#1
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¢é͹ÕéËÒ y' ¡Ñº y" ãªéÇÔ¸Õä˹¤ÃѺ
y = $ \frac{1}{\sqrt{x+1} } + \frac{1}{\sqrt{x-1} } $ äÁèá¹èã¨ÇèҨѴÃÙ»áÅéÇãªé $ \left (\,\frac{u}{v}\right)' $ ¢Íº¤Ø³ÁÒ¡¤ÃѺ |
#2
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ãªé $\dfrac{dx^n}{dx}=nx^{n-1}$
$y=(x+1)^{1/2}+(x-1)^{1/2}$
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site:mathcenter.net ¤Ó¤é¹ |
#3
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¹èÒ¨Ð
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I think you're better than you think you are. |
#4
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$\frac{dy}{dx} = \frac{1}{2} (x + 1)^-\frac{1}{2} + \frac{1}{2}(x - 1)^-\frac{1}{2}
= \frac{1}{2\sqrt{x + 1} } + \frac{1}{2\sqrt{x - 1} } $ ¶Ù¡µéͧÁÑé¤ÃѺ |
#5
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$\frac{dy}{dx} = -{\frac{1}{2\sqrt{(x+1)^3}}}-{\frac{1}{2\sqrt{(x-1)^3}}}$
à¼ÍÔ¢é͹ÕéÁÕ $x$ ¡¡ÓÅѧ˹Öè§áÅÐÊ»Ê.˹éÒ $x$ ¡çà»ç¹Ë¹Öè§ (·Ñé§ $(x+1), (x-1)$) ¶éÒ ¿Ñ§¡ìªÑ¹¢éÒ§ã¹Ç§àÅçºà»ÅÕè¹ä»à»ç¹áººÍ×è¹ ¨Ðµéͧ diff. ã¹Ç§àÅ纴éǹèФÃѺ (·ÕèàÃÕ¡ÇèÒ diff. äÊé ¹èФÃѺ ) 16 µØÅÒ¤Á 2009 09:11 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Kowit Pat. à˵ؼÅ: á¡éä¢ÊÑÅѡɳì͹ؾѹ¸ì |
#6
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Áѹ¡ç¤×Í¡®ÅÙ¡â«è䧤ÃѺ
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