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Math Help - Differentiability

  1. #1
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    Differentiability

    Hi there. I got next problem, and I must say if it is differentiable in all of its domain.
    f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}

    So, I thought trying with the partial derivatives.

    f_x=\begin{Bmatrix} 2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}

    f_y=\begin{Bmatrix} 2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}

    And here is the deal. As you see, I've calculated by definition that for all points of the form (x_0,-x_0) the derivative is zero. But the limit of the derivative tending to that kind of points doesn't exists because of the cosine. Is there something wrong with this? I mean is there any kind of contradiction with this, or its alright and just the derivative isn't continuous at that kind of points?

    Bye and thanks.
    Last edited by Ulysses; September 21st 2010 at 07:20 PM.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Ulysses View Post
    1. The problem statement, all variables and given/known data
    Hi there. I got next problem, and I must say if it is differentiable in all of its domain.
    f(x,y)=\begin{Bmatrix} (x+y)^2\sen(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq{-x}\\0 & \mbox{si}& y=-x\end{matrix}

    So, I thought trying with the partial derivatives.

    f_x=\begin{Bmatrix} 2(x+y)\sen(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq{-x}\\0 & \mbox{si}& y=-x\end{matrix}

    f_y=\begin{Bmatrix} 2(x+y)\sen(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq{-x}\\0 & \mbox{si}& y=-x\end{matrix}

    And here is the deal. As you see, I've calculated by definition that for all points of the form (x_0,-x_0) the derivative is zero. But the limit of the derivative tending to that kind of points doesn't exists because of the cosine. Is there something wrong with this? I mean is there any kind of contradiction with this, or its alright and just the derivative isn't continuous at that kind of points?

    Bye and thanks.
    how did you get that cos ? your partial derivation on x should be something like  \frac {df}{dx} = \pi ( y'(x) +1)
    or there is more functions in starting f(x,y) but miss type them ?

    or lol it's just me


    Edit: if function f(X) in region of some point A have partial derivations on all variables and if those derivations are continuous in point A, than function f(X) is differentiable in point A
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  3. #3
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    So sorry, theres a mistake, theres a sine there. I'll correct it right now. I'm sorry, I wanted to post before leaving some hours ago, so I did it hurried and then I made some mistakes, but its ok now. Thanks.
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