I am working on a homework problem that requires me to differentiate $$\frac{\partial}{\partial t} \psi(\vec r-\vec a(t)),$$
where $\vec r$ is the position vector and $\vec a$ is a vector that varies with time $t$.
I tried using the chain rule and arrived at $$\frac{\partial}{\partial t} \psi(\vec r-\vec a(t)) = \frac{\partial \psi}{\partial (\vec r- \vec a(t))} \frac{\partial (\vec r- \vec a(t))}{\partial t}.$$
How do I continue? The answer given was $$\frac{\partial}{\partial t} \psi(\vec r-\vec a(t))=-\frac{\partial \vec a}{\partial t} \cdot\nabla \psi (\vec r-\vec a(t)).$$
I can't see how my attempt can match this answer. Perhaps I was wrong regarding how to use the chain rule for vector calculus?
$r$ is presumably a constant, so your answer and the book's answer agree (up to being sure you understand what the derivative of a function $f:\mathbb{R}^n\to \mathbb{R}$.
Recall that the chain rule for scalar valued functions is $$ \frac{d}{d{t}}f(x(t))=\nabla f(x(t))\cdot x'(t) $$ So, $$ \frac{d}{dt}\psi(\vec{r}-\vec{a(t)})=\nabla\psi(\vec{r}-\vec{a(t)})\cdot -\frac{d\vec{a}(t)}{dt} $$