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2nd partial derivative

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Given the function

    [itex]f(x,y)=\frac{1}{2x^2 + y}[/itex]

    Find the partial derivative fxx(x,y)

    2. Relevant equations


    3. The attempt at a solution

    Seems pretty straight forward, just treat y as a constant and differentiate twice. But I keep getting the answer wrong and I have no idea why. Here's what I did:

    [itex]\frac{∂f(x,y)}{∂x}=\frac{-4x}{(2x^2+y)^2}[/itex]

    Then I differentiate with respect to x again using the quotient rule

    [itex]\frac{∂^{2}f(x,y)}{∂x^{2}}=\frac{-4(2x^2+y)^2 + 4x(2(4x(2x^2 + y))}{(2x^2+ y)^4}[/itex]

    I've also tried to do it by re-arranging and using the product rule, but this fails also. It's driving me mad. Have I done something wrong, or could the supposed correct answer actually be wrong?
     
  2. jcsd
  3. Feb 12, 2013 #2

    vela

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    Your work looks fine to me. What was the supposed answer?
     
  4. Feb 12, 2013 #3

    HallsofIvy

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    Rather than use the quotient rule, you could also write [itex]f(x,y)= (2x^2+y)^{1/2}[/itex] so that [itex]f_x= (1/2)(2x^2+y)^{-1/2}(2x)= x(2x+ y)^{-1/2}[/itex]. Now take the derivative, with respect to x, again, using the product rule.
     
  5. Feb 12, 2013 #4

    Char. Limit

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    Um, shouldn't that be [itex]f(x,y) = (2 x^2 + y)^{-1}[/itex], and not 1/2 as you put?
     
  6. Feb 12, 2013 #5
    Thanks for the replies, as stated, I've tried rearranging into a form where I can apply the product rule, but the answer is also wrong. I've also tried expanding the brackets and then differentiation, but wrong again. I don't know the correct answer as it's entered onto a program which only tells me if my answer is correct or incorrect. I've e-mailed the course coordinator about it, as at this point I'm almost certain that my answer is correct and that there's a problem with the program...
     
  7. Feb 12, 2013 #6

    Mark44

    Staff: Mentor

    Try this version:
    $$ \frac{4(6x^2 - y)}{(2x^2 + y)^3}$$

    All I did was find the common factor for the terms in the numerator, and then simplify. Many times when the quotient rule is involved, the textbook answers will to this kind of simplification.
     
  8. Feb 12, 2013 #7
    I finally figured it out, I was accidentally placing the minus sign in front of the bracketed numerator. There were so many brackets that it slipped by me. Thanks for all the input though guys :)
     
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