# Christoffel symbol and partial derivative of the metric

It may seem like a dumb question but I'm trying to solve a problem involving coordinate transformations on the Christoffel symbol and to solve it they do the product rule $$\partial_\alpha g_{\beta ' \gamma '} = \frac{\partial}{\partial x^{\alpha}} \left(\frac{\partial x^{\gamma}}{\partial x^{\gamma '}} \frac{\partial x^{\beta}}{\partial x^{\beta '}} \right)g_{\beta \gamma} + \frac{\partial x^{\gamma}}{\partial x^{\gamma '}} \frac{\partial x^{\beta}}{\partial x^{\beta '}} \frac{\partial x^{\nu}}{\partial x^{\alpha }} \partial_{\nu} g_{\beta \gamma}.$$

I understand that we need to take the product rule, but I don't understand why the second term is $$\frac{\partial x^{\gamma}}{\partial x^{\gamma '}} \frac{\partial x^{\beta}}{\partial x^{\beta '}} \frac{\partial x^{\nu}}{\partial x^{\alpha }} \partial_{\nu} g_{\beta \gamma}$$ and not $$\frac{\partial x^{\gamma}}{\partial x^{\gamma '}} \frac{\partial x^{\beta}}{\partial x^{\beta '}} \frac{\partial}{\partial x^{\alpha }} g_{\beta \gamma}$$. Where does the $$\frac{\partial x^{\nu}}{\partial x^{\alpha }}$$ come from?

• $\frac{\partial x^\nu}{\partial x^\alpha}=\delta^\nu_\alpha$ Mar 13, 2023 at 10:07

The two expressions are actually the same. Notice that $$\frac{\partial x^{\nu}}{\partial x^{\alpha }} \partial_{\nu} = \frac{\partial x^{\nu}}{\partial x^{\alpha }} \frac{\partial}{\partial x^{\nu}} = \frac{\partial}{\partial x^{\alpha}}.$$