# Post History

##
**#3: Post edited**

- How to derive the Lagrangian differential force?
- $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
- I was trying to do something.
- $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
- $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
- $$\frac{\partial L}{\partial x}=-kx$$
- $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
- So, I can write that
- $$m\ddot{x}(t)+kx$$
~~But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?~~

- How to derive the Lagrangian differential force?
- $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
- I was trying to do something.
- $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
- $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
- $$\frac{\partial L}{\partial x}=-kx$$
- $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
- So, I can write that
- $$m\ddot{x}(t)+kx$$
- But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?
**>![Lagrangian](https://physics.codidact.com/uploads/z26XyqKjrj2xqwbS9PnAqnhB) ~ https://people.umass.edu/~bvs/601_Lagrange.pdf <br/> Which Euler's equation?**

##
**#2: Post edited**

- How to derive the Lagrangian differential force?
- $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
- I was trying to do something.
- $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
- $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
- $$\frac{\partial L}{\partial x}=-kx$$
- $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
- So, I can write that
~~$$m\ddot{x}(t)~~**-**kx$$- But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?

- How to derive the Lagrangian differential force?
- $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
- I was trying to do something.
- $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
- $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
- $$\frac{\partial L}{\partial x}=-kx$$
- $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
- So, I can write that
- $$m\ddot{x}(t)
**+**kx$$

##
**#1: Initial revision**

Prove differential form of Lagrangian