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

sjerman1 Group Title

find the derivative of x(x-1)^2

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    my advice for you is..don't use product rule. expand (x-1)^2 first

    • 2 years ago
  2. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    alright so \[x^2-2x+1\]

    • 2 years ago
  3. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Why not to use product rule?

    • 2 years ago
  4. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    right. don't forget the x x(x^2 - 2x + 1) now distribute x into the trinomial

    • 2 years ago
  5. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @SheldonEinstein because expanding it will be a lot simpler

    • 2 years ago
  6. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    if you expand it, you just get a polynomial (then you can derive the terms individually)

    • 2 years ago
  7. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh! Right, I thought you were referring that "Product rule will not work here" . It's my bad, sorry :)

    • 2 years ago
  8. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    then \[x^3-2x^2+x\] so f'(x) = \[3x^2-4x+1\]

    • 2 years ago
  9. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    yes

    • 2 years ago
  10. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    can you show product rule with that quickly please?

    • 2 years ago
  11. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Though, I would like to show how product rule will work in this way : \[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]

    • 2 years ago
  12. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    put u = x and v = (x-1)^2

    • 2 years ago
  13. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Now? @sjerman1 can you do it?

    • 2 years ago
  14. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    are you required to solve it by product rule or are you just interested @sjerman1 ?

    • 2 years ago
  15. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i am just interested, that was a much easier way to solve it but I am actually in the process of finding points guaranteed to exist by rolle's theorem

    • 2 years ago
  16. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait I am typing :(

    • 2 years ago
  17. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i have no idea what that theorem is.....but let's see what @SheldonEinstein is typing

    • 2 years ago
  18. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large{x \frac{d (x-1)^2}{dx} + (x-1)^2 \frac{dx}{dx} }\] \[\large{x \frac{d(x^2+1-2x)}{dx} + (x-1)^2}\] \[\large{x [ \frac{d(x^2)}{dx} + \frac{d(1)}{d(x)} - \frac{d(2x)}{dx} ] +(x-1)^2}\] \[\large{x [ 2x + 0 - 2 ] + x^2 +1 - 2x}\] \[\large{ 2x^2 - 2x + x^2 + 1 - 2x}\] \[\large{3x^2-4x+1}\]

    • 2 years ago
  19. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry for late answer :( I was checking it :)

    • 2 years ago
  20. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Amazing job! Thank you very much!

    • 2 years ago
  21. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You're welcome @sjerman1 ... Thanks for the patience friends :) Also , good job @lgbasallote

    • 2 years ago
  22. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    you can also do it without expanding (x-1)^2 in the derivative x(x-1)^2 x[2(x-1)] + (x-1)^2 x[2x -2] + x^2 - 2x + 1 2x^2 - 2x + x^2 - 2x + 1 3x^2 - 4x + 1 just showing an alternative

    • 2 years ago
  23. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Right ..... but usually I prefer to just use d(x^n) / dx and let the students know that x can be anything. like x can be (x-1) and similarly it can be (x+1) ...

    • 2 years ago
  24. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i wonder, is there anyway to view these questions after they are closed?

    • 2 years ago
  25. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes @sjerman1

    • 2 years ago
  26. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Save the link of the question in your computer databse.

    • 2 years ago
  27. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *database. This will help you to reopen it.

    • 2 years ago
  28. sjerman1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you both!

    • 2 years ago
  29. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    or...you can just click your username on the upper right corner and then press questions asked in the profile page...

    • 2 years ago
  30. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Here the message says all.

    • 2 years ago
    1 Attachment
  31. SheldonEinstein Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    There in the message has "reopen" word but that means like you can not get in "OPEN QUESTIONS SECTION" again but still the other users can help you if they regularly see closed quest. section.

    • 2 years 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.