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Partial Differentiation - The Chain Rule

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate ∂f/∂s + ∂f/∂t at s = 2, t = -1.

    Given:
    f = f(x,y)
    x = s - t
    y = s2 + t2
    ∂f/∂x (3,5) = 0.06170
    ∂f/∂y (3,5) = 0.06170

    2. Relevant equations

    ∂f/∂s = ∂f/∂x * ∂x/∂s + ∂f/∂y * ∂y/∂s

    ∂f/∂t = ∂f/∂x * ∂x/∂t + ∂f/∂y * ∂y/∂t


    3. The attempt at a solution

    I calculated:

    ∂x/∂s = s = 2
    ∂x/∂t = t = -1
    ∂y/∂s = 2s = 4
    ∂y/∂t = 2t = -2

    I know I haven't gotten far, but I don't know how to get ∂f/∂x or ∂f/∂y from the given equations.
     
  2. jcsd
  3. Oct 17, 2011 #2

    SammyS

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    What is are the values of x & y when s = 2 and t = -1 ?
     
  4. Oct 17, 2011 #3

    Dick

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    They gave you ∂f/∂x and ∂f/∂y at the point (x,y)=(3,5). What might the point (x,y)=(3,5) have to do with the point (s,t)=(2,-1)? And I don't think ∂x/∂s = s = 2 is right either, recheck that and ∂x/∂t.
     
  5. Oct 17, 2011 #4

    Mark44

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    No, this isn't right.
    x = s - t,
    so ∂x/∂s = 1 and ∂x/∂t = -1

    y = s2 + t2,
    so ∂y/∂s = 2s and ∂y/∂t = 2t

    Now, when s = 2 and t = -1, what are the values of the four partials?

    You are confusing a function (such as ∂x/∂s = s) with its value at a particular number in its domain (such as ∂x/∂s(2) = 2).
    You don't need the formulas for ∂f/∂x and ∂f/∂y. You are given the values of these functions at the point (3, 5) in their domains.
     
  6. Oct 17, 2011 #5
    Wow...I can't believe I missed that one. x = 3 and y = 5, so the given values can be used to calculate it. Thanks a lot!

    Edit: yeah, i definitely screwed up ∂x/∂s, and ∂x/∂t...Thanks!
     
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