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anonymous

  • one year ago

let f(x)=cubed root of x If a does not =0 find f'(a) using the definition of a derivative.

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    definition, not the power rule right?

  3. anonymous
    • one year ago
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    yeah, I am having some trouble because of the cubed root

  4. misty1212
    • one year ago
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    i bet i can show you the gimmick did you do it with the square root ever?

  5. anonymous
    • one year ago
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    yeah I have

  6. misty1212
    • one year ago
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    ok the idea with the square root is to multiply by the conjugate, because \[a^2-b^2=(a+b)(a-b)\] so the radical goes away

  7. misty1212
    • one year ago
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    but for the cubed root you have to use the difference of two cubes, not the difference of two squares that will get rid of the cubed root i.e. \[a^3-b^3=(a-b)( a^2+ab+b^2)\]

  8. anonymous
    • one year ago
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    ohhh I was doing it as though it was a squared .-.

  9. anonymous
    • one year ago
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    okay so how exactly will that look for this problem

  10. misty1212
    • one year ago
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    \[\frac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}\] multiply top and bottom by \[\sqrt[3]{(a+h)^2}+\sqrt[3]{a}\sqrt[3]{a+h}+\sqrt[3]{a^2}\]

  11. misty1212
    • one year ago
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    don't really multiply it out the numerator will be \[a+h-a=h\]

  12. misty1212
    • one year ago
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    leave the denominator in factored form cancel the \(h\) top and bottom and then replace \(h\) by \(0\)

  13. misty1212
    • one year ago
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    you are using this \[(a-b)(a^2+ab+b^2)=a^3-b^3\]with \[a=\sqrt[3]{a+h},b=\sqrt[3]{a}\]

  14. anonymous
    • one year ago
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    I am still a a but confused..

  15. anonymous
    • one year ago
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    *a bit

  16. anonymous
    • one year ago
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    @misty1212

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