Linking to a question that I previously asked (Differentiate expressions involving symmetric matrix $\mathbf{D}=\mathrm{diag}(\tau)\Omega\mathrm{diag}(\tau)$ with respect to element of $\tau$), I now find that I need to find the value of the following:
$$ \frac{d(\mathbf{b}^{T}\mathbf{D}^{-1}\mathbf{b})}{db_{g}} $$
In the above, $\mathbf{D}^{-1}$ is the inverse of a symmetric $q$ by $q$ covariance matrix, and $\mathbf{b}$ is a vector of length $q$. I want to differentiate the expression with respect to the $g$th element of $\mathbf{b}$.
I can see (e.g. using equation 85 in matrix cookbook https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf how to find the value of the above if differentiating by $\mathbf{b}$, but cannot see a clear step-by-step explanation for a single element $b_g$. Any guidance / help / pointers would be greatly appreciated.
Background: This stems from manipulating a multivariate normal distribution for parameter $\mathbf{b}$ with mean vector equal to vector of zeros and covariance matrix equal to $\mathbf{D}$.
Take $f(b) = b^TD^{-1}b$. From the equation you mention, we have $$ \frac{\partial f}{\partial b} = 2D^{-1}b. $$ Note that this assumes that $D$ is symmetric (which applies in your case because $D$ is a covariance matrix) and that $D$ is independent of $b$.
Since we're taking a derivative with respect to a column-vector, extracting the individual partial derivative is simple. $\frac{\partial f}{\partial b_g}$ is just the $g$th entry of $\frac{\partial f}{\partial b}$. In particular: let $e_g$ denote the $g$th column of the size $q$ identity matrix. We have $$ \frac{\partial f}{\partial b} = e_g^T(2D^{-1}b) = 2(e_g^TD^{-1})b. $$ Note that $e_g^T D^{-1}$ is the $g$th row of $D^{-1}$.