In Ciarlet`s book on Mathematical Elasticity (vol. 1: Three-dimensional elasticity) on pages 39-40 we read:
Counting the indices modulo 3, the elements of the matrix $\operatorname{cof}\nabla\varphi$ are given by: $$ \left(\operatorname{cof}\nabla\varphi\right)_{ij}=\partial_{j+1}\varphi_{i+1}\partial_{j+2}\varphi_{i+2}-\partial_{j+2}\varphi_{i+1}\partial_{j+1}\varphi_{i+2}~~ {\rm (no~ summation)}, $$ and a direct computation shows that $$ \partial_{j}\left(\operatorname{cof}\nabla\varphi\right)_{ij}=0. $$
Above $\varphi:\mathbb{R}^3\to\mathbb{R}^3$, $\partial_j$ denotes its partial derivative with respect to $j$th argument.
The first element of $\operatorname{cof}\nabla\varphi$ is $\partial_2\varphi_2\partial_3\varphi_3-\partial_3\varphi_2\partial_2\varphi_3$, the derivative $\partial_1$ of which is $$ \partial_{1,2}\varphi_2\partial_3\varphi_3+\partial_2\varphi_2\partial_{1,3}\varphi_3-\partial_{1,3}\varphi_2\partial_2\varphi_3-\partial_3\varphi_2\partial_{1,2}\varphi_3. $$ I have a difficulty to understand why is it 0. Any hint or idea?
The point is, that in $$ \partial_{j}\left(\operatorname{cof}\nabla\varphi\right)_{ij}=0 $$ there is summation over $j$, so, actually, instead of $$ \partial_{1,2}\varphi_2\partial_3\varphi_3+\partial_2\varphi_2\partial_{1,3}\varphi_3-\partial_{1,3}\varphi_2\partial_2\varphi_3-\partial_3\varphi_2\partial_{1,2}\varphi_3, $$ we have $$ \partial_1\left(\operatorname{cof}\nabla\varphi\right)_{i1}+\partial_2\left(\operatorname{cof}\nabla\varphi\right)_{i2}+\partial_3\left(\operatorname{cof}\nabla\varphi\right)_{i3}=0. $$