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Math Help - [SOLVED] Partial derivative

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Partial derivative

    f(x,y)=\left\{\begin{array}{cc}y^2arctan\frac{x}{y  },&\mbox{ if } <br />
y\neq 0\\0, & \mbox{ if } y=0\end{array}\right.

    Show that f \in C^1.

    y = 0: \frac{\partial{f}}{\partial{x}} = \lim_{h\to 0} \frac{f(a+h, 0) - f(a,0)}{h} = \frac{0-0}{h} = 0\  \forall x

    Correct?

    y \neq 0: \frac{\partial{f}}{\partial{x}} = \frac{y^3}{y^2 + x^2}, \frac{\partial{f}}{\partial{x}} \in C^n
    Last edited by Spec; April 10th 2008 at 07:21 AM.
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  2. #2
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    Quote Originally Posted by Spec View Post
    f(x,y)=\left\{\begin{array}{cc}y^2arctan\frac{x}{y  },&\mbox{ if } <br />
y\neq 0\\0, & \mbox{ if } y=0\end{array}\right.

    Show that f \in C^1.
    There is still one more thing you need to show. Remember the theorem says that if \partial_1 f,\partial_2 f exist and are continous and f is continous then f is differenciable. You want to show that f is continous at any point (a,b). Say that b=0 then f(a,b) = 0. And so |f(x,y) - f(a,b)| = |f(x,y)|. We want to show we can make this sufficiently small in the punctured disk 0<\sqrt{(x-a)^2+(y-b)^2} < \delta. We can safely ignore the case when y=b because then |f(x,y)| = 0 < \epsilon for any \epsilon > 0. Note that |y|=\sqrt{ 0 + (y-0)^2} \leq \sqrt{(x-a)^2+(y-b)^2} < \delta, so, |f(x,y)| = \left| y^2 \tan^{-1} \frac{y}{x} \right| \leq \frac{\pi}{2}y^2 < \frac{\pi}{2}\delta^2, and this quantity can be made sufficiently small. A similar argument applies when b\not = 0. And so f is continous at any point.
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