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Partial Derivative Problem

  1. Jul 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Define f(x,y) = x+2y, w = x+y. What is ∂f / ∂w?

    2. Relevant equations


    3. The attempt at a solution
    f = w+y so:

    ∂f/∂w = ∂(w+y)/∂w = ∂w/∂w + ∂y/∂w = 1 + ∂y/∂w. But I'm really not sure if this is right and if it right so far, I can't figure out what ∂y/∂w should be...
     
  2. jcsd
  3. Jul 5, 2015 #2
    This is right.

    Now partial derivative means the derivative of given parameter.The other parameters will be assumed constant.Cause they are not changing respet to that parameter.I mean If I set two parameter a and b and make a fuction using them (It can be anything).Lets call it f(a,b)=a+b then ∂f/∂a means derivative of function respet to a not b.So we will assume b is constant and ∂f/∂a=1 so.Do the same thing.
     
  4. Jul 5, 2015 #3

    mfb

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    @RyanH42: the problem here: which other variable is supposed to be constant? Why should it be y (as you seem to suggest), not x, or x-y? Those would lead to different answers.
     
  5. Jul 5, 2015 #4
    But w depends implicitly on y so can I really take y as constant? If I get y in terms w (y = w-x) and continue this way (so 1 + ∂y/∂w = 1 + ∂(w-x)/∂w). I just get an endless chain of 1+1+1+1+1... That's why I think what I'm doing is not right.
     
  6. Jul 5, 2015 #5
    I dont know but he answer might be this ∂f/∂w=∂f/∂y.∂y/∂w+∂f/∂x.∂x/∂w look this rule.
     
  7. Jul 5, 2015 #6
    I think this is the chain rule but we haven't learned that yet. I'll read ahead and come back to this and make sense of it but apparently there should be a way to do this without directly making use of the chain rule.
     
  8. Jul 5, 2015 #7
    I have an idea ∂f/∂w=1 + ∂y/∂w now ∂f/∂w=1 + 1/∂w/∂y then ∂f/∂w=2 Its a trick but I dont know its ture or not.I know it was stupid idea
     
    Last edited: Jul 5, 2015
  9. Jul 5, 2015 #8

    verty

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    This problem can't be solved, it is underspecified.

    So ∂y/∂w + 1/∂w/∂y = 0? It wasn't actually a stupid idea, you just made an algebra mistake.
     
  10. Jul 5, 2015 #9
    Hmm, ok. I think I should have posted the full problem because I think it's more open ended than what my original post conveys:

    problem.PNG

    I'm just treating w as the variable and going from there but maybe that's not the right definition? So is there some way to do this problem that makes sense?
     
  11. Jul 5, 2015 #10

    verty

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    I still maintain this is a flawed question and you should move on. It doesn't contain enough information to answer it.
     
  12. Jul 5, 2015 #11
    Is this true ? or You mean ∂w/∂w + 1/∂w/∂y ? I am confused
     
  13. Jul 5, 2015 #12
    If you use chain rule you get 2 again.I think answer is 2.
    ##∂f/∂w=∂f/∂y.∂y/∂w+∂f/∂x.∂x/∂w##
    ##∂f/∂y=2##
    ##∂y/∂w=1/2##
    ##∂f/∂x=1##
    ##∂x/∂w=1##
    So answer is 2 I guess.

    Whats your's idea ?
     
    Last edited: Jul 5, 2015
  14. Jul 5, 2015 #13

    verty

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    ##f(x,y) = x + 2y##
    ##{∂f \over ∂w} = {∂x \over ∂w} + 2 {∂y \over ∂w}##

    The problem happens because we don't know what ##{∂x \over ∂w}## and ##{∂y \over ∂w}## are, we don't have enough information to determine them. If this isn't clear, be sure to look again at partial derivatives and what they mean.
     
    Last edited: Jul 5, 2015
  15. Jul 5, 2015 #14

    mfb

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    w=x+y. Why should the two derivatives be different?

    Also, the rule doesn't work like that with partial derivatives. You could introduce arbitrary new parameters (e. g. z=x) and add more and more terms that would change the result.
     
  16. Jul 5, 2015 #15
    Here my last idea then.The question ask's us for general solution and we do the general solution : ##{∂f \over ∂w} = {∂x \over ∂w} + 2 {∂y \over ∂w}## or
    ##{∂f \over ∂w} = {∂f \over ∂x}{∂x \over ∂w} +{∂f \over ∂w}{∂y \over ∂w}##
    Part b asks solve these equation with spesific parameters.I think there must be some difference between question a and b so I thought we can think x and y like numbers or actually constant parameters.So I mean f=x+2y is a constant cause x and y is contant so the answer is zero.
     
  17. Jul 6, 2015 #16
    Forgive me but Why you guys stop answering the question.Theres a problem and you are avoiding to answer.If one of you find the answer he can tell here cause I am curios.
     
  18. Jul 6, 2015 #17

    ChrisVer

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    The question (a) asks for the definition of the derivative of a general function f(x,y)...The definition of the derivatives is with limits I guess...even with or without limits, you can write your 2nd expression in p#15, and try to find the dx/dw, dy/dw.

    (b) asks for the given function: f=x+2y
     
  19. Jul 6, 2015 #18

    mfb

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    Maybe we are overthinking this. There is no explicit dependence on w in f...
     
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