Question

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

Answer 1

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

Answer 2

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

Answer 3

Why not to use product rule?

Answer 4

right. don't forget the x

x(x^2 - 2x + 1)

now distribute x into the trinomial

Answer 5

@SheldonEinstein because expanding it will be a lot simpler

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

yes

Answer 10

can you show product rule with that quickly please?

Answer 11

Though, I would like to show how product rule will work in this way :

\[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]

Answer 12

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

Answer 13

Now? @sjerman1 can you do it?

Answer 14

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

Answer 15

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

Answer 16

Wait I am typing :(

Answer 17

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

Answer 18

\[\large{x \frac{d (x-1)^2}{dx} + (x-1)^2 \frac{dx}{dx} }\]
\[\large{x \frac{d(x^2+1-2x)}{dx} + (x-1)^2}\]
\[\large{x [ \frac{d(x^2)}{dx} + \frac{d(1)}{d(x)} - \frac{d(2x)}{dx} ] +(x-1)^2}\]
\[\large{x [ 2x + 0 - 2 ] + x^2 +1 - 2x}\]
\[\large{ 2x^2 - 2x + x^2 + 1 - 2x}\]
\[\large{3x^2-4x+1}\]

Answer 19

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

Answer 20

Amazing job!
Thank you very much!

Answer 21

You're welcome @sjerman1 ...

Thanks for the patience friends :) Also , good job @lgbasallote

Answer 22

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

Answer 23

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) ...

Answer 24

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

Answer 25

Yes @sjerman1

Answer 26

Save the link of the question in your computer databse.

Answer 27

*database. This will help you to reopen it.

Answer 28

Thank you both!

Answer 29

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

Answer 30

Here the message says all.

Answer 31

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.