anonymous one year ago find the derivative and second derivative of (x^2-9)^2

1. amoodarya

$y=(x^2-9)^2=\\x^4-18x^2+81\\$it is easier now to find f',f''

2. anonymous

Wouldn't it just be a chain rule from step 1?

3. amoodarya

|dw:1437519674083:dw|

4. amoodarya

|dw:1437519721063:dw|

5. anonymous

That's too confusing lol could you just do the chain rule?

6. triciaal

|dw:1437519701313:dw|

7. anonymous

8. amoodarya

that was chain rule

9. anonymous

Sorry amood it was just your hand writing haha no offense

10. amoodarya

what 's f'= ? simplify and writ it here plz

11. anonymous

$4x(x^2-9)$

12. triciaal

|dw:1437519790039:dw|

13. amoodarya

distribute 4x in (x^2-9 ) then take y''

14. anonymous

Done, thank you sir Triciaal.

15. anonymous

Thanks Amood

16. triciaal

@MBilbeisi either way you will get the same result

17. triciaal

I am not a Sir

18. anonymous

@Triciaal are you good at extreme values?

19. anonymous

20. triciaal

ok

21. anonymous

You are?

22. triciaal

maximum and minimum

23. anonymous

Yes!!

24. anonymous

It's a really hard question i can't seem to get it right

25. anonymous

Maybe not so hard for you

26. triciaal

when 2nd derivative at critical point is negative you have max when positive then you have minimum

27. triciaal

the 2nd derivative is showing the change in the slope the slope is the 1st derivative

28. anonymous

So if the question says Find the extreme values of the function and where they occur, i find f'(x) and f''(x)?

29. triciaal

out of practice but I had learnt it well enough

30. anonymous

$\left(\begin{matrix}1 \\ x^2-1\end{matrix}\right)$

31. triciaal

find the critical points use in 2nd derivative

32. triciaal

sorry have to leave but post question will check back later if still need help

33. anonymous

Okay i got you, so for that function i do f', by quotient rule, and then critical values, then f''?

34. anonymous

okay i will, thanks

35. triciaal

x not = 1 or -1 for this because do not want to divide by 0