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DLS

y=sinx^cosx+cosx^sinx dy/dx=?

  • one year ago
  • one year ago

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  1. DLS
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    My attempt:

    • one year ago
  2. DLS
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    \[\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (-sinx) + (cosx)(-cosec^{2}x) + \frac{sinx}{cosx} \] \[\LARGE \times (cosx)+(sinx)(\sec^{2}x)\]

    • one year ago
  3. DLS
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    last term hidden is sinx/cos then continued below

    • one year ago
  4. hartnn
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    what was that ?!! y= u+v find du/dx and dv/dx using logarithmic differentiation then add...

    • one year ago
  5. DLS
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    lol

    • one year ago
  6. DLS
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    I used chain rule for logsinx then UV rule with (cosx)logsinx

    • one year ago
  7. DLS
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    \[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]

    • one year ago
  8. hartnn
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    log (A+B) cannot be simplified to log A +log B

    • one year ago
  9. DLS
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    where is log(A+B) :O

    • one year ago
  10. DLS
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    oh

    • one year ago
  11. DLS
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    NEVERMIND

    • one year ago
  12. hartnn
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    y= u+v find du/dx and dv/dx using logarithmic differentiation then add... thats the only way...

    • one year ago
  13. abb0t
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    Use chain rule.

    • one year ago
  14. DLS
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    \[\LARGE logy=\log(cosx^{cosx} \times -sinx + (-sinx^{sinx} \times cosx)\]

    • one year ago
  15. DLS
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    ?

    • one year ago
  16. hartnn
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    u =sin x^cos x du/dx =... ?

    • one year ago
  17. DLS
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    again log? O_O

    • one year ago
  18. DLS
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    okay no

    • one year ago
  19. hartnn
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    again ? this is the 1st time, you use log...

    • one year ago
  20. DLS
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    :/

    • one year ago
  21. DLS
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    cosxlogsinx?

    • one year ago
  22. DLS
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    CONFUSED

    • one year ago
  23. hartnn
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    log u = cos x log sin x now diff. use product rule.

    • one year ago
  24. DLS
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    thats what I did.. I got: cotx cosx

    • one year ago
  25. hartnn
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    did u forget product rule ?

    • one year ago
  26. DLS
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    :/ no :///

    • one year ago
  27. hartnn
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    (uv)' = uv'+u'v

    • one year ago
  28. DLS
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    \[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (-sinx)\]

    • one year ago
  29. DLS
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    isnt it :/..

    • one year ago
  30. hartnn
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    1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

    • one year ago
  31. DLS
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    I guess I studied too much :P

    • one year ago
  32. DLS
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    thats what I did >.<

    • one year ago
  33. hartnn
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    extra cos x come due to chain rule is it ?

    • one year ago
  34. DLS
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    cosxlogsinx =>cosx(1/sinx * cosx)(chain rule ) cosxcotx now product rule whats wrong with it :/

    • one year ago
  35. hartnn
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    because there was no 2nd term in there....thats why i asked u about product rule...

    • one year ago
  36. DLS
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    stil confused :|

    • one year ago
  37. hartnn
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    ok, just tell me what u get finallu for du/dx .. ?

    • one year ago
  38. hartnn
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    u get this, right ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

    • one year ago
  39. DLS
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    let me write

    • one year ago
  40. hartnn
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    don't write in latex, takes time unnecessarily...

    • one year ago
  41. DLS
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    yes,im solving last time :o only du/dx

    • one year ago
  42. DLS
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    i get

    • one year ago
  43. hartnn
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    yes, only du/dx....dv/dx will be very similar....

    • one year ago
  44. DLS
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    -cosx cosec^2 x -sinxcotx :///////

    • one year ago
  45. DLS
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    on solving cosxcotx by product rule thats it

    • one year ago
  46. hartnn
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    :( didn't u get this ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

    • one year ago
  47. DLS
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    I dont think so loll :/

    • one year ago
  48. hartnn
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    log u = cos x log sin x this ?

    • one year ago
  49. DLS
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    yes

    • one year ago
  50. DLS
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    cos x (1/sinx * cosx)

    • one year ago
  51. DLS
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    log dissapeared

    • one year ago
  52. hartnn
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    (log u)' = (cos x log sin x )' 1/u u' = cos x (log sin x)' + log sin x (cos x)' did u get this step ?

    • one year ago
  53. hartnn
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    just simple product rule.

    • one year ago
  54. DLS
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    okay

    • one year ago
  55. hartnn
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    want to continue ? or should i ?

    • one year ago
  56. DLS
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    i will

    • one year ago
  57. DLS
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    logsinx(-sinx)+cosx(logsinx) okay?

    • one year ago
  58. DLS
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    DUH

    • one year ago
  59. hartnn
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    my next step 1/u u' = cos x (log sin x)' + log sin x (cos x)' 1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x) get that ? ^

    • one year ago
  60. hartnn
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    use of chain rule.

    • one year ago
  61. DLS
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    1minute and its over!

    • one year ago
  62. DLS
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    give me

    • one year ago
  63. DLS
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    lol been 30 minutes:/

    • one year ago
  64. hartnn
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    let me do it...

    • one year ago
  65. hartnn
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    u just see.

    • one year ago
  66. DLS
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    ivedone

    • one year ago
  67. DLS
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    \[\LARGE logsinx(-sinx)+cosx(\frac{1}{sinx} \times cosx)\] RIGHT?

    • one year ago
  68. hartnn
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    1/u u' = cos x (log sin x)' + log sin x (cos x)' 1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x) 1/u u' = cos x (1/ sin x)(cos x) - sin x log sin x du/dx = u [cos x . cot x - sin x log sin x] finally du/dx = [sin x^cos x][cos x . cot x - sin x log sin x] any doubts in any step ?

    • one year ago
  69. DLS
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    \[FINAL:logsinx(-sinx)+cosxcot\]

    • one year ago
  70. DLS
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    x

    • one year ago
  71. DLS
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    this is what u get out of cosx log sinx right? ^ and im doing it myself...ill check from urs,this is very confusingnow just tell me that^ though i checked with WF

    • one year ago
  72. hartnn
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    du/dx = [sin x^cos x][cos x . cot x - sin x log sin x] in same manner find dv/dx.

    • one year ago
  73. hartnn
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    and to get final answer, just add them

    • one year ago
  74. DLS
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    YES! correct!

    • one year ago
  75. DLS
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    that was one hell of a question -_-'

    • one year ago
  76. hartnn
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    but easy.

    • one year ago
  77. DLS
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    for u!

    • one year ago
  78. hartnn
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    sum of 2 eaSY Q's

    • one year ago
  79. DLS
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    it was new for me..so yeah..next time easy for me too...(hopefully)

    • one year ago
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