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sjerman1

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

  • one year ago
  • one year ago

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

    • one year ago
  2. sjerman1
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    alright so \[x^2-2x+1\]

    • one year ago
  3. SheldonEinstein
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    Why not to use product rule?

    • one year ago
  4. lgbasallote
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    right. don't forget the x x(x^2 - 2x + 1) now distribute x into the trinomial

    • one year ago
  5. lgbasallote
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    @SheldonEinstein because expanding it will be a lot simpler

    • one year ago
  6. lgbasallote
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    if you expand it, you just get a polynomial (then you can derive the terms individually)

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

    • one year ago
  8. sjerman1
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    then \[x^3-2x^2+x\] so f'(x) = \[3x^2-4x+1\]

    • one year ago
  9. lgbasallote
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    yes

    • one year ago
  10. sjerman1
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    can you show product rule with that quickly please?

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

    • one year ago
  12. SheldonEinstein
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    put u = x and v = (x-1)^2

    • one year ago
  13. SheldonEinstein
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    Now? @sjerman1 can you do it?

    • one year ago
  14. lgbasallote
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    are you required to solve it by product rule or are you just interested @sjerman1 ?

    • one year ago
  15. sjerman1
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    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

    • one year ago
  16. SheldonEinstein
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    Wait I am typing :(

    • one year ago
  17. lgbasallote
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    i have no idea what that theorem is.....but let's see what @SheldonEinstein is typing

    • one year ago
  18. SheldonEinstein
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    \[\large{x \frac{d (x-1)^2}{dx} + (x-1)^2 \frac{dx}{dx} }\] \[\large{x \frac{d(x^2+1-2x)}{dx} + (x-1)^2}\] \[\large{x [ \frac{d(x^2)}{dx} + \frac{d(1)}{d(x)} - \frac{d(2x)}{dx} ] +(x-1)^2}\] \[\large{x [ 2x + 0 - 2 ] + x^2 +1 - 2x}\] \[\large{ 2x^2 - 2x + x^2 + 1 - 2x}\] \[\large{3x^2-4x+1}\]

    • one year ago
  19. SheldonEinstein
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    Sorry for late answer :( I was checking it :)

    • one year ago
  20. sjerman1
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    Amazing job! Thank you very much!

    • one year ago
  21. SheldonEinstein
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    You're welcome @sjerman1 ... Thanks for the patience friends :) Also , good job @lgbasallote

    • one year ago
  22. lgbasallote
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    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

    • one year ago
  23. SheldonEinstein
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    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) ...

    • one year ago
  24. sjerman1
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    i wonder, is there anyway to view these questions after they are closed?

    • one year ago
  25. SheldonEinstein
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    Yes @sjerman1

    • one year ago
  26. SheldonEinstein
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    Save the link of the question in your computer databse.

    • one year ago
  27. SheldonEinstein
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    *database. This will help you to reopen it.

    • one year ago
  28. sjerman1
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    Thank you both!

    • one year ago
  29. lgbasallote
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    or...you can just click your username on the upper right corner and then press questions asked in the profile page...

    • one year ago
  30. SheldonEinstein
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    Here the message says all.

    • one year ago
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  31. SheldonEinstein
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    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.

    • one year ago
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