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

Calculus1
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my advice for you is..don't use product rule. expand (x-1)^2 first
alright so \[x^2-2x+1\]
Why not to use product rule?

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Other answers:

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