What is the derivative of $x \times y$ with respect to $x$

Question

In physics class, we saw that for an unit vector $e$, and a random vector $p$, $$\frac{\partial}{\partial \vec x}\hat e \cdot(\vec x \times \vec p) = \hat e \times \vec p.$$ It seems to be taken for granted and I don't know how to reach the solution. Is there a general formula for this? And how can we derive it?

Answer 1

What you seem to mean is that if we fix two 3D vectors $\vec{v}$ and $\vec{p}$ and define the function of a 3D vector $\vec{x}$ by $f( \vec{x})= \vec{v}\cdot (\vec{x}\times \vec{p})$ then we have for the gradient of $f$ the formula $\vec{\nabla} f(\vec{x})=\vec{p}\times \vec{v}$ (note the ordering!). This is indeed true, the high-brow reason being that since $f$ is linear, the (row vector) derivative of $f$ is $f$, meaning

$$\vec{\nabla} f\cdot \vec{\delta}=Df(\vec{\delta})=f(\vec{\delta})= \vec{v}\cdot (\vec{\delta}\times \vec{p})=\vec{\delta}\cdot (\vec{p}\times\vec{v})$$

for all vectors $\vec{\delta}$ and so indeed $\vec{\nabla} f(\vec{x})=\vec{p}\times \vec{v}$, a constant vector field.

You can also deduce this from the (corrected) equality for $\hat{e}$ that you have by writing $\vec{v}=|\vec{v}|\hat{v}$ and observing that compared with formulas for $\hat{v}$, both the left and the right hand side get multiplied by the same constant $|\vec{v}|$ and so remain equal.