anonymous
  • anonymous
find the derivative of x(x-1)^2
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lgbasallote
  • lgbasallote
my advice for you is..don't use product rule. expand (x-1)^2 first
anonymous
  • anonymous
alright so \[x^2-2x+1\]
anonymous
  • anonymous
Why not to use product rule?

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lgbasallote
  • lgbasallote
right. don't forget the x x(x^2 - 2x + 1) now distribute x into the trinomial
lgbasallote
  • lgbasallote
@SheldonEinstein because expanding it will be a lot simpler
lgbasallote
  • lgbasallote
if you expand it, you just get a polynomial (then you can derive the terms individually)
anonymous
  • anonymous
Oh! Right, I thought you were referring that "Product rule will not work here" . It's my bad, sorry :)
anonymous
  • anonymous
then \[x^3-2x^2+x\] so f'(x) = \[3x^2-4x+1\]
lgbasallote
  • lgbasallote
yes
anonymous
  • anonymous
can you show product rule with that quickly please?
anonymous
  • anonymous
Though, I would like to show how product rule will work in this way : \[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]
anonymous
  • anonymous
put u = x and v = (x-1)^2
anonymous
  • anonymous
Now? @sjerman1 can you do it?
lgbasallote
  • lgbasallote
are you required to solve it by product rule or are you just interested @sjerman1 ?
anonymous
  • anonymous
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
anonymous
  • anonymous
Wait I am typing :(
lgbasallote
  • lgbasallote
i have no idea what that theorem is.....but let's see what @SheldonEinstein is typing
anonymous
  • anonymous
\[\large{x \frac{d (x-1)^2}{dx} + (x-1)^2 \frac{dx}{dx} }\] \[\large{x \frac{d(x^2+1-2x)}{dx} + (x-1)^2}\] \[\large{x [ \frac{d(x^2)}{dx} + \frac{d(1)}{d(x)} - \frac{d(2x)}{dx} ] +(x-1)^2}\] \[\large{x [ 2x + 0 - 2 ] + x^2 +1 - 2x}\] \[\large{ 2x^2 - 2x + x^2 + 1 - 2x}\] \[\large{3x^2-4x+1}\]
anonymous
  • anonymous
Sorry for late answer :( I was checking it :)
anonymous
  • anonymous
Amazing job! Thank you very much!
anonymous
  • anonymous
You're welcome @sjerman1 ... Thanks for the patience friends :) Also , good job @lgbasallote
lgbasallote
  • lgbasallote
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
anonymous
  • anonymous
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) ...
anonymous
  • anonymous
i wonder, is there anyway to view these questions after they are closed?
anonymous
  • anonymous
Yes @sjerman1
anonymous
  • anonymous
Save the link of the question in your computer databse.
anonymous
  • anonymous
*database. This will help you to reopen it.
anonymous
  • anonymous
Thank you both!
lgbasallote
  • lgbasallote
or...you can just click your username on the upper right corner and then press questions asked in the profile page...
anonymous
  • anonymous
Here the message says all.
1 Attachment
anonymous
  • anonymous
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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