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A partial derivative problem?

  1. Mar 2, 2009 #1
    Find par(z)/par(t) at s=1, t=0
    when z= ln(x+y), x=s+t, y=s-t

    Not sure how to approach cause if i plug in s's and t's i get an answer of 0 because taking the partial with respect to t yields a zero. Can someone shed some light on how to correctly solve?

    par(z)/par(t) = partial derivative of z with respect to t
     
  2. jcsd
  3. Mar 2, 2009 #2

    djeitnstine

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    Gold Member

    find [tex]\frac{\partial{z}}{\partial{x}} \frac{\partial{x}}{\partial{t}}}[/tex] and [tex]\frac{\partial{z}}{\partial{y}} \frac{\partial{y}}{\partial{t}}[/tex]
     
  4. Mar 2, 2009 #3

    HallsofIvy

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    The chain rule for partial derivatives is
    [tex]\frac{\partial z}{\partial t}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}[/tex]
     
  5. Mar 2, 2009 #4
    I just don't know how to deal with it in the form its in. The s and t are what are confusing me, can u give me some sort of an example?

    The way i did it is substitute s and t for y and x so i get ln(2s) but when u take partial with respect to t you get 0? is this correct?
     
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