Let B be a tensor-valued variable, taking values from the set of second order tensors on the vector space naturally associated with Euclidean 3-space.
It is given that B is invertible.
I am looking for a proof of the following statement:
$$ \frac{\partial (\log \det \mathbf{B})}{\partial \mathbf{B}} = \mathbf{B^{-T}} $$
Also, it seems one would need an assumption on the positivity of the determinant of B, so I am guessing this is allowed too.
Thanks.
By the chain rule $$\frac{\partial (\log \det B)}{\partial B} = \frac{1}{\det B} \frac{\partial (\det B)}{\partial B} = \frac{1}{\det B} \det (B) B^{-T}$$
Apart from the last equality this should be clear, the last equality follows from Cramers rule. A reference is this wikipedia page and this section on the same page.
(Note: it may be not completely clear what is exactly meant by $\frac{\partial}{\partial B} f(B) = A $. The interpreation of this expression is that $$\frac{\partial}{\partial B_{ij}} f(B) = A_{ij} $$ for every pair of indices $i,j$).
