ to $\frac{\partial f}{\partial x}x$, but I am unsure what this is.

If, for example, the operator tooked $f$ to $\frac{\partial f}{\partial x}$ then I understand that this operator derives by $x$, $\frac{\partial f}{\partial x}x$ derives by $x$ and then what ?

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$f \mapsto \frac{\partial f}{\partial x}x$ derives by $x$ and multiplies by $x$ afterwards. –  martini Apr 23 '12 at 19:09
For example, if $f(x,y) = x^2+2xy$, then $\frac{\partial f}{\partial x} = 2x+2y$, and $\frac{\partial f}{\partial x}x = (2x+2y)x = 2x^2+2xy$. –  Arturo Magidin Apr 23 '12 at 19:10
Let $D$ be a (first order) differential operator (i.e. a derivation) and let $h$ be a function. (In your example $h$ is the function $x$ and $D$ is $\frac{\partial}{\partial x}$). I claim that there is a differential operator $hD$ given by $$f \mapsto hDf.$$ (This means multiply $h$ by $Df$ pointwise.) For this claim to hold, we just need it to be true that the Leibniz (product) rule holds, i.e. $$hD(fg) = f(hD)g + g(hD)f.$$ Since $D$ itself satisfies the Leibniz rule, this is true. (We also need the operator $hD$ to take sums to sums, which it does.)
Put less formally, to compute $x \frac{\partial}{\partial x}$ on a function $f$, take the partial with respect to $x$, then multiply what you get by $x$.