Equation of Geodesic From Lagranian Equations of Motion

From Schutz (A First Course in General Relativity, Second Edition) we have:
Eq. 6.7 and Eq. 6.8

$l=\int_{\lambda_{0}}^{\lambda _{1}}\left | g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda}\frac{\mathrm{d} x^{\beta }}{\mathrm{d}\lambda }\right |^{1/2}d\lambda$

$l=\int_{\lambda_{0}}^{\lambda _{1}}\left | \vec{V}\cdot \vec{V}\right |^{1/2}d\lambda$

We'll also need Eq. 6.32 from Schutz:

$\Gamma ^{\alpha }_{\mu \beta }=\frac{1}{2}g^{\alpha \beta }\left ( g_{\beta \mu ,\upsilon } +g_{\beta \upsilon ,\mu }-g_{\mu \upsilon ,\beta }\right )$

We need to end up with Schutz, Eq. 6.51 as our result (geodesic equation):

$\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda } \right )+\Gamma ^{\alpha }_{\mu \beta }\frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }=0$

We'll start with the Euler-Lagrange equations:

$\frac{\partial L}{\partial q_{i}}-\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial L}{\partial \dot{q}_{i}} \right )=0$,

where L is the integrand fuction of Eq. 6.7, ${q}_{i}$ are the generalized coordinates of L, and t =$\lambda$

$L=\left | g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }\right |^{1/2}$, and let
$F= g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }$

switching from Leibniz notation to Newton dot notation... $\frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda }=\dot{x}^{\mu }$ and calculating derivatives:

$\frac{\partial L}{\partial x^{\mu }}=\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu }\dot{x}^{\alpha }\dot{x}^{\beta }$

$\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\partial L}{\partial \dot{x}^{\mu }} \right )=\frac{\mathrm{d} }{\mathrm{d} \lambda }\left [\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta }\left ( \frac{\partial \dot{x}^{\alpha }}{\partial \dot{x}^{\mu }} \dot{x}^{\beta }+\frac{\partial \dot{x}^{\beta }}{\partial \dot{x}^{\mu }}\dot{x}^{\alpha }\right )\right ]$

using $\frac{\partial \dot{x}^{\alpha }}{\partial \dot{x}^{\mu }}=\delta ^{\alpha }_{\mu }$ and making the appropriate substitutions into the Euler-Lagrange equation gives:

$\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu}\dot{x}^{\alpha }\dot{x}^{\beta }-\frac{\mathrm{d} }{\mathrm{d} \lambda }\left [\frac{F}{2\left | F \right |^{3/2}}\left ( g_{\mu \beta } \dot{x}^{\beta }+g_{\alpha \mu }\dot{x}^{\alpha }\right )  \right ]=0$

the derivative on the L.H.S. includes two subterms with a factor of $\frac{\mathrm{d} F}{\mathrm{d} \lambda }$, but since $F=\vec{V}\cdot \vec{V}$, then $\frac{\mathrm{d} F}{\mathrm{d} \lambda }=0$ and we have:

$\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu }\dot{x}^{\alpha }\dot{x}^{\beta }-\frac{F}{2\left | F \right |^{3/2}}\left ( g_{\mu \beta ,\alpha } \dot{x}^{\alpha }\dot{x}^{\beta }+g_{\alpha \mu ,\beta }\dot{x}^{\alpha }\dot{x}^{\beta }+g_{\mu \beta }\ddot{x}^{\beta }+g_{\alpha \mu }\ddot{x}^{\alpha }\right )=0$

Almost there... canceling like factors in each term of the L.H.S., using the symmetry of $g_{\alpha \beta }$, and relabeling dummy indices, we end up with:

$g_{\mu \beta ,\alpha }\dot{x}^{\alpha }\dot{x}^{\beta }+g_{\alpha \mu .\beta  }\dot{x}^{\alpha }\dot{x}^{\beta }-g_{\alpha \beta ,\mu  }\dot{x}^{\alpha }\dot{x}^{\beta }+2g_{\alpha \beta }\ddot{x}^{\mu }=0$

Making a substitution into this equation using Schutz Eq. 6.32 and going back to Leibniz notation exclusively for the derivatives gives:

$2g_{\alpha \beta }\Gamma ^{\mu }_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }+2g_{\alpha \beta }\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda } \right )=0$

which obviously becomes:

$\Gamma ^{\mu }_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }+\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda } \right )=0$    Q.E.D.