If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, show by direct substitution that
$L` = L + \frac{dF(q_1,...,q_n,t)}{dt}$ also satisfies Lagrange's equation where F is any arbitrarily, but differentiable function of its arguments.
So I think I got everything correct except I don't know why the last step vanishes ?
$\frac{d}{dt}\frac{\partial M}{\partial \dot{q_j}} - \frac{\partial M}{\partial q_j} = \frac{d}{dt}*(\frac{\partial(L + dF(q_1,...,q_n,t)/dt)}{\partial \dot q_j}) - \frac{\partial(L + dF(q_1,...,q_n,t)/dt)}{\partial \dot q_j}$
=$ (\frac{d}{dt}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j}) +\frac{d}{dt}(\frac{\partial}{\partial \dot q_j}(\frac{dF(q_1,...,q_n,t)}{dt})) -(\frac{\partial}{\partial q_j}(\frac{dF(q_1,...,q_n,t)}{dt})) $ (Eq 1)
By chain rule we have $\frac{dF}{dt} = \Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t}$
Notice since L satisfies E-L equation the first two components of equation 1 vanishes. So equation 1 becomes
$\frac{d}{dt}(\frac{\partial}{\partial \dot q_j}(\Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t})) -(\frac{\partial}{\partial q_j}(\Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t})) $ (Eq 2)
So why is (Eq 2) = 0 ? Can someone explain ?
Perform the partial derivative with respect to $\dot q_i$. Since $F$ doesn't depend on $\dot q_i$, the first term is simply
$$ \frac{\mathrm d}{\mathrm dt}\frac{\partial F}{\partial q_j}\;. $$
You can apply the chain rule just like you applied it for $\frac{\mathrm dF}{\mathrm dt}$ to find that this corresponds to what's subtracted in the second term. (Note that partial derivatives commute.)