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sjerman1

  • 2 years ago

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

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

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

  3. SheldonEinstein
    • 2 years ago
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    Why not to use product rule?

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

  5. lgbasallote
    • 2 years ago
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    @SheldonEinstein because expanding it will be a lot simpler

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

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

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

  9. lgbasallote
    • 2 years ago
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    yes

  10. sjerman1
    • 2 years ago
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    can you show product rule with that quickly please?

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

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

  13. SheldonEinstein
    • 2 years ago
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    Now? @sjerman1 can you do it?

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

  15. sjerman1
    • 2 years ago
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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

  16. SheldonEinstein
    • 2 years ago
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    Wait I am typing :(

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

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

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

  20. sjerman1
    • 2 years ago
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    Amazing job! Thank you very much!

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

  22. lgbasallote
    • 2 years ago
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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

  23. SheldonEinstein
    • 2 years ago
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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) ...

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

  25. SheldonEinstein
    • 2 years ago
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    Yes @sjerman1

  26. SheldonEinstein
    • 2 years ago
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    Save the link of the question in your computer databse.

  27. SheldonEinstein
    • 2 years ago
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    *database. This will help you to reopen it.

  28. sjerman1
    • 2 years ago
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    Thank you both!

  29. lgbasallote
    • 2 years ago
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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...

  30. SheldonEinstein
    • 2 years ago
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    Here the message says all.

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  31. SheldonEinstein
    • 2 years ago
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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.

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