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

1. lgbasallote

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

2. sjerman1

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

3. SheldonEinstein

Why not to use product rule?

4. lgbasallote

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

5. lgbasallote

@SheldonEinstein because expanding it will be a lot simpler

6. lgbasallote

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

7. SheldonEinstein

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

8. sjerman1

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

9. lgbasallote

yes

10. sjerman1

can you show product rule with that quickly please?

11. SheldonEinstein

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

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

13. SheldonEinstein

Now? @sjerman1 can you do it?

14. lgbasallote

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

15. 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

16. SheldonEinstein

Wait I am typing :(

17. lgbasallote

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

18. SheldonEinstein

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

19. SheldonEinstein

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

20. sjerman1

Amazing job! Thank you very much!

21. SheldonEinstein

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

22. 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

23. SheldonEinstein

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

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

25. SheldonEinstein

Yes @sjerman1

26. SheldonEinstein

27. SheldonEinstein

28. sjerman1

Thank you both!

29. lgbasallote

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

30. SheldonEinstein

Here the message says all.

31. SheldonEinstein

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.