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Chain Rule for Functions of Two Variables Partial Differentiation Question

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Let x=ts^2 -1 and y=ln(s)-t

    Use the chain rule for functions of two variables to determine ∂f/∂t at (s,t)=(1,1)





    3. The attempt at a solution

    y=ln(s)-t

    ∂f/∂t= ∂f/∂s X ∂s/∂t -1

    t=x+1/s^2

    ∂t/∂s= -2(x+1)/s^3

    ∂s/∂t=s^3/-2(x+1)

    ∴ ∂f/∂t= s^2/-2(x+1) -1

    = ∂f/∂t= s^2/-2(ts^2) -1

    ∂f/∂t (1,1) = 1/-2 +1

    =0.5

    However, the answer is 1 and I have spent two hours trying to work out why I am wrong. All help will be met with great appreciation. Thanks <3
     
  2. jcsd
  3. May 26, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What is f? Are we to assume that f is some differentiable function of x and y?


    [tex]\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}[/tex]

    [tex]\frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}(2st)+ \frac{\partial f}{\partial y}\frac{1}{s}[/tex]
    At (s, t)= (1, 1), that would be
    [tex]\frac{\partial f}{\partial s}(1, 1)= 2\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}[/tex]

    I have no clue where this is coming from. Before, x and y were functions of s and t. Now t is a function of x and s? Is this a different problem?

    Given f a function of x and y and [itex]x=ts^2 -1[/itex] and [itex]y=ln(s)-t[/itex] as before
    [tex]\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}= s^2\frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}[/tex]
    At (s, t)= (1, 1) that is
    [tex]\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}[/tex]

    It's impossible to give specific numbers, as you seem to have, because you have not given f as a function of x and y.
     
  4. May 26, 2012 #3
    Hi, I'll upload the link to the question. I highly doubt here is an error in the actual question.
     
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