I am given a Laplacian in Cartesian coordinates $(x^1, x^2, x^3)$ $$ \Delta = \frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}\delta^{ij}. $$ Using the chain rule, I express my Cartesian derivatives in terms of sperical ones $$ \frac{\partial}{\partial x^i} = \frac{\partial r^k}{\partial x^i}\frac{\partial}{\partial r^k}, $$ where my spherical coordinates are $(r^1, r^2, r^3)$ or $(r, \theta, \varphi).$ Using the previous equation, I simply find the second order partial derivatives and hence the Laplacian is $$ \delta^{ij}\frac{\partial^2}{\partial x^i\partial x^j} = \delta^{ij} \left(\frac{\partial^2 }{\partial r^m\partial r^k} \frac{\partial r^m}{\partial x^j} \frac{\partial r^k}{\partial x^i} + \frac{\partial }{\partial r^k} \frac{\partial^2 r^k}{\partial x^j \partial x^i}\right). \tag{1} $$ To actually compute all first and second order derivatives are quite cumbersome, so I thought to ignore all the mixed derivatives, hence only compute those that are in $$ \sum_{i,j,m,k} \delta^{ij}\delta^{mk} \left(\frac{\partial^2 }{\partial r^m\partial r^k} \frac{\partial r^m}{\partial x^j} \frac{\partial r^k}{\partial x^i} + \frac{\partial }{\partial r^k} \frac{\partial^2 r^k}{\partial x^j \partial x^i}\right). $$ This resulted in the correct Laplacian $$ \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial }{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta}\frac{\partial^2}{\partial \varphi^2}. $$
My question is: Why can I simply ignore mixed derivatives in (1)?