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Partial Derivatives

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data

    If V=x[tex]^{3}[/tex]f(y/x) show that x[tex]^{2}[/tex]Vxx + 2xyVxy + y[tex]^{2}[/tex]Vyy = 6V


    3. The attempt at a solution
    i would normally just use the chain rule to differenciate this with respect to x and then so on but the f(y/x) is throwing me. Do i just treat the f like a constant or is it a whole new function which i have to differenciate first?
    thanks
    gf
     
  2. jcsd
  3. Aug 18, 2008 #2
    Use the chain rule.
     
  4. Aug 18, 2008 #3
    ok,
    let u = f(y/x)
    then du/dx = -f(y/x2)

    am i doing this right?
     
  5. Aug 18, 2008 #4
    I'm a bit confused now, your f appears to be a function of x and y i.e. f(x,y) but yet you have f(x/y), are you sure that's the problem word for word?
     
  6. Aug 18, 2008 #5
    yes its definately f(y/x) thats whats throwing me...
     
  7. Aug 18, 2008 #6
    can you write out the question exactly as it's worded?
     
  8. Aug 18, 2008 #7
    its worded exactly like i worded it at the start...do you think the question is wrong?
     
  9. Aug 18, 2008 #8

    Dick

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    No, use the chain rule, du/dx=f'(y/x)*d(y/x)/dx=-y*f'(y/x)/x^2.
     
  10. Aug 18, 2008 #9

    HallsofIvy

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    No. You were told to use the chain rule
    [tex]\frac{\partial f(y/x)}{\partial x}= f'(y/x)\frac{\partial (y/x)}{\partial x}[/tex]
    [tex]= f'(y/x)\frac{\partial yx^{-1}}{\partial x}= - f'(y/x)yx^{-2}= -f'(y/x)\frac{y}{x^2}[/tex]
    where "f'(y/x)" is the ordinary derivative of f(x) evaluated at y/x.
     
  11. Aug 19, 2008 #10
    thanks lads,
    got it worked out, when i looked at the first line in my log tables it all became clear. Thanks for the help
     
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