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Second partial derivatives

  1. Apr 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the second partial derivatives.
    z= x/(x+y)



    3. The attempt at a solution
    I solved the correct df/dx, d^2f/dx^2, df/dy, and d^2f/dy^2, however I can't seem to get the correct answer for d^2f/dydx and d^2f/dxdy.

    My df/dx is y/(x+y)^2 which I changed to y((x+y)^-2)

    Differentiating with respect to y, using the product rule: (1)(x+y)^-2 + y(-2)((x+y)^-3) = (x+y)^-2 - 2y((x+y)^-3)

    My df/dy is -x((x+y)^-2)

    Differentiating with respect to x, using the product rule: (-1)(x+y)^-2 + (-x)(-2)((x+y)^-3) = -(x+y)^-2 + 2x((x+y)^-3)

    I know they aren't right b/c 1) they don't equal each other (Clairaut's Thm.) and 2) it doesn't match the book's answer =)

    Can anyone catch the mistake I'm making? Thank you.
     
    Last edited: Apr 26, 2008
  2. jcsd
  3. Apr 26, 2008 #2
    I think you're right. You seem to have done everything ok. Just try to simplifiy both result solving the newton binomials. Probably everything will reduce and you'll have your right awnser.
    Remember if you get tired check the trigger happy videos at tubepolis.com They're really really funny and stress reliefers.
     
  4. Apr 26, 2008 #3
    Hm, I don't think it works out. I just tried to simplify it and I still didn't get the right answer. Also, my d^2f/dydx answer has a y variable in it, while my d^2f/dxdy answer has an x variable in it. And simplifying either expression doesn't get rid of them.

    The correct answer is supposed to be (x-y)((x+y)^-3)
     
  5. Apr 26, 2008 #4

    olgranpappy

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    Homework Helper

    you can simplify df/dx to
    [tex]
    \frac{y}{(x+y)^2}
    [/tex]
    and you can simplify df/dy to
    [tex]
    \frac{-x}{(x+y)^2}
    [/tex]
    Taking d/dx on the above gives
    [tex]
    (-1)(x+y)^{(-2)}+2x/(x+y)^3
    =\frac{x-y}{(x+y)^3}\;.
    [/tex]
    You can also show that d^2f/dydx is the same
     
  6. Apr 26, 2008 #5
    Isn't the denominator supposed to be [g(x)]^2 so why isn't the denominator (x+y)^4?

    And, can you see what I did wrong when I used the product rule?
     
  7. Apr 26, 2008 #6

    olgranpappy

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    Homework Helper

    you didn't do anything wrong. df/dx and df/dy are *not* "supposed to equal each other". It is
    [tex]
    \frac{d^2 f}{dxdy}
    [/tex]
    which is supposed to equal
    [tex]
    \frac{d^2 f}{dydx}
    [/tex]

    The expression you gave for d^2f/dxdy is
    [tex]
    -\frac{1}{(x+y)^2} + \frac{2x}{(x+y)^3}
    [/tex]
    which can be simplified to the book's answer using algebra (multiply the first term by (x+y)/(x+y) and then the terms have the same denominators and you can add the numerators)
     
  8. Apr 26, 2008 #7
    I can't see your latex image for some reason--it says it is invalid. Also, my answers are not the same. One has a y variable, the other has an x variable. Also there are different signs within each expression.
     
  9. Apr 26, 2008 #8

    olgranpappy

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    Homework Helper

    hmm... I don't know why you cant see the latex... maybe try reloading.
     
  10. Apr 26, 2008 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, there is no mistake- that's what everyone is trying to tell you. Yes, If you had done this using the "quotient" rule, the denominator would have (x+ y)4. But the numerator would have a factor of (x+ y). Try it!

    So [tex]\frac{\partial^2 f}{\partial x\partial y}= \frac{1}{(x+y)^2}+ \frac{-2y}{(x+y)^3}[/tex]
    [tex]= \frac{x+ y}{(x+y)^3}+ \frac{-2y}{(x+y)^3}= \frac{x- y}{(x+y)^3}[/tex]

    So [tex]\frac{\partial^2 f}{\partial y\partial x}= \frac{-1}{(x+ y)^2}+ \frac{2x}{(x+y)^3}[/tex]
    [tex]= \frac{-(x+y)}{(x+y)^3}+ \frac{2x}{(x+y)^3}= \frac{x- y}{(x+y)^3}[/tex]

    Looks to me like they are exactly the same!
     
  11. Apr 26, 2008 #10
    I see it!! Thank you all =)

    I did not catch the multiplication of the expression by (x+y)/(x+y)
     
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