How is $\frac{1}{|\vec{r} - \vec{r}'|} = \frac{1}{r} - \vec{r} \cdot \triangledown \frac{1}{r} + \ldots$ a Taylor expansion? From pg. 112 of No-Nonsense Electrodynamics, the author uses multivariable Taylor expansion to assert:



In case it matters, the author is also assuming that $|\vec{r}| \gg | \vec{r}'|$ (stated elsewhere in the text).  Also, $\vec{r}, \vec{r}'$ are both 3-dimensional vectors.
Question: How does this identity follow from Taylor?  According to Wikipedia

A second-order Taylor series expansion of a scalar-valued function of more than one variable can be written compactly as
$$T(\mathbf{x}) = f(\mathbf{a}) + (\mathbf{x} - \mathbf{a})^\mathsf{T} D f(\mathbf{a})  + \cdots$$

where $D$ in this context denotes the gradient $\triangledown$ operator.  If we plug this into to our context we get
$$
f(\mathbf{a}) + (\mathbf{x} - \mathbf{a})^T Df(\mathbf{a}) = \frac{1}{r} + ( \vec{r} - \vec{r}') \cdot \triangledown \frac{1}{| \vec{r} - \vec{r}' |}
$$
which doesn't obviously resemble the formula the author derived, unless I'm missing something?
 A: HINT:
We can write the term of interest as
$$\begin{align}
\frac1{|\vec r-\vec r'|}&=\frac{1}{\sqrt{|\vec r|^2+|\vec r'|^2-2\vec r'\cdot\vec r}}\\\\
&=\frac1{|\vec r|}\left(1-2\frac{\vec r'\cdot\vec r}{|\vec r|^2}+\left(\frac{|\vec r'|}{|\vec r|}\right)^2\right)^{-1/2}
\end{align}$$

Now applying the binomial theorem (which is a special case of Taylor's Theorem) reveals
$$\begin{align}
\frac1{|\vec r-\vec r'|}&=\frac1{|\vec r|}\left(1+\frac{\vec r'\cdot\vec r}{|\vec r|^2}+O\left(\frac{|\vec r'|^2}{|\vec r|^2}\right)\right)\\\\
&=\frac1{|\vec r|}+\vec r'\cdot \frac{\vec r}{|\vec r|^3}+O\left(\frac{|\vec r'|^2}{|\vec r|^3}\right)
\end{align}$$
and we see that the Equation $(4.30)$ in the OP is incorrect by a factor of $-1$ on the second term of the expansion.  In fact, $(4.30)$ is inconsistent with the equation that precedes it since $\nabla \frac1r=-\frac{\vec r}{|\vec r|^3}$ and not $\nabla \frac1r=+\frac{\vec r}{|\vec r|^3}$

NOTE:
Application of Taylor's Theorem to the function $\frac1{|\vec r-\vec r'|}$ that led to the expression written in the OP is $o(|\vec r-\vec r'|)$ and does not lead, therefore, to a useful expansion when $|\vec r|>>|\vec r'|$.
Instead, we have used Taylor's Theorem (which is also application of the binomial theorem in this case) to expand the term $\displaystyle \left(1-2\frac{\vec r'\cdot\vec r}{|\vec r|^2}+\left(\frac{|\vec r'|}{|\vec r|}\right)^2\right)^{-1/2}$.
