How do I calculate the derivative of a derivative w.r.t. another derivative? The question is a bit of a mouthful. In Classical Mechanics by Goldstein I have seen the use of the following:
$$ \frac{d\dot{F}}{d\dot{q}_i} = \frac{dF}{dq_i} $$
where $F = F(q_1,q_2,...,q_n;t)$, $ \dot{F} = \frac{dF}{dt} $ and $ \dot{q_i} = \frac{dq_i}{dt} $ . How would I show this?
The use of this result is used on page 17 of the PDF solutions guide (Page 10 within the derivations chapter) by showing that the transformation $L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q,t)}{dt}$ keeps the form of the Euler-Lagrange equation invariant:
http://www.slideshare.net/venuatsrr/solution-manual-classical-mechanics-goldstein
 A: In general, this is not true.  For instance, suppose the curve is parametrized by $x=t^2$, $y=t$.  Then
$$
    \frac{dy}{dx} = \frac{\dot y}{\dot x} = \frac{1}{2t} = \frac{1}{2y}
$$
But $\dot y$ is constant, so
$$
    \frac{d\dot y}{d\dot x} = 0
$$
I looked at your reference, and as far as I can tell, your question is about this statement: If $F = F(q_1,q_2,\dots,q_n;t)$, then
$$
    \frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}
$$
The reason for this is the chain rule.
For each $i$, the expression $\dfrac{\partial \dot F}{\partial \dot q_i}$ stands for $\dfrac{\partial}{\partial \dot q_i}\dfrac{dF}{dt}$.  By the chain rule:
\begin{align*}
    \frac{dF}{dt} &= \frac{\partial F}{\partial q_1}\frac{dq_1}{dt}+\dots
                   +\frac{\partial F}{\partial q_n}\frac{dq_n}{dt} 
                   +\frac{\partial F}{\partial t} \\
&=  \frac{\partial F}{\partial q_1}\dot q_1+\dots
                   +\frac{\partial F}{\partial q_n}\dot q_n 
                   +\frac{\partial F}{\partial t} \\
\end{align*}
Then take $\dfrac{\partial}{\partial \dot q_i}$ of each side.  
