# Derivative of vector matrix multipication by element of vector

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 $$D^{-1}$$.