MAP estimation (for stationary iid gaussian environment)
$\newcommand{\bd}{\mathbf{d}}\newcommand{\bw}{\mathbf{w}}\newcommand{\bX}{\mathbf{X}}$Your observation is correct--it appears to be several typos. To see why let's re-write the likelihood equation to something a bit more suitable for differentiating using matrix/vector notation. $$\text{NLL}(w) = \frac{1}{2}||\bd - \bX\bw||^2_2 + \frac{\lambda}{2}||\bw||^2_2,$$

where $\bX$ is our matrix of observations, $\bd$ is our vector of labels and $\bw$ is our vector parameter of interest. Taking the derivative yields,
$$\frac{\partial \text{NLL}}{\partial \bw} = -\bX^t(\bd - \bX\bw) + \lambda\bw.$$

Setting this equal to 0 and solving gives,
$$ \begin{align*}\lambda \bw &amp;= \bX^t(\bd - \bX\bw) \\
&amp;= \bX^t\bd - \bX^t\bX\bw \iff\\
\lambda\bw + \bX^t\bX\bw &amp;= \bX^t\bd \iff \\
(\lambda\mathbf{I} + \bX^t\bX)\bw &amp;= \bX^t\bd \iff \\
\hat{\bw}_{\text{MAP}} &amp;= (\lambda\mathbf{I} + \bX^t\bX)^{-1}(\bX^t\bd).
\end{align*}$$

Notice that $$\bX^t\bX = \sum_i \mathbf{x_i} \mathbf{x_i}^t = R_{xx}$$ and $$\bX^t\bd = \sum_i \mathbf{x_i} d_i = r_{dx}.$$