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onegirl Group Title

Find the indicated derivative of the function f(x) = cosx^2, find f''(x)

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    \[2\cos(x)\sin(x)\]is your first derivative take the derivative of that one, using the product rule

    • one year ago
  2. onegirl Group Title
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    @satellite73 it will be -2cos(x)sin(x)

    • one year ago
  3. onegirl Group Title
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    @genius12 so the derivative of -2cos(x) sin(x) , and i got -2 cos(2x)

    • one year ago
  4. onegirl Group Title
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    @satellite73 ?

    • one year ago
  5. onegirl Group Title
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    @jazy should i now take the derivative of -2 cos(2x) ? or

    • one year ago
  6. RadEn Group Title
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    which one : |dw:1362427439289:dw|

    • one year ago
  7. onegirl Group Title
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    cosx^2

    • one year ago
  8. RadEn Group Title
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    so, the first one in my diagram ?

    • one year ago
  9. onegirl Group Title
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    yes

    • one year ago
  10. RadEn Group Title
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    oke, to get the first derivative of f(x) we can use the cain rule hint : derivative of x^2 = 2x derivative of cos = - sin

    • one year ago
  11. onegirl Group Title
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    yea i got it the first derivative will be -2 cos(x) sin (x) right?

    • one year ago
  12. RadEn Group Title
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    so, derivative of f(x) = cosx^2 ---> f ' (x) = 2x(-sinx^2) = -2x sinx^2

    • one year ago
  13. onegirl Group Title
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    yes

    • one year ago
  14. onegirl Group Title
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    so then i take the derivative of -2x sinx^2 ?

    • one year ago
  15. RadEn Group Title
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    yes

    • one year ago
  16. onegirl Group Title
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    okay after i take the derivative of -2x sinx i get -2 cos(2x) so then i take the derivative of -2 cos(2x) right?

    • one year ago
  17. RadEn Group Title
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    use the product rule to get the 2nd derivative : use the formula : f '' = u'v + uv' assuming that u=-2x ---> u' = ... ? v = sinx^2 ---> v ' = ... ?

    • one year ago
  18. onegirl Group Title
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    how would i do it?

    • one year ago
  19. onegirl Group Title
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    can u show it?

    • one year ago
  20. RadEn Group Title
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    do u know derivative of -2x ?

    • one year ago
  21. onegirl Group Title
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    yea

    • one year ago
  22. onegirl Group Title
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    its -2 right?

    • one year ago
  23. RadEn Group Title
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    yes, that's right also, sinx^2 ?

    • one year ago
  24. onegirl Group Title
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    okay hold on

    • one year ago
  25. onegirl Group Title
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    but where does sinx^2 come from i got -2 cos(x) sin(x) ??

    • one year ago
  26. onegirl Group Title
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    for my derivative of cosx^2

    • one year ago
  27. RadEn Group Title
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    f(x) = cosx^2 ---> f ' (x) = 2x(-sinx^2) = -2x sinx^2

    • one year ago
  28. onegirl Group Title
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    okay

    • one year ago
  29. RadEn Group Title
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    it just the first derivative of f(x), now we have to deriv again to get f " then, use the product rule

    • one year ago
  30. onegirl Group Title
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    so now i need to find the derivative of sin^x right? since we found -2x which we got -2

    • one year ago
  31. onegirl Group Title
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    and i got cos (x) sin(x)

    • one year ago
  32. onegirl Group Title
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    so it will be -2 cos(x) sin(x) right?

    • one year ago
  33. RadEn Group Title
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    -2x sinx^2 let u=-2x ---> u' = -2 v = sinx^2 ----> v ' = 2xcoscx^2

    • one year ago
  34. RadEn Group Title
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    sorry, it should v' = 2xcosx^2

    • one year ago
  35. onegirl Group Title
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    okay so how would i set it up in the formula ?

    • one year ago
  36. RadEn Group Title
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    f " = u'v + uv' f " = (-2)sinx^2 + (-2x)2xcosx^2 = -2sinx^2 - 4x^2cosx^2

    • one year ago
  37. onegirl Group Title
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    ok

    • one year ago
  38. onegirl Group Title
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    so what do i do next?

    • one year ago
  39. RadEn Group Title
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    nothing, it is clear :)

    • one year ago
  40. onegirl Group Title
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    okay so the f'''(x) will be -2sinx^2 = 4x^2 cosx^2 ?

    • one year ago
  41. onegirl Group Title
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    i meant f''(x)

    • one year ago
  42. onegirl Group Title
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    are you there?

    • one year ago
  43. onegirl Group Title
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    @RadEn so thats the final answer?

    • one year ago
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