Given an invertible coordinate transform between a set of coordinates $\{y^1, ..., y^n \}$ and $\{x^1, ..., x^n \}$ where $y^i = y^i(x^1,...,x^n)$ and $x^i = x^i(y^1,...,y^n)$ for each $i \in \{1,...,n\}$, I am trying to prove the following formula for the gradient of the Jacobian determinant:
$$\dfrac{\partial}{\partial x^l} \left|\dfrac{\partial y^i}{\partial x^j}\right| = \dfrac{\partial^2y^\alpha}{\partial x^l \partial x^\beta}\dfrac{\partial x^\beta}{\partial y^\alpha} \left|\dfrac{\partial y^i}{\partial x^j}\right|$$
For reference, this is from a problem in p.87 of Tensor Analysis: Theory and Application by Ivar Stephen Sokolnikoff. I've tried using Jacobi's identity but I'm not sure whether it's applicable here, and using the co-factor expansion definition of the determinant but it hasn't led me anywhere.
Let $A:U\subset\Bbb{R}^n\to GL_n(\Bbb{R})\subset M_{n\times n}(\Bbb{R})$ be a smooth invertible-matrix-valued function defined on an open subset $U$ of $\Bbb{R}^n$. Then, the composite mapping $\det\circ A$, or simply $\det A$ is also smooth and we can investigate its partial derivatives. Well, fix a point $p\in U$. Then, \begin{align} \frac{\partial (\det A)}{\partial x^l}\bigg|_p&=\frac{d}{dt}\bigg|_{t=0}\det A(p+te_l), \end{align} where $e_l=(0,\dots, 1, \dots, 0)$ with the $1$ in the $l^{th}$ spot. Now, we calculate: \begin{align} \det\left(A(p+te_l)\right)&=\det\left(A(p)+ t\frac{\partial A}{\partial x^l}(p)+ \mathcal{O}(t^2)\right)\\ &=\det\left(A(p)\right)\cdot \det\left(I+t\frac{\partial A}{\partial x^l}(p)\cdot A(p)^{-1}+ \mathcal{O}(t^2)\right)\\ &= \det(A(p))\cdot \left(1+ t\,\text{trace}\left(\frac{\partial A}{\partial x^l}(p)\cdot A(p)^{-1}\right)+ \mathcal{O}(t^2)\right). \end{align} So, if you calculate the derivative with respect to $t$ at $t=0$, we find that \begin{align} \frac{\partial (\det A)}{\partial x^l}\bigg|_p&=\det(A(p))\cdot \text{trace}\left(\frac{\partial A}{\partial x^l}(p)\cdot A(p)^{-1}\right), \end{align} or $\frac{\partial (\det A)}{\partial x^l}=\det A\cdot \text{trace}\left(\frac{\partial A}{\partial x^l}\cdot A^{-1}\right)$ for short if you suppress the point of evaluation. Can you specialize to the case of $A=\frac{\partial y}{\partial x}$?
Btw the formula $\det(I+tB)=1+t\cdot\text{trace}(B)+\mathcal{O}(t^2)$ is easily proved if $B$ is triangular. In the general case, you can perform a similarity transformation (by passing to the Jordan canonical form over $\Bbb{C}$ for instance) and then deduce the result. Or... you can directly apply the Leibniz formula/cofactor expansion and induction or any of the other brute-force methods.