Let $A=[a_{ij}(x)]$ be a non singular matrix values function with inverse $A^{-1}=B=[b_{ij}(x)]$.
Use the chain rule to justify
$$\frac{\partial}{\partial x^i}(log|det A|)= \frac{(cof A)_{rs}}{det A} \frac{\partial_{rs}}{\partial x^i}= b_{sr} \frac{\partial a_{rs}}{\partial x^i}$$
The proof follows by noting the expansion of determinant by rows $$det A=\sum^n _{r=1} A_{ir} (cof A)_{ir}$$ for any fixed $1 \leq i \leq n$ and then we use the chain rule.
How do you apply the chain rule to this? I would like to see the step by step working.
$$\frac{d}{dt} \log\ {\rm det}\ A = \frac{\frac{d}{dt} {\rm det}\ A}{{\rm det}\ A} = \sum_{i,j} (\frac{d}{dt} a_{ij} ) b_{ij} $$