Derivative of a function with respect to its first derivative?

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  • #1
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.
 

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  • #2
arildno
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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.
 
  • #3
Isn't q a function of q'? q= integral of q' dt + C, right? Is that not enough?
 
  • #4
pasmith
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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
[tex]
S[q] = \int_a^b L(q(t),q'(t),t)\,\mathrm{d}t
[/tex]
we are using [itex]q[/itex] to stand for a differentiable real-valued function of a single real variable, with [itex]q'[/itex] being its derivative (which is of course a function of [itex]q[/itex]). But when we write
[tex]
\frac{\partial L}{\partial q}\qquad\mbox{and}\qquad
\frac{\partial L}{\partial q'}
[/tex]
we are instead using [itex]q[/itex] and [itex]q'[/itex] to indicate independent real variables (the first and second arguments of L respectively). And since they are independent variables, we have
[tex]
\frac{\partial q}{\partial q'} = \frac{\partial q'}{\partial q} = 0.
[/tex]
 
  • #5
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.
 
  • #6
Office_Shredder
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Because they don't literally mean "how L changes when q changes" when they write [itex] \partial L/\partial q[/itex]. 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
 

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