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bmelyk

  • 2 years ago

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

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

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

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

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

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

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

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

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

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

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

  11. cwrw238
    • 2 years ago
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    yes

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

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

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

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

  16. bmelyk
    • 2 years ago
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    okay.

  17. bmelyk
    • 2 years ago
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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.

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

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

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

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

  22. bmelyk
    • 2 years ago
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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.

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

  24. satellite73
    • 2 years ago
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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

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

  26. nphuongsun93
    • 2 years ago
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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

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

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

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

  30. bmelyk
    • 2 years ago
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    no idea

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

  32. bmelyk
    • 2 years ago
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    ???

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

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

  35. nphuongsun93
    • 2 years ago
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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 \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]

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

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