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I partial derivatives

  1. Aug 1, 2017 #1

    dyn

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    Hi.
    If I have a function f ( x , t ) = x - 6t with x ( t ) = t2 and I take the partial derivative of f with respect to x I get the answer 1 as t acts as a constant so its derivative is zero. But if I substitute t with x1/2 I get the answer 1 - 3x-1/2 which is obviously different and wrong , I think ! So , why can't I do this method ?
    Thanks
     
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  3. Aug 1, 2017 #2

    FactChecker

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    If t is constant, then so is x=t2. Better is t = x0.5 and df/dx = 1 - 6*0.5*x-0.5
     
  4. Aug 1, 2017 #3

    dyn

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    with f(x,t) = x-6t and x(t ) = t2 then ∂f/∂x = 1 but if I substitute t for x1/2 I get a different answer. I think my first answer is correct but I don't understand why it doesn't work if I make the substitution ?
     
  5. Aug 1, 2017 #4

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    No, your first answer is wrong. You can't say that x = t2, that t is constant, and that x changes (since dx ≠ 0 ). Since x and t are related, you have to say that df/dx = dx/dx -6 dt/dx and figure it out from there. t is not constant.
     
  6. Aug 2, 2017 #5
    Well, when you take a partial derivative of a function of several variables, what you get will depend, of course, on the dependencies you decide to impose on these variables. For example, you can just use a partial derivative chain rule:
    [tex]\frac{\partial f(u,v)}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}[/tex]
    Now you'll have to decide how these variables (abstract first and second variables of the function of two variables ##f##; I even made a different notation for these variables - ##u## and ##v##, to underscore that) depend on your ##x##. Since ##u## is equal to ##x##, then ##\frac{\partial u}{\partial x} = 1##, and you only have to decide whether you take ##v=t## independent of ##x## (then you'll basically get a partial derivative of ##f## with respect to the first variable, since ##\frac{\partial u}{\partial x} = \frac{\partial t}{\partial x} = 0##) or not:
    [tex]\frac{\partial f}{\partial u} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial x}[/tex]
     
  7. Aug 2, 2017 #6

    dyn

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    I will quote the full question and answer
    Calculate ∂f/∂x for f(x,y,t) = 3x2 +2xy + y1/2t - 5xt . With x(t) = t3 and y(t) = 2t5
    The answer is given as ∂f/∂x = 6x + 2y -5t.
    This looks to me like differentiating with respect to x while holding y and t constant even though x depends on t
     
  8. Aug 2, 2017 #7
    Probably because by ##\partial f / \partial x## they meant a partial derivative of the function ##f## with respect to the first argument.
    Now that I think of it, it would make more sense to call the other case (differentiate ##f## over all of its variables with the use of the chain rule) taking a full derivative with respect to ##x## (when we look at ##f## as a function over ##x## only).
     
  9. Aug 2, 2017 #8

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    Obviously this is a different example, but it certainly looks like you are correct about what they are doing. I don't think that I agree, but maybe I am missing a trick. Is there any more explanation or context given for this example? What subject is this supposed to illustrate? There are techniques for handling constraints where I could understand their partial derivatives.
     
  10. Aug 2, 2017 #9

    dyn

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    It was just a question I found on partial derivatives. I understand that there is a difference between a partial derivative and a full/total derivative. My understanding is that a partial derivative means you treat all the other variables as constants but now i'm confused if one of the other variables is related to t ie. x(t) = t3
     
  11. Aug 2, 2017 #10
    If we took all dependencies into consideration ##f## would be a function of just one variable, kind of like ##f(x,y(x),t(x))##, where each of ##y## and ##t## is itself a function of ##x##, and we would have to find their derivatives with respect to ##x##. That wouldn't be exactly a partial derivative.
     
    Last edited: Aug 3, 2017
  12. Aug 3, 2017 #11

    dyn

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    I'm getting a bit confused now. Is the stated answer given above correct ? And if it is , why is it correct ?
     
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