The action for a relativistic particle of mass $m$ in a curved $D$-dimensional is $$\tilde{S}_0=\frac{1}{2}\int d\tau (\dot{X}^2-m^2)$$ for particular gauge and $\dot{X}^2=g_{\mu\nu}(X)\dot{X}^{\mu}\dot{X}^{\nu}$ where $g_{\mu\nu}$ has Minkowski signature $(- + \cdots +)$. I want to get the equation of motion for $X^{\mu}$ but there is a problem.
My try: The equation of motion is $\frac{\delta \tilde{S}_0}{\delta X^{\alpha}}=0$. Then we have: $$\frac{1}{2}[(\partial_{\alpha}g_{\mu\nu})\dot{X}^{\mu}\dot{X}^{\nu}+g_{\mu\nu}\partial_{\alpha}(\dot{X}^{\mu})\dot{X}^{\nu}+g_{\mu\nu}\dot{X}^{\mu}\partial_{\alpha}(\dot{X}^{\nu})]=0.$$ But $$\partial_{\alpha}(\dot{X}^{\mu})=\partial_{\alpha}\frac{dX^{\mu}}{d\tau}=\frac{d}{d\tau}(\partial_{\alpha}X^{\mu})=\frac{d}{d\tau}\eta^{\mu}_{\alpha}$$ and $$\partial_{\alpha}(\dot{X}^{\nu})=\frac{d}{d\tau}\eta_{\alpha}^{\nu}$$. Therefore, $$\frac{1}{2}[(\partial_{\alpha}g_{\mu\nu})\dot{X}^{\mu}\dot{X}^{\nu}+g_{\mu\nu}(\frac{d}{d\tau}\eta_{\alpha}^{\mu})\dot{X}^{\nu}+g_{\mu\nu}\dot{X}^{\mu}\frac{d}{d\tau}\eta_{\alpha}^{\nu}]=0.$$
But how can I get the geodesic equation?
