## anonymous 3 years ago find the derivative of x(x-1)^2

1. anonymous

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

2. anonymous

alright so $x^2-2x+1$

3. anonymous

Why not to use product rule?

4. anonymous

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

5. anonymous

@SheldonEinstein because expanding it will be a lot simpler

6. anonymous

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

7. anonymous

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

8. anonymous

then $x^3-2x^2+x$ so f'(x) = $3x^2-4x+1$

9. anonymous

yes

10. anonymous

can you show product rule with that quickly please?

11. anonymous

Though, I would like to show how product rule will work in this way : $\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}$

12. anonymous

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

13. anonymous

Now? @sjerman1 can you do it?

14. anonymous

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

15. 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

16. anonymous

Wait I am typing :(

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

Amazing job! Thank you very much!

21. anonymous

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

22. anonymous

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. 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) ...

24. anonymous

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

25. anonymous

Yes @sjerman1

26. anonymous

27. anonymous

28. anonymous

Thank you both!

29. anonymous

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

30. anonymous

Here the message says all.

31. 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.