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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- katieb

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- lgbasallote

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

- anonymous

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

- anonymous

Why not to use product rule?

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## More answers

- lgbasallote

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

- lgbasallote

@SheldonEinstein because expanding it will be a lot simpler

- lgbasallote

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

- anonymous

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

- anonymous

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

- lgbasallote

yes

- anonymous

can you show product rule with that quickly please?

- 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

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

- anonymous

Now? @sjerman1 can you do it?

- lgbasallote

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

- 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

Wait I am typing :(

- lgbasallote

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

- 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

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

- anonymous

Amazing job! Thank you very much!

- anonymous

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

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

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

- anonymous

Yes @sjerman1

- anonymous

Save the link of the question in your computer databse.

- anonymous

*database. This will help you to reopen it.

- anonymous

Thank you both!

- lgbasallote

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

- anonymous

Here the message says all.

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- 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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