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Math Help - First principles

  1. #1
    Member
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    First principles

    hi all,

    Find \frac{dy}{dx} for y=\sqrt{x}

    = \frac{1}{2}\displaystyle{x}^{-\frac{1}{2}}\equiv\frac{1}{2\sqrt{x}}

    i then try to get this result from first principles:

    f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow0}\frac{(x+h)^{\frac{1}  {2}} - x^{\frac{1}{2}}}{h}

    then cancel the fractional powers by raising both numerator and denominator by reciprocal power, so that

    \lim_{h\rightarrow0}\frac{h}{h^2}=0

    what am i doing wrong?

    thanks sammy.
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  2. #2
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    Anchorage, AK
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    (\sqrt{x+h}-\sqrt{x})^2\ne(x+h)-x. To find the limit, we rationalize the numerator
    \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+  \sqrt{x}}
    =\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}
    =\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}
    =\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x})},

    --Kevin C.
    Last edited by TwistedOne151; October 30th 2009 at 02:40 PM. Reason: LaTeX error
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  3. #3
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    thanks a lot for that .
    i should have tried expanding the numerator and would have seen it was wrong!
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