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Math Help - find the derivative

  1. #1
    Junior Member
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    find the derivative

    i'm having some problem differentiating this function. the answer says (-1/3)x^(-5/3) but the answer i'm getting is (-1/3)(square root 2)(x)^(-5/3)


    the function is this:

    1/ cube root of 8x^2

    thank you!
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  2. #2
    Super Member

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    Hello, finalfantasy!

    i'm having some problem differentiating this function.

    The answer says: -\frac{1}{3}x^{-\frac{5}{3}}
    but the answer i'm getting is: -\frac{1}{3}\sqrt{2}\,x^{-\frac{5}{3}}

    The function is: . f(x) \;=\;\frac{1}{\sqrt[3]{8x^2 }}
    I can't see how you got \sqrt{2} in there . . .


    We have: . f(x) \;=\;\frac{1}{\sqrt[3]{8x^2}} \;=\;\frac{1}{\sqrt[3]{8}\cdot\sqrt[3]{x^2}} \;=\; \frac{1}{2\cdot x^{\frac{2}{3}}} \;=\;\frac{1}{2}x^{-\frac{2}{3}}

    . . f'(x)\;=\;\left(\frac{1}{2}\right)\left(-\frac{2}{3}\right)x^{-\frac{5}{3}} \;=\;-\frac{1}{3}x^{-\frac{5}{3}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    If we didn't simplify first, we can use the Chain Rule . . .

    We have: . f(x) \;=\;\frac{1}{\sqrt[3]{8x^2}} \;=\;\frac{1}{\left(8x^2\right)^{\frac{1}{3}}} \;=\;\left(8x^2\right)^{-\frac{1}{3}}

    Then: . f'(x) \;=\;-\frac{1}{3}\left(8x^2\right)^{-\frac{4}{3}}(16x) \;=\;-\frac{1}{3}\left(8^{-\frac{4}{3}}\right)\left(x^2\right)^{-\frac{4}{3}}(16x)

    . . = \;-\frac{1}{3}\left(2^{-4}\right)x^{-\frac{8}{3}}(16x) \;=\;-\frac{1}{3}\cdot\frac{1}{2^4}\cdot x^{\frac{8}{3}}\cdot 16 \cdot x

    . . = \;-\frac{1}{3}\cdot\frac{1}{16}\cdot 16 \cdot x^{-\frac{8}{3}} \cdot x \;=\;-\frac{1}{3}x^{-\frac{5}{3}}

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