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Partial differential with respect to y

  1. Feb 13, 2008 #1
    [SOLVED] Partial differential with respect to y

    1. The problem statement, all variables and given/known data
    Given the equation f(x,y) = (x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{1/3}[/tex]

    Show that [tex]f_{y}[/tex](0,0) = 1

    2. Relevant equations
    Basic chain rule..

    3. The attempt at a solution
    Based on the chain rule...I believe that

    [tex]f_{y}[/tex] = 1/3(x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex] 3y[tex]^{2}


    f_{y}[/tex] = y[tex]^{2}[/tex](x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex]

    However, plugging in (0,0) does make [tex]f_{y}[/tex] equivalent to 1. Where is my mistake? Any help would be greatly appreciated.. thanks
  2. jcsd
  3. Feb 13, 2008 #2


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    Staff Emeritus
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    The problem is that the function f, itself, is not differentiable at (0,0) even though the partial derivatives exist at (0,0). (But the partial derivatives are not continuous at (0,0).)

    Go back to the original definition of the derivative:
    At (0,0), f(x,y)= f(0,0)= 0.

    For h not 0, f(0,0+h)= f(0,h)= ((h)3)1/3= h.
    [tex]\frac{\partial f}{\partial y}(0,0)= \lim_{h\rightarrow 0}\frac{f(0,h)- f(0,0)}{h}= \lim_{h\rightarrow 0}\frac{h- 0}{h}= 1[/tex]
  4. Feb 13, 2008 #3
    Yah, you can tell right away something is fishy because the work you did, if you actually plug in x and y = 0, not only doesn't yield 1, it gives you 0/0

    A bit of thinking and you'll see the original f(x,y) is undefined for x and y less than 0(one of them can be negative, but x^3+y^3 must be positive)

    So at the point (0,0) the function kinda "comes into existence", so it's like a sharp sudden point where it begins, instead of being [math words that I forget], so you know you need to do a slightly more detailed analysis to find the derivative
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