Total Derivative Insights?

  • #1

Main Question or Discussion Point

I’ve always been confused by the formula for the Total Derivative of a function. $$\frac{df(u,v)}{dx}= \frac{\partial f}{\partial x}+\frac{\partial f }{\partial u}\frac{\mathrm{d}u }{\mathrm{d} x}+\frac{\partial f}{\partial v}\frac{\mathrm{d}v }{\mathrm{d} x}$$
Any insight would be greatly appreciated!

What I do understand:
  1. It’s kind of like a multivariable chain rule
  2. The partials terms tell us how the function changes with respect to a parameter and the full derivatives tell us how the parameters themselves change
  3. The result is a scalar function that represents how much the original function changes at every point

What I don't understand:
  1. If it is a multivariable chain rule then why is it written with partials and not as $$\frac{df(u,v)}{dx}= \frac{\mathrm{d}f }{\mathrm{d} u}\frac{\mathrm{d}u }{\mathrm{d} x}+\frac{\mathrm{d}f }{\mathrm{d} v}\frac{\mathrm{d}v }{\mathrm{d} x}$$
  2. It seems there is no necessary reason for terms to be summed instead of combined in some other functional form (multiplied for example)
  3. What are the conditions that allow us to multiply by the denominator and arrive at the total differential? $$df = \frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv$$Is it really ok to be so cavalier manipulating differentials and derivatives?
  4. Is there a connection between the total derivative and functional derivative that makes the expressions look so similar, or is it just a cosmetic similarity?
 

Answers and Replies

  • #2
Charles Link
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I don't agree with your first line. It should read: ## \frac{df(u,v)}{dx}=\frac{\partial{f}}{\partial{u}}\frac{du}{dx}+\frac{\partial{f}}{\partial{v}}\frac{dv}{dx} ##. You only have the first term ## \frac{\partial{f}}{\partial{x}}##, along with the others, if ## f ## has the form ## f(x,u,v) ##.
 
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  • #3
kuruman
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I don't agree with your first line: ## \frac{df(u,v)}{dx}=\frac{\partial{f}}{\partial{u}}\frac{du}{dx}+\frac{\partial{f}}{\partial{v}}\frac{dv}{dx} ##. You only have the first term if ## f ## has the form ## f(x,u,v) ##.
I think OP is considering ##f[u(x),v(x)]##.
 
  • #4
Charles Link
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I think OP is considering ##f[u(x),v(x)]##.
Yes. I agree. ## u=u(x) ## and ## v=v(x) ##. But unless ## f=f(x,u,v) ## the first ## \frac{\partial{f}}{\partial{x}} ## (for constant ## u ## and ## v ##) does not belong there. ## \\ ## I have seen quite often ## f=f(x,y,z,t) ## for a function describing the density as a function of position and time. Meanwhile if the system is sampled at a location ## (x,y,z) ## that is time dependent, (so that ## x=x(t) ##, ## y=y(t) ## and ## z=z(t) ##), then ## \frac{df}{dt} ## will contain a term ## \frac{\partial{f}}{\partial{t}} ## in addition to terms of the form ## \frac{\partial{f}}{\partial{x}}\frac{dx}{dt} ##, etc.## \\ ## In the function of the OP, he doesn't have this extra parameter in the description of the function. All he has is ## f=f(u,v)=f(u(x),v(x)) ##.
 
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  • #5
Yes, you are right, I should have explicitly included a dependence of x in my function.
 
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  • #6
jambaugh
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It is the (possible) occurrence of the explicit x dependency in your original function that makes what you are doing a "Total Derivative" rather than just "a derivative utilizing the multivariable chain rule".

As an insight into the difference between the partial and general derivative notation you can think of "the derivative" of a function of many variables in terms of vectors. I like the Gâteaux definition which boils down to: "A derivative is a linear mapping between differentials":
[tex] dy = f'(x)dx = \lim_{h\to 0} \frac{f(x+h dx)-f(x)}{h}[/tex]
This reads as the derivative is just a multiplier if x and y are scalars, but it becomes a linear operator when x and/or y are vectors. In this context:
[itex]f'(u,v)[/itex] is the operator mapping the differential vector [itex]\langle du, dv\rangle[/itex] to the differential scalar [itex] dy[/itex] as:
[tex] dy = f'(u,v)\bullet \langle du,dv\rangle = \frac{\partial f(u,v)}{\partial u} du + \frac{\partial f(u,v)}{\partial v}dv[/tex]

The beauty of the differential notation and context is that the chain rule is built in. So for the very original context:

[tex] dy = \frac{\partial f(u,v,x)}{\partial u} du + \frac{\partial f(u,v,x)}{\partial v} dv + \frac{\partial f(u,v,x)}{\partial x} dx =[/tex]
[tex] = \frac{\partial f(u,v,x)}{\partial u} \frac{du}{dx} dx+ \frac{\partial f(u,v,x)}{\partial v} \frac{dv}{dx}dx + \frac{\partial f(u,v,x)}{\partial x} dx [/tex]
and thence [tex]dy/dx[/tex] is as stated in the OP.
 
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  • #7
So we can interpret the partials and differentials differently, thinking of differentials as little vectors and the partials as components of a transformation operator? If this is the right way to think about things, how should I interpret dy/dx? What is the division of 2 vectors?
 

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