Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bmelyk Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  2. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    just not sure how to simplify from here.

    • one year ago
  3. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    sqrtx * 6x^2 + (1/2) x^(-1/2) * (2x^3 - 1)

    • one year ago
  4. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    √x(6x^2)+(2x^3−1)+(x^2) is what i got.

    • one year ago
  5. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    derivative of sqrtx or x^(1/2) is (1/2) x^(-1/2)

    • one year ago
  6. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    the deriviate of \[\sqrt{x} = x ^{2}\]

    • one year ago
  7. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i think? lol, that's what my prof wrote down.

    • one year ago
  8. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    the rule is if f(x) = ax^n , f'(x) = an x^(n-1)

    • one year ago
  9. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    if n = 1/2, n-1 = -1/2

    • one year ago
  10. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so it's \[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] ??

    • one year ago
  11. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

    • one year ago
  12. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay im still not sure how to further simplfy from there.

    • one year ago
  13. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, i don't even need the deriviate of the square root of x ..

    • one year ago
  14. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    nevermind i do lol anyways,

    • one year ago
  15. cwrw238 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    which can be written as 1 / 2x^(1/2) sorry can't help anymore - gotta go out

    • one year ago
  16. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay.

    • one year ago
  17. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so im left with \[x ^{\frac{ 1 }{ 2 }}(6x ^{2})+(2x-1)(\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\] i need help further simplifying.

    • one year ago
  18. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\huge 6x^2\sqrt x + \frac{2x^3-1}{2\sqrt x}\]add them ~

    • one year ago
  19. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    why 2 root x?

    • one year ago
  20. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    because\[\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\]

    • one year ago
  21. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  22. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i got all that, im just not sure how to further simply once im at the last step with all deriviates completed.

    • one year ago
  23. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i have everything over 2 root x

    • one year ago
  24. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  25. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  26. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\huge 6x^2 \sqrt x + \frac{2x^3-1}{2 \sqrt x}\] you got to here, now just make the denominators common and add them

    • one year ago
  27. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i have \[(\sqrt{x})(6x ^{2})(2\sqrt{x})+2x ^{3}-1 all \over 2\sqrt{x}\]

    • one year ago
  28. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i cant get any further than that.

    • one year ago
  29. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\huge 6x^2 \sqrt x \times 2 \sqrt x = ?\]

    • one year ago
  30. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    no idea

    • one year ago
  31. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    12x^2 root x ?

    • one year ago
  32. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ???

    • one year ago
  33. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\huge \sqrt x \times \sqrt x = x\]\[\huge 6x^2 \sqrt x \times 2 \sqrt x = 12 x ^3\]

    • one year ago
  34. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    why x cubed and not x squared?

    • one year ago
  35. nphuongsun93 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    because~\[\huge 6x^2 \times 2 = 12x^2\]\[\huge \sqrt x \times \sqrt x = x\]\[\huge 12x^2 \times x = 12 x^3\] then \[\huge \frac{12x^3 + 2x^3 + 1}{2 \sqrt x}=\frac{ 14x^3 + 1}{2\sqrt x}\]

    • one year ago
  36. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thank-you! then i can just rationalize the denominator and ill be done.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.