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  1. #1
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    derivative

    yea so my tutor screwed this one up and she did it wrong.. can someone show a step by step for this derivative please.

    (x^3-2x)/(sqrt(2x+1))
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  2. #2
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    Quote Originally Posted by tbenne3 View Post
    yea so my tutor screwed this one up and she did it wrong.. can someone show a step by step for this derivative please.

    (x^3-2x)/(sqrt(2x+1))
    \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}}}
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  3. #3
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    thanks but i have a question.. did you chain rule the derivative of sqrt(2x+1)
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  4. #4
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    Quote Originally Posted by tbenne3 View Post
    thanks but i have a question.. did you chain rule the derivative of sqrt(2x+1)
    sure did ...

     <br />
\frac{d}{dx} (2x+1)^{\frac{1}{2}} = \frac{1}{2}(2x+1)^{-\frac{1}{2}} \cdot 2 = (2x+1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x+1}}<br />
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