# D/dx g(df/dx)

1. Oct 10, 2009

### lolgarithms

for example, isn't the derivative of the first derivative squared:

d/dt (y')^2 = 2y'y''? why does susskind claim it is 2y'', in his classical lecture 3?

Last edited: Oct 10, 2009
2. Oct 11, 2009

### Hepth

Ahh, he's doing lagrangian mechanics.
He is NOT doing
$$\frac{d}{dt} y'^2$$

He is using euler's equations and doing instead:
$$\frac{d}{dt} \frac{\partial}{\partial y'} (y'^2) = \frac{d}{dt} 2 y' = 2 y''$$
The two dissappears from the kinetic energy term because its (1/2) m v^2.