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7.1 ¶éÒ $y \ne 0$ áÅéÇáÊ´§ÇèÒ $0 + y \ne 0$ ¹Ñ蹤×͵éͧãªéÍѹº¹

$D_1f(x, y) = \frac{\partial f}{\partial x}$ = ͹ؾѹ¸ì¢Í§ $\frac{x^2-xy}{x+y}$ â´ÂÁͧÇèÒ $x$ à»ç¹µÑÇá»Ã $y$ à»ç¹¤èÒ¤§µÑÇ

ãªéÊٵôԿ¼ÅËÒÃä´é $D_1f(x, y) = \frac{(x+y)(2x-y)-(x^2-xy)(1)}{(x+y)^2}$

´Ñ§¹Ñé¹ $D_1f(0, y) = \frac{(0+y)(0-y)-(0-0)}{(0+y)^2} = - 1$

¶éÒ¨ÐËÒ $D_1f(0, 0)$ ¨ÐàËç¹ÇèÒà¹×èͧ¨Ò¡ $0+0=0$ ´Ñ§¹Ñé¹ ãªé $f(x,y) = 0$ «Öè§ä´é $D_1f(x,y) = 0$ áÅéÇ $D_1f(0, 0) = 0$

¢éÍ 7.2 ¡ç¤ÅéÒ¡ѹ¤ÃѺ áµèÁͧÇèÒ y à»ç¹µÑÇá»Ã
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Old 29 àÁÉÒ¹ 2016, 12:48
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¢éͤÇÒÁà´ÔÁà¢Õ¹â´Â¤Ø³ gon View Post
7.1 ¶éÒ $y \ne 0$ áÅéÇáÊ´§ÇèÒ $0 + y \ne 0$ ¹Ñ蹤×͵éͧãªéÍѹº¹

$D_1f(x, y) = \frac{\partial f}{\partial x}$ = ͹ؾѹ¸ì¢Í§ $\frac{x^2-xy}{x+y}$ â´ÂÁͧÇèÒ $x$ à»ç¹µÑÇá»Ã $y$ à»ç¹¤èÒ¤§µÑÇ

ãªéÊٵôԿ¼ÅËÒÃä´é $D_1f(x, y) = \frac{(x+y)(2x-y)-(x^2-xy)(1)}{(x+y)^2}$

´Ñ§¹Ñé¹ $D_1f(0, y) = \frac{(0+y)(0-y)-(0-0)}{(0+y)^2} = - 1$

¶éÒ¨ÐËÒ $D_1f(0, 0)$ ¨ÐàËç¹ÇèÒà¹×èͧ¨Ò¡ $0+0=0$ ´Ñ§¹Ñé¹ ãªé $f(x,y) = 0$ «Öè§ä´é $D_1f(x,y) = 0$ áÅéÇ $D_1f(0, 0) = 0$

¢éÍ 7.2 ¡ç¤ÅéÒ¡ѹ¤ÃѺ áµèÁͧÇèÒ y à»ç¹µÑÇá»Ã

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