I want to show that the directional covariant derivative of the velocity vector yields the geodesic equation but I get stuck at a partial derivative. To make it clear what I want to show, the statement is:
Suppose $\gamma$ is a path on a manifold $M$ and $\varphi$ is a coordinate map. Let $x^\mu (\lambda) = (\varphi \circ \gamma)^\mu(\lambda)$ be the component functions of the path in $\mathbb{R}^n$, then $$ 0 = \frac{D}{d\lambda} \frac{dx^\mu}{d \lambda}= \frac{d^2x^\mu}{d \lambda^2} + \Gamma^{\mu}_{\rho \sigma} \frac{d x^\rho}{d\lambda}\frac{d x^\sigma}{d\lambda}. $$ This is my attempt. I get stuck at the end: Suppose $\frac{dx^\mu}{d \lambda} = U^\mu$, then unwinding definitions $$ 0 = \frac{D}{d\lambda} \frac{dx^\nu}{d \lambda}= \frac{D}{d\lambda} U^\nu = \frac{d x^\mu}{d \lambda} \nabla_\mu \left( U^\nu \right) = \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu + \frac{d x^\mu}{d \lambda} \Gamma^\nu_{\mu \rho} U^\rho = \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu + \Gamma^\nu_{\mu \rho} \frac{d x^\mu}{d \lambda}\frac{d x^\rho}{d \lambda} $$ It remains to show that $$ \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu = \frac{d^2x^\mu}{d \lambda^2} $$ Using the definition $\partial_\mu f = \frac{\partial}{\partial x^\mu}(f \circ \phi^{-1})$, I cannot see that it is correct: $$ \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu = \frac{d x^\mu}{d \lambda} \partial_\mu \frac{d x^\nu}{d \lambda} = \frac{d x^\mu}{d \lambda} \frac{\partial}{\partial(\phi \circ \gamma)^\mu} \left(\frac{d (\phi \circ \gamma)^\nu}{d \lambda} \circ \phi^{-1} \right) = ? $$ I am confused. Any help? It would be great if you could be precise about what you mean about each step.