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Math Help - Derivative help

  1. #1
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    Derivative help

    Pretty sure I got this right, my online math homework program says it's wrong though

    Find the derivative of the function
    y=(x^2+2)(\sqrt[3]{x^2+3})

    After working through it using chain rule and product rule, I found

    \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}

    What did I do wrong?
    Thanks!
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  2. #2
    MHF Contributor harish21's Avatar
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    without looking at your work, its not possible to tell where you went wrong.

    Note: You have to use the product rule here.

    u=(x^2+2) \; and\; v = (x^2+3)^{\frac{1}{3}}
    Last edited by harish21; February 17th 2011 at 07:56 PM.
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  3. #3
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    Quote Originally Posted by beanus View Post
    Pretty sure I got this right, my online math homework program says it's wrong though

    Find the derivative of the function
    y=(x^2+2)(\sqrt[3]{x^2+3})

    After working through it using chain rule and product rule, I found

    \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}

    What did I do wrong?
    Thanks!
    wolfram alpha agrees with your result (it just has the numerator factored).

    derivative of (x^2+2)*(x^2+3&#41 ;^(1/3) - Wolfram|Alpha
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  4. #4
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    (x^2+2)\frac{d}{dx} [x^2+3]^\frac{1}{3} + \frac{d}{dx} [x^2+2] (x^2+3)^\frac{1}{3}

    y= (x^2+3)^\frac{1}{3}
    u= (x^2+3)

    \frac{dy}{du} = \frac{1}{3u^\frac{2}{3}}
    \frac{du}{dx} = 2x

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

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

    \frac{2x^3+4x}{3(x^2+3)^\frac{2}{3}} + \frac{6x^3+18x}{3(x^2+3)^\frac{2}{3}}

    \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by beanus View Post

    \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}

    that actually is the correct answer.
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