Q:

Deriving the Geodesic Equation

I found a derivation of the geodesic equation that includes this step as I write it:
$$
\frac{d (g_{ab}\dot{x}^b)}{dt}=\frac{1}{2}\partial_ag_{bc}\dot{x}^b\dot{x}^c \Rightarrow
\\
\\
\partial_cg_{ab}\dot{x}^b\dot{x}^c+g_{ab}\ddot{x}^b=\frac{1}{2}\partial_ag_{bc}\dot{x}^b\dot{x}^c
\\$$
How exactly did we get that? Please help I really need to understand it. If this is wrong please tell me.
Also how did we get that first bit from the Euler Lagrange equations? should it not have a factor of $\frac{1}{2}$ on the LHS and be:
$$
\frac{1}{2}\frac{d (g_{ab}\dot{x}^b)}{dt}=\frac{1}{2}\partial_ag_{bc}\dot{x}^b\dot{x}^c 
$$?

A:

$$\dfrac{d(g_{ab})}{dt}=\dfrac{\partial(g_{ab})}{\partial x^c}\cdot\dfrac{\partial x^c}{\partial t}=\partial_cg_{ab}\cdot\dot{x}^c$$
$$\dfrac{d(g_{ab}\dot{x}^b)}{dt}=\dfrac{d(g_{ab})}{dt}\cdot\dot{x}^b+g_{ab}\cdot\dfrac{d(\dot{x}^b)}{dt}=\partial_cg_{ab}\dot{x}^c\dot{x}^b+g_{ab}\ddot{x}^b$$

