## sjerman1 Group Title find the derivative of x(x-1)^2 one year ago one year ago

1. lgbasallote Group Title

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

2. sjerman1 Group Title

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

3. SheldonEinstein Group Title

Why not to use product rule?

4. lgbasallote Group Title

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

5. lgbasallote Group Title

@SheldonEinstein because expanding it will be a lot simpler

6. lgbasallote Group Title

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

7. SheldonEinstein Group Title

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

8. sjerman1 Group Title

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

9. lgbasallote Group Title

yes

10. sjerman1 Group Title

can you show product rule with that quickly please?

11. SheldonEinstein Group Title

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. SheldonEinstein Group Title

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

13. SheldonEinstein Group Title

Now? @sjerman1 can you do it?

14. lgbasallote Group Title

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

15. sjerman1 Group Title

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. SheldonEinstein Group Title

Wait I am typing :(

17. lgbasallote Group Title

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

18. SheldonEinstein Group Title

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

19. SheldonEinstein Group Title

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

20. sjerman1 Group Title

Amazing job! Thank you very much!

21. SheldonEinstein Group Title

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

22. lgbasallote Group Title

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. SheldonEinstein Group Title

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. sjerman1 Group Title

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

25. SheldonEinstein Group Title

Yes @sjerman1

26. SheldonEinstein Group Title

27. SheldonEinstein Group Title

28. sjerman1 Group Title

Thank you both!

29. lgbasallote Group Title

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

30. SheldonEinstein Group Title

Here the message says all.

31. SheldonEinstein Group Title

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.