# What will be the derivative of $x$ with respect to $\frac{dx}{dt}$ in physics when conducting a Taylor expansion?

As far as I know, there are two solutions to this problem. The first solution is to transform the equation as $$\dfrac{dx}{d\dot{x}} = \dfrac{dx}{dt}\dfrac{dt}{d\dot{x}} = \dfrac{\dot{x}}{\ddot{x}}$$ Yet when I was viewing lecture 15 of the engineering dynamics lectures of MIT here, when talking about the Lagrange method, I saw the lecturer simply evaluate this derivative to 0 because of the displacement, x is apparently not a function of speed, $\dot{x}$. (it may have been in regards to the angle, $\theta$)

I have followed the second approach for the most part yet I was required to Taylor expand a certain differential equation(which contains x) with respect to $\dot{x}$ while doing a physics equation. In this case, should I evaluate $\dfrac{dx}{d\dot{x}}$ to 0 or $\dfrac{\dot{x}}{\ddot{x}}$? If the latter is the case, was I ever correct in evaluating $\dfrac{dx}{d\dot{x}}$ to 0?

As a reference, $\ddot{x}$ is a function of $x$ and $\dot{x}$ in my differential equation.

• Before you can evaluate it, you need to define it. You need more context on what is actually going on. What you even mean by $$\frac{dx}{d\dot{x}}$$ – GEdgar Jan 12 '18 at 2:24
• At what minute in the lecture does this happen? I suspect it could be a partial derivative, in which case your lecturer's reasoning would be correct. – John Doe Jan 12 '18 at 2:25
• + John Doe it occurred at 1:02:06. And it was a partial! Thank you! You can post that as the answer and I will accept it. – Isamu Isozaki Jan 12 '18 at 2:40
• GEdgar Oh, I'm sorry. I was trying to Taylor expand the differential equation given in my past question here which was given by applying the Lagrange method to a system where a cylinder is pushed. $\theta$ is the angle the cylinder tips. The reason I was taylor expanding it was – Isamu Isozaki Jan 12 '18 at 2:46
• because in the question here I was advised to conduct a multivariate taylor expansion on it so that I can approximate $\theta$ – Isamu Isozaki Jan 12 '18 at 2:51

As stated in the comments, it is a partial derivative $$\frac{\partial x}{\partial\dot x}$$If $x$ does not have any explicit dependence on $\dot x$, then this is $0$, which is what happens in this case. If it was a total derivative $\frac{dx}{d\dot x}$ where $x$ was just a function of $\dot x$, then yes, it would be $$\frac{dx}{d\dot x}=\frac{\dot x}{\ddot x}$$