Invariance of lagrangian under point transformation A lagrangian $L(q,\dot q, t)$ is invariant under the point transformation $$q_i=q_i(s_1,...,s_n,t)$$
To prove this I show that
$$\frac{d}{dt} \frac{\partial L}{\partial \dot s_i} - \frac{\partial L}{\partial s_i} = 0$$
where 
$$\frac{\partial L}{\partial s_i} = \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial s_i}$$
$$\frac{\partial L}{\partial \dot s_i} = \frac{\partial L}{\partial \dot q_j}\frac{\partial \dot q_j}{\partial \dot s_i}$$
$$\frac{\partial \dot q_j}{\partial \dot s_i}= \frac{\partial q_j}{\partial s_i}$$
which gives 
$$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\frac{\partial q_j}{\partial s_i} \right) - \frac{\partial L}{\partial q_j}\frac{\partial  q_j}{\partial s_i} = 0$$
but in order to get into the form
$$\left(\frac{d}{dt} \frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j}\right)\frac{\partial  q_j}{\partial s_i} = 0$$
I need to prove that $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\frac{\partial q_j}{\partial s_i} \right) = \frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\right)\frac{\partial q_j}{\partial s_i} $$
What is the logic behind this? Any help would be greatly appreciated!
 A: In case anyone wants the whole proof.
Invariance of Lagrange's Equations under Point Transformations
Let $q_1,q_2,\cdots,q_n$ be a set of independent generalised coordinates for a system on $n$ degrees of freedom, with Lagrangian $L(q,\dot{q},t)$. Suppose we transform to another set of independent coordinates $s=s_1,s_2,\cdots,s_n$ be means of transformation equations:
\begin{equation}
 q_i=q_i(s,t) \qquad i=1,2,\cdots,n \qquad (1)
\end{equation}
and
\begin{align}
 \dot{q}_i=\frac{dq_i}{dt}=\sum_j \frac{\partial q_i}{\partial s_j}\dot{s}_j+\frac{\partial q_i}{\partial t} \nonumber \\
 \dot{q}_i=\dot{q_i}(s,\dot{s},t)\qquad i=1,\cdots,n \qquad (2)
\end{align}
Such a transformation is called a point transformation. Then:
\begin{align}
 \frac{\partial L}{\partial s_j}=\sum_i\left[\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial s_j} +\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial s_j}\right]\qquad (3)
\end{align}
\begin{align}
 \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\frac{d}{dt}\left[\sum_i\left(\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial \dot{s}_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial \dot{s}_j}\right) \right] \qquad (4)
\end{align}
but we also know that $\partial q_i/\partial \dot{s}_j=0$ and also:
\begin{align}
  dq_i=\sum_j\frac{\partial q_i}{\partial s_j}ds_j+\frac{\partial q_i}{\partial t}dt \nonumber \\
  \frac{dq_i}{dt}=\sum_j\frac{\partial q_i}{\partial s_j}\frac{ds_j}{dt}+\frac{\partial q_i}{\partial t} \nonumber \\
  \dot{q}_i=\sum_j\frac{\partial q_i}{\partial s_j}\dot{s}_j+\frac{\partial q_i}{\partial t} \nonumber 
 \end{align}
\begin{equation}
  \frac{\partial \dot{q}_i}{\partial \dot{s}_j}=\frac{\partial q_i}{\partial s_j} \qquad (5)
 \end{equation}
Making these changes to (4) we get:
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\frac{d}{dt}\left[\sum_i \frac{\partial L}{\partial \dot{q}_i}\frac{\partial q_i}{\partial s_j}\right] =\nonumber \\
&=\sum_i \left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)\frac{\partial q_i}{\partial s_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{d}{dt}\left(\frac{\partial q_i}{\partial s_j}\right)\right] \qquad (6)
\end{align}
But we know that $L$ satisfies the Lagrangian equations for the $q$ set of coordinates. Hence:
\begin{equation}
 \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)=\frac{\partial L}{\partial q_i} \qquad (7)
\end{equation}
as well as the property:
\begin{align}
 \frac{d}{dt}\left(\frac{\partial q_i}{\partial s_j}\right)&=\sum_k\frac{\partial}{\partial q_k}\left(\frac{\partial q_i}{\partial s_j}\right)\dot{q_k}+\frac{\partial^2q_i}{\partial t\partial s_j}=\nonumber \\
 &=\frac{\partial}{\partial s_j}\left[\sum_k\frac{\partial q_i}{\partial q_k}\dot{q}_k+\frac{\partial q_i}{\partial t}\right]\nonumber \\
 &=\frac{\partial}{\partial s_j}\left(\frac{d q_i}{dt}\right)=\nonumber \\
 &=\frac{\partial \dot{q}_i}{\partial s_j}\qquad (8)
\end{align}
and (6) becomes
\begin{align}
 \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\sum_i\left[\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial  s_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial s_j}\right] \qquad(9)
\end{align}
Now making the substitutions (3) and (9), we get the important result that:
\begin{align}
 \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)-\frac{\partial L}{\partial s_j}=0 \qquad(10)
\end{align}
