# What is the derivative of the inverse square root of the determinant?

I am looking for the expression for the derivative of the inverse square root of the determinant, $$\frac{\partial}{\partial \Sigma} \det(\Sigma)^{-1/2},$$ where $\Sigma$ a symmetric matrix. Is it correct to first apply chain rule as in...

$$\frac{\partial}{\partial \Sigma} \det(\Sigma)^{-1/2} = -\frac{1}{2} \det(\Sigma)^{-3/2} \frac{\partial}{\partial \Sigma} \det(\Sigma) = -\frac{1}{2} \det(\Sigma)^{-3/2} \det(\Sigma) tr(\Sigma^{-1}) \\ = -\frac{1}{2} \det(\Sigma)^{-1/2} tr(\Sigma^{-1})$$

The second equation then follows from matrix cookbook, i.e.

$$\frac{\partial \det(Y)}{\partial x}=\det(Y) tr(Y^{-1} \frac{\partial Y}{\partial x}).$$

The background of my question is taking the derivative of the multivariate normal distribution in which $\det(\Sigma)^{-1/2}$ occurs as parameter. Note I do not intend to derive the log of the MV-normal as done in ML estimation.

Given the function $$f=(\det\Sigma)^{-1/2}$$ Reversing the relationship yields an expression whose RHS-derivative is known and whose LHS is easy to calculate \eqalign{ f^{-2}&=\det\Sigma \cr\cr -2f^{-3}\frac{\partial f}{\partial\Sigma}&=\frac{\partial\det\Sigma}{\partial\Sigma} \cr\cr \frac{\partial f}{\partial\Sigma}&= -\frac{f^{3}}{2}\,\frac{\partial\det\Sigma}{\partial\Sigma}\cr &= -\Big(\frac{f^{3}\det\Sigma}{2}\Big)\,\,\Sigma^{-T}\cr &= -\frac{f}{2}\,\,\Sigma^{-T}\cr }