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Math Help - derivative question

  1. #1
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    derivative question

    Quote Originally Posted by skeeter View Post
    \frac{d}{dx} \frac{x^3-2x}{\sqrt{2x+1}} = \frac{\sqrt{2x+1} \cdot (3x^2-2) - (x^3-2x) \cdot \frac{1}{\sqrt{2x+1}}}{2x+1}

    multiply by \frac{\sqrt{2x+1}}{\sqrt{2x+1}} to clear the inner fraction ...

    \frac{(2x+1)(3x^2-2) - (x^3-2x)}{(2x+1)^{\frac{3}{2}}}<br />

    \frac{6x^3+3x^2-4x-2 - x^3 + 2x}{(2x+1)^{\frac{3}{2}}}

    \frac{5x^3 + 3x^2 - 2x - 2}{(2x+1)^{\frac{3}{2}}}
    did he use the chain rule to come up with 1/sqrt(2x+1)
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  2. #2
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    Quote Originally Posted by tbenne3 View Post
    did he use the chain rule to come up with 1/sqrt(2x+1)

    If you mean in the first line's numerator the answer is no: that is simply the derivative of \sqrt{2x+1}

    Tonio
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  3. #3
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    Yes tbenne3,

    skeeter calculated

    \frac{d}{dx}\sqrt{2x+1}=\frac{d}{dx}\left(2x+1\rig  ht)^\frac{1}{2}

    =\frac{du}{dx}\frac{d}{du}u^{\frac{1}{2}}
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  4. #4
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    no i was talking about on the very first line to the far right where it has 1/sqrt(2x+1)
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  5. #5
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    how did he get 1/sqrt(2x+1) in the very first step??????
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  6. #6
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    The Quotient Rule is

    \frac{d}{dx}\left(\frac{a(x)}{b(x)}\right)=\frac{b  \frac{da}{dx}-a\frac{db}{dx}}{b^2}

    The \frac{1}{\sqrt{2x+1}} is \frac{db}{dx}

    as I described above by letting u=2x+1
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