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Partial derivative, I'm not sure if my solution is right

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, it looks simple, but I'm not sure If the answer I'm giving is right.

    The function is:

    [tex]f(x,y)=\sin|y|[/tex]
    And it asks for the partial derivatives, so:

    [tex]\displaystyle\frac{\partial f}{\partial x}=0[/tex]

    And
    [tex]\displaystyle\frac{\partial f}{\partial y}=\begin{Bmatrix}{ \cos|y|}&\mbox{ if }& y>0\\\not{\exists & \mbox{if}& y=0\\-\cos|y| & \mbox{if}& y<0\end{matrix}[/tex]

    Is this right?
     
  2. jcsd
  3. Sep 20, 2010 #2

    rock.freak667

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    When you are defining the domains like that, you do not need to rewrite the |y|. But if y=0, then f(x,y)=0 so that ∂f/∂y = 0.
     
  4. Sep 20, 2010 #3

    statdad

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    "But if y=0, then f(x,y)=0 so that ∂f/∂y = 0."

    No. think about [itex] f(x) = |x| [/itex] from ordinary calculus. it is true [itex] f(0) = 0 [/itex] but remember f is not differentiable @ x = 0; similar situation in the case given by the OP
     
  5. Sep 20, 2010 #4
    Its continuous but not differentiable at that point.
     
  6. Sep 20, 2010 #5

    statdad

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    The absolute value function is continuous at x =0: is your partial derivative continuous in y at y = 0?
     
  7. Sep 20, 2010 #6
    Yes, it is. It can be demonstrated that Lim f(x,y)->(0,0)=f(0,0). So, what you say, the solution I've given is right?
     
  8. Sep 20, 2010 #7

    statdad

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    yes indeed - i switched my "thinking" to the partial in my previous comment.
     
  9. Sep 20, 2010 #8
    Thanks.
     
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