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I Partial and Full derivatives

  1. Feb 25, 2017 #1

    joshmccraney

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    Hi PF!

    Regarding derivatives, suppose we have some function ##f = y(t)x +x^2## where ##y## is an implicit function of ##t## and ##x## is independent of ##t##. Isn't the following true, regarding the difference between a partial and full derivative?
    $$ \frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial y}\frac{dy}{d t} + \frac{\partial f}{\partial x}\frac{d x}{d t} = \frac{\partial f}{\partial y}\frac{dy}{d t}$$
     
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  3. Feb 25, 2017 #2
    Yes, and [tex]\frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}} = x\frac{{dy}}{{dt}}[/tex]
     
  4. Feb 25, 2017 #3

    PeroK

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    This makes no sense to me. If ##x## is independent of ##t##, then ##f## is not a function of ##t## and ##\frac{df}{dt}## makes no sense.
     
  5. Feb 25, 2017 #4
    y is a function of t.
     
  6. Feb 25, 2017 #5

    PeroK

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    But not ##f##.
     
  7. Feb 25, 2017 #6
    Not sure where you're going. f is a function of y, y is a function of t.
     
  8. Feb 25, 2017 #7

    PeroK

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    If ##f(t, x)## is a function of two independent variables, then ##\frac{df}{dt}## is not defined. It's ##\frac{\partial f}{\partial t}##.

    If ##x## is implicitly a function of ##t##, then ##f(t, x(t))## is now a function of only one independent variable and ##\frac{df}{dt}## is defined.
     
  9. Feb 25, 2017 #8

    joshmccraney

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    I thought ##f = f(y(t),x,t)##; is this incorrect? As ##t## changes, wouldn't that imply ##y## would change, and also ##f##?
    In this case are you saying ##\frac{\partial f}{\partial t} = \frac{\partial f}{\partial y}\frac{dy}{d t}##?
     
  10. Feb 25, 2017 #9

    PeroK

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    This is where you must be careful. You have two different functions there. First, suppose we have a function of 3 variables: ##f(x, y, t)##.

    Here we take ##x, y, t## to be independent variables and ##f## is defined on a 3D space and we have partial derivatives of ##f## wrt each of the variables.

    Now, suppose we take a trajectory of a particle, say, where ##x(t)## and ##y(t)## are now functions of ##t##. Most physics books will continue using ##f## but actually what we have now is a new function ##g## of one variable ##t##, defined by:

    ##g(t) = f(x(t), y(t), t)##

    Now, you can have the "total derivative" of ##f##, which is really just the normal derivative of ##g## and can be calculated using the chain rule:

    ##\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t}##

    As I said, many physics book will omit introducing ##g## and write the mathematically rather imprecise:

    ##\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t}##
     
  11. Feb 25, 2017 #10

    joshmccraney

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    Cool, this matches what I've learned! But how do we deal with ##f = y(t) x+x^2##? Are you saying ##\frac{df}{dt}## is not defined because we have two independent variables, namely, ##t## and ##x##? If so, then what is ##\frac{\partial f}{\partial t}##?
     
  12. Feb 25, 2017 #11

    PeroK

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    If we are dealing with a particle's trajectory then everything essentially depends on time along the trajectory. But, mathematically, you could have a function where not all the variables are functions of just one. To take an example, suppose we have ##f(x, y, z)## but we have a relationship between ##x## and ##y##, given by the function ##x(y)##. Now, we have a new function:

    ##g(y, z) = f(x(y), y, z)##

    And we have a partial derivative of ##g## wrt ##y##:

    ##\frac{\partial g}{\partial y} = \frac{\partial f}{\partial x} \frac{dx}{dy} + \frac{\partial f}{\partial y}##
     
  13. Feb 25, 2017 #12

    PeroK

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    In this case, it's simply:

    ##\frac{\partial f}{\partial t} = \frac{dy}{dt} x##
     
  14. Feb 25, 2017 #13

    joshmccraney

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    This is very helpful! So given ##g(y, z) = f(x(y), y, z)##, are you saying ##df## does not exist, so that ##df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz##, and equivalently, that ##dg=\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz## are both meaningless statements?
     
  15. Feb 25, 2017 #14

    PeroK

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    ##df## is a differential, which is something different again.
     
  16. Feb 25, 2017 #15

    joshmccraney

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    Shoot, I thought the full differential was intimately related to the full derivative. So to recap, ##\frac{d g}{dy}## is a meaningless statement but ##\frac{\partial g}{\partial y}## is meaningful?

    Could you explain the differential ##dg## or ##df##, whichever you prefer?
     
  17. Feb 25, 2017 #16

    PeroK

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    For a function of two variables, say, the differential relates the "amount" the function changes as its variables change. Note that, for small ##\Delta x, \Delta y##:

    ##\Delta f \approx \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y##

    This leads to the equation for differentials:

    ##df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy##

    You have to be careful with differentials. For example, if you blindly divided by ##dx## you would get:

    ##\frac{df}{dx} = \frac{\partial f}{\partial x} \frac{dx}{dx} + \frac{\partial f}{\partial y} \frac{dy}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}##

    Which is true, as long as ##y## is a function of ##x##. But, if ##x## and ##y## are independent, then it leads to the essentially nonsensical:

    ##\frac{df}{dx} = \frac{\partial f}{\partial x}##

    Hopefully you can see in what sense that is true and in what sense that is meaningless.
     
  18. Feb 25, 2017 #17

    joshmccraney

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    It seems this is only true when ##f## is a function of only ##x##?
     
  19. Feb 25, 2017 #18

    PeroK

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    It's really notational. If you defined the partial derivative as it is but made it apply also when ##f## is a function of one variable, then we'd use the partial derivative notation all the time. But, we draw an important (but not essentially fundamental) distinction between the notation of a derivative when ##f## is a single valued function: ##f', \frac{df}{dx}## and the derivative when ##f## is a function of more than one variable: ##f_x, \frac{\partial f}{\partial x}##.

    This is a convention that must be respected, which is why ##\frac{df}{dx} = \frac{\partial f}{\partial x}## is notationally (by convention) nonsensical.
     
  20. Feb 25, 2017 #19

    joshmccraney

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    Thanks a lot! I really appreciate your clarification here!
     
  21. Mar 6, 2017 #20

    joshmccraney

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    Are you sure this is correct? Today my professor wrote ##F(x,y(x),y'(x)) = y\sqrt{1+y'^2}## and then that ##F_x = 0 \implies \frac{\partial F}{\partial x}=0##. This was working with the calculus of variations, the Euler-Lagrange equations.
     
  22. Mar 7, 2017 #21

    PeroK

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    How does that contradict anything about the chain rule? If ##F## is independent of the variable ##x##, then ##F_x = \frac{\partial F}{\partial x} = 0##.

    If you are studying Euler-Lagrange, my advice is that you need to be really on top of functions, derivatives, what they mean and how they are defined.
     
    Last edited: Mar 7, 2017
  23. Mar 7, 2017 #22

    joshmccraney

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    It does not contradict the chain rule, it contradicts your statement, where you wrote
    ##\frac{\partial g}{\partial y} = \frac{\partial f}{\partial x} \frac{dx}{dy} + \frac{\partial f}{\partial y}## where ##g = f(x(y),y,z)##. Now let ##f = y \sqrt{1+y'^2}:y=y(x)##. According to what you wrote ##\frac{\partial g}{\partial x} = \frac{\partial f}{\partial y} \frac{dy}{dx} + \frac{\partial f}{\partial x} = y'\sqrt{1+y'^2}\neq 0##. We see that ##f_y = \sqrt{1+y'}## and ##dy/dx = y'##, where I am referencing the first part of what you wrote, namely, ##\frac{\partial f}{\partial y} \frac{dy}{dx}##.
     
  24. Mar 7, 2017 #23

    PeroK

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    ##F_x## is the partial derivative of ##F## with respect to the first argument. It is not the total derivative of ##F## with respect to ##x##. This is where you have to be very careful with what derivatives you are talking about. If we have that ##y## is a function of ##x##, then we have:

    ##F_x = \frac{\partial F}{\partial x}## = the "normal" partial derivative of ##F## with respect to its first argument - denoted by ##x##. In this case ##F_x = 0##.

    But, we also have:

    ##\frac{dF}{dx}## = the total derivative of ##F## with respect to the variable ##x##, which involves all partial derivatives of ##F##, with respect to all its arguments. In this case ##\frac{dF}{dx} \ne 0##.
     
  25. Mar 7, 2017 #24

    PeroK

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    PS in E-L and related situations, the trick I use is this. When I see ##F_x##, say, I look at the function, see that ##x## is the first argument in the function and read this not as "d-f-by-d-x" or "the partial derivative of ##F## wrt ##x##", but as "the partial derivative of F wrt its first argument".
     
  26. Mar 7, 2017 #25

    joshmccraney

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    I'm a little confused. You said:
    1) ##f = y(t) x+x^2 \implies \frac{\partial f}{\partial t} = \frac{dy}{dt} x##
    2)##f=y(x)\sqrt{1+y'(x)^2} \implies \frac{\partial f}{\partial x}=0##.
    I'm confused because in both 1) and 2) the independent variable we are differentiating with respect to does not explicitly appear, yet in each case we have a non-zero and then zero partial derivative. Can you explain the difference in 1) and 2)?
     
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