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anonymous

  • one year ago

find the derivative and second derivative of (x^2-9)^2

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  1. amoodarya
    • one year ago
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    \[y=(x^2-9)^2=\\x^4-18x^2+81\\\]it is easier now to find f',f''

  2. anonymous
    • one year ago
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    Wouldn't it just be a chain rule from step 1?

  3. amoodarya
    • one year ago
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    |dw:1437519674083:dw|

  4. amoodarya
    • one year ago
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    |dw:1437519721063:dw|

  5. anonymous
    • one year ago
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    That's too confusing lol could you just do the chain rule?

  6. triciaal
    • one year ago
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    |dw:1437519701313:dw|

  7. anonymous
    • one year ago
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    Thank you. What about f''(x)?

  8. amoodarya
    • one year ago
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    that was chain rule

  9. anonymous
    • one year ago
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    Sorry amood it was just your hand writing haha no offense

  10. amoodarya
    • one year ago
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    what 's f'= ? simplify and writ it here plz

  11. anonymous
    • one year ago
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    \[4x(x^2-9)\]

  12. triciaal
    • one year ago
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    |dw:1437519790039:dw|

  13. amoodarya
    • one year ago
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    distribute 4x in (x^2-9 ) then take y''

  14. anonymous
    • one year ago
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    Done, thank you sir Triciaal.

  15. anonymous
    • one year ago
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    Thanks Amood

  16. triciaal
    • one year ago
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    @MBilbeisi either way you will get the same result

  17. triciaal
    • one year ago
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    I am not a Sir

  18. anonymous
    • one year ago
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    @Triciaal are you good at extreme values?

  19. anonymous
    • one year ago
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    Sorry, madam.

  20. triciaal
    • one year ago
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    ok

  21. anonymous
    • one year ago
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    You are?

  22. triciaal
    • one year ago
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    maximum and minimum

  23. anonymous
    • one year ago
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    Yes!!

  24. anonymous
    • one year ago
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    It's a really hard question i can't seem to get it right

  25. anonymous
    • one year ago
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    Maybe not so hard for you

  26. triciaal
    • one year ago
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    when 2nd derivative at critical point is negative you have max when positive then you have minimum

  27. triciaal
    • one year ago
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    the 2nd derivative is showing the change in the slope the slope is the 1st derivative

  28. anonymous
    • one year ago
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    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
    • one year ago
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    out of practice but I had learnt it well enough

  30. anonymous
    • one year ago
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    \[\left(\begin{matrix}1 \\ x^2-1\end{matrix}\right)\]

  31. triciaal
    • one year ago
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    find the critical points use in 2nd derivative

  32. triciaal
    • one year ago
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    sorry have to leave but post question will check back later if still need help

  33. anonymous
    • one year ago
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    Okay i got you, so for that function i do f', by quotient rule, and then critical values, then f''?

  34. anonymous
    • one year ago
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    okay i will, thanks

  35. triciaal
    • one year ago
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    x not = 1 or -1 for this because do not want to divide by 0

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