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Thread: derivative cube root

  1. #1
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    derivative cube root

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

    $\displaystyle
    8 \sqrt[3]{x}
    $
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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

    $\displaystyle
    8 \sqrt[3]{x}
    $
    Remember that $\displaystyle \sqrt[n]{x}$ can be rewritten $\displaystyle x^{\tfrac 1n}$

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

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

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

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

    Then remember some basic rules of differentiation :
    $\displaystyle (a \cdot f(x))'=a \cdot f'(x)$, a a constant
    $\displaystyle (x^n)'=nx^{n-1}$
    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

    $\displaystyle
    8 \sqrt[3]{x}
    $
    Please see here,

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

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

    Differentiate:

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

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