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Let $u:D~(\subset\mathbb R^2)\to\mathbb R$ be a function. By the partial derivative of $u$ w.r.t. $x$ at $(x_0,y_0)\in D$ we mean the limit $$\lim_{h\to 0}\frac{u(x_0+h,y_0)-u(x_0,y_0)}{h}$$

I've interpreted it in the following way:

  • The partial derivative of $u$ w.r.t. $x$ at $(x_0,y_0)\in D$ is the $\lim_{~h\to {x_0}} g(h)$ where $$g:\{h\in\mathbb R:(h,y_0)\in D\}-\{x_0\}\to\mathbb R:h\mapsto\frac{u(x,y_0)-u(x_0,y_0)}{x-x_0}$$

    Of course for the limit to be valid, $x_0$ must belongs to $A^d$ where $A=\{h\in\mathbb R:$$(h,y_0)\in D\}$

Am I right? How can I show that $(x_0,y_0)\in D^d\iff x_0\in A^d?$

share|improve this question
    
$D^d$ being what ?? –  Halil Duru Apr 26 '13 at 8:03
    
@HalilDuru: Most probably the derived set of $D.$ –  Sugata Adhya Apr 26 '13 at 13:25

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