Derivative of a function with respect to its first derivative?

[LYSpaceman]

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]

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.

 

[LYSpaceman]

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

 

[pasmith]

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
$$\frac{\partial L}{\partial q}\qquad\mbox{and}\qquad<br /> \frac{\partial L}{\partial q'}$$
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
$$\frac{\partial q}{\partial q'} = \frac{\partial q'}{\partial q} = 0.$$

 

[LYSpaceman]

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]

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