According to my solutions guide, if I am given:
$$g = \sqrt{1+\dot{x}^2}$$
where $\dot{x}$ is a function of time, then the following equation:
$$\frac{\partial g}{\partial x} - \frac{d}{dt} \left[ \frac{\partial g}{\partial \dot{x}} \right] = 0$$
should simplify to:
$$\ddot{x} = 0$$
but that is not what I'm getting...here are my steps
I ignore the left term since g is not a function of x, and the inner partial becomes:
$$\frac{\partial g}{\partial \dot{x}} = \frac{\dot{x}}{\sqrt{\dot{x}^2+1}}$$
Taking the time derivative calls for the quotient rule:
$$\frac{\dot{f}p - f\dot{p}}{p^2}$$
where $f = \dot{x}, \dot{f} = \ddot{x}, p = \sqrt{\dot{x}^2+1}$
I use the chain rule to find $\dot{p}$
$$\frac{\partial}{\partial \dot{x}}\sqrt{\dot{x}^2+1}\frac{\partial \dot{x}}{\partial t} = \frac{\ddot{x}\dot{x}}{\sqrt{\dot{x}^2+1}}$$
My quotient rule becomes:
$$\frac{\ddot{x}\sqrt{\dot{x}^2+1}-\ddot{x}\dot{x}^2\ (\dot{x}^2+1)^{\frac{-1}{2}}}{\dot{x}^2+1}$$
which is not reducing to $\ddot{x}$. Did I make a mistake or is my solutions guide mistaken?
Note that the mess you derived is set equal to zero. Pull out the $\ddot{x}$ and you get:
$\ddot{x}\frac{\sqrt{\dot{x}^2+1}-\dot{x}^2\ (\dot{x}^2+1)^{\frac{-1}{2}}}{\dot{x}^2+1}=0$
Now, note that all the terms in the second factor are squared, so $\frac{\sqrt{\dot{x}^2+1}-\dot{x}^2\ (\dot{x}^2+1)^{\frac{-1}{2}}}{\dot{x}^2+1}\neq 0\;\;\forall x$
Therefore, the only way to make the equality work out is if $\ddot{x}=0$