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Math Help - derivative cube root

  1. #1
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    derivative cube root

    hey guys, just wondering what the derivative of this is... thank u

    <br />
8 \sqrt[3]{x}<br />
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  2. #2
    Moo
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    Hello !
    Quote Originally Posted by jvignacio View Post
    hey guys, just wondering what the derivative of this is... thank u

    <br />
8 \sqrt[3]{x}<br />
    Remember that \sqrt[n]{x} can be rewritten x^{\tfrac 1n}

    So here, \sqrt[3]{x}=\dots

    Then remember some basic rules of differentiation :
    (a \cdot f(x))'=a \cdot f'(x), a a constant
    (x^n)'=nx^{n-1}
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello !

    Remember that \sqrt[n]{x} can be rewritten x^{\tfrac 1n}

    So here, \sqrt[3]{x}=\dots

    Then remember some basic rules of differentiation :
    (a \cdot f(x))'=a \cdot f'(x), a a constant
    (x^n)'=nx^{n-1}
    thank u! got it
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    hey guys, just wondering what the derivative of this is... thank u

    <br />
8 \sqrt[3]{x}<br />
    Please see here,

    f(x)= 8 \sqrt[3]{x}

    f(x)= 8 x^{\frac{1}{3}}

    Differentiate:

    f'(x)= 8 (\frac{1}{3}) x^{\frac{-2}{3}}

    f'(x)= \frac{8}{\sqrt[3]{x^2}}
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