# Derivative of a function with respect to its first derivative?

Derivative of a function with respect to its first derivative??

First, I will apologize for my inability to do Latex. Second, I am providing some background on my confusion. Recently, I have been learning functional derivatives and ran into the Euler-Lagrange equation. When I was reading the derivation, I had some confusion in the use of chain rule.

With my notation being, S = Integral of L( q, q', t) dt, when the partial derivatives are taken to find the change in functional S, and we end up with dS/de= Integral of ( dL/dq * dq/de + dL/dq' * dq'/de ) dt.

Where the d's are replaced by the appropriate partial or functional derivative. Now, my question about the chain rule is, is it possible to turn dL /dq' into dL/dq * dq/dq'?

Does dq/dq' exist? And if it does, is it a total derivative, partial derivative, or functional derivative?
Sorry if this seems a little bit confusing or if I am formulating my question incorrectly.

arildno
Homework Helper
Gold Member
Dearly Missed
In general, you can't. Regard them as independent functions, until you apply the added condition that one of them is the derivative of the other.

Isn't q a function of q'? q= integral of q' dt + C, right? Is that not enough?

pasmith
Homework Helper
First, I will apologize for my inability to do Latex. Second, I am providing some background on my confusion. Recently, I have been learning functional derivatives and ran into the Euler-Lagrange equation. When I was reading the derivation, I had some confusion in the use of chain rule.

With my notation being, S = Integral of L( q, q', t) dt, when the partial derivatives are taken to find the change in functional S, and we end up with dS/de= Integral of ( dL/dq * dq/de + dL/dq' * dq'/de ) dt.

Where the d's are replaced by the appropriate partial or functional derivative. Now, my question about the chain rule is, is it possible to turn dL /dq' into dL/dq * dq/dq'?

No. There is an abuse of notation. When we write
$$S[q] = \int_a^b L(q(t),q'(t),t)\,\mathrm{d}t$$
we are using $q$ to stand for a differentiable real-valued function of a single real variable, with $q'$ being its derivative (which is of course a function of $q$). But when we write
$$\frac{\partial L}{\partial q}\qquad\mbox{and}\qquad \frac{\partial L}{\partial q'}$$
we are instead using $q$ and $q'$ to indicate independent real variables (the first and second arguments of L respectively). And since they are independent variables, we have
$$\frac{\partial q}{\partial q'} = \frac{\partial q'}{\partial q} = 0.$$

How are q and q' independent variables if a change in q' is a change for q as well? The arguments cannot be independent unless I am making a mistake.

Office_Shredder
Staff Emeritus
Because they don't literally mean "how L changes when q changes" when they write $\partial L/\partial q$. They mean "L is a function of three variables, and take the partial derivative of L with respect to the first variable". The fact that you then later plug in values into the first and second variable of L which are related is irrelevant to that calculation