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bmelyk Group Title

\[\sqrt{x}(2x ^{3}-1)\] Find the derivative using the product rule.

  • 2 years ago
  • 2 years ago

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  1. bmelyk Group Title
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    i have gotten down to: \[\sqrt{x}(6x ^{2})+(2x ^{3}-1)+(x ^{2})\]

    • 2 years ago
  2. bmelyk Group Title
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    just not sure how to simplify from here.

    • 2 years ago
  3. cwrw238 Group Title
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    sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

    • 2 years ago
  4. bmelyk Group Title
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    √x(6x^2)+(2x^3−1)+(x^2) is what i got.

    • 2 years ago
  5. cwrw238 Group Title
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    derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)

    • 2 years ago
  6. bmelyk Group Title
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    the deriviate of \[\sqrt{x} = x ^{2}\]

    • 2 years ago
  7. bmelyk Group Title
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    i think? lol, that's what my prof wrote down.

    • 2 years ago
  8. cwrw238 Group Title
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    the rule is if f(x) = ax^n , f'(x) = an x^(n-1)

    • 2 years ago
  9. cwrw238 Group Title
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    if n = 1/2, n-1 = -1/2

    • 2 years ago
  10. bmelyk Group Title
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    so it's \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] ??

    • 2 years ago
  11. cwrw238 Group Title
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    yes

    • 2 years ago
  12. bmelyk Group Title
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    okay im still not sure how to further simplfy from there.

    • 2 years ago
  13. bmelyk Group Title
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    wait, i don't even need the deriviate of the square root of x ..

    • 2 years ago
  14. bmelyk Group Title
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    nevermind i do lol anyways,

    • 2 years ago
  15. cwrw238 Group Title
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    which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out

    • 2 years ago
  16. bmelyk Group Title
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    okay.

    • 2 years ago
  17. bmelyk Group Title
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    so im left with \[x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x-1)(\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] i need help further simplifying.

    • 2 years ago
  18. nphuongsun93 Group Title
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    \[\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}\]add them ~

    • 2 years ago
  19. bmelyk Group Title
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    why 2 root x?

    • 2 years ago
  20. nphuongsun93 Group Title
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    because\[\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\]

    • 2 years ago
  21. satellite73 Group Title
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    \[(fg)'=f'g+g'f\] with \[f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},g(x)=2x^3-1,g'(x)=6x^2\]

    • 2 years ago
  22. bmelyk Group Title
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    yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.

    • 2 years ago
  23. bmelyk Group Title
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    i have everything over 2 root x

    • 2 years ago
  24. satellite73 Group Title
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    plug and play (algebra) not to butt in, but \(\sqrt{x}\) is a very very common function, plus math teachers like to put it on tests and quizzes, so rather than use the power rule each time it would help your cause out to memorize \[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] like you know \(8\times 9=72\) for example it will save you time on exams if nothing else

    • 2 years ago
  25. bmelyk Group Title
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    i know all that but what im having trouble with is simplyfying \[\sqrt{x}(6x ^{2})+2x ^{3}-1 /2\sqrt{x}\]

    • 2 years ago
  26. nphuongsun93 Group Title
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    \[\huge 6x^2 \sqrt x + \frac{2x^3-1}{2 \sqrt x}\] you got to here, now just make the denominators common and add them

    • 2 years ago
  27. bmelyk Group Title
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    i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}\]

    • 2 years ago
  28. bmelyk Group Title
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    i cant get any further than that.

    • 2 years ago
  29. nphuongsun93 Group Title
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    \[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]

    • 2 years ago
  30. bmelyk Group Title
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    no idea

    • 2 years ago
  31. bmelyk Group Title
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    12x^2 root x ?

    • 2 years ago
  32. bmelyk Group Title
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    ???

    • 2 years ago
  33. nphuongsun93 Group Title
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    \[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]

    • 2 years ago
  34. bmelyk Group Title
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    why x cubed and not x squared?

    • 2 years ago
  35. nphuongsun93 Group Title
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    because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\] then \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]

    • 2 years ago
  36. bmelyk Group Title
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    thank-you! then i can just rationalize the denominator and ill be done.

    • 2 years ago
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