Algebra step for Noether theorem In Tong’s preliminaries for QFT, he makes the following step while showing Noether’s theorem:
$(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta \dot q) = \left(\frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q} \right)\right)\delta q + \frac{\partial L}{\partial \dot q}\delta q$
How do you get algebraically from the left to the right side there?
 A: The equality you asked about is wrong, and I will show why. Since $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} =\frac{\partial L}{\partial\dot{q}}\left ( \frac{d}{dt}\delta q \right )$$ we can write using the product rule: $$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right ) = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}$$ and so $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right )-\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q$$ and so the equality would be: $$\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \left ( \frac{\partial L}{\partial q} - \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\right )\delta q + \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \delta q\right )$$ which is the same as your equality, except it has a time derivative on the last term of the RHS. Notice that the first two terms in the RHS vanish by the Euler-Lagrange equation.
