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DLS

  • 3 years ago

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

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  1. DLS
    • 3 years ago
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    My attempt:

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

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

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

  5. DLS
    • 3 years ago
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    lol

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

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

  8. hartnn
    • 3 years ago
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    log (A+B) cannot be simplified to log A +log B

  9. DLS
    • 3 years ago
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    where is log(A+B) :O

  10. DLS
    • 3 years ago
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    oh

  11. DLS
    • 3 years ago
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    NEVERMIND

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

  13. abb0t
    • 3 years ago
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    Use chain rule.

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

  15. DLS
    • 3 years ago
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    ?

  16. hartnn
    • 3 years ago
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    u =sin x^cos x du/dx =... ?

  17. DLS
    • 3 years ago
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    again log? O_O

  18. DLS
    • 3 years ago
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    okay no

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

  20. DLS
    • 3 years ago
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    :/

  21. DLS
    • 3 years ago
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    cosxlogsinx?

  22. DLS
    • 3 years ago
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    CONFUSED

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

  24. DLS
    • 3 years ago
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    thats what I did.. I got: cotx cosx

  25. hartnn
    • 3 years ago
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    did u forget product rule ?

  26. DLS
    • 3 years ago
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    :/ no :///

  27. hartnn
    • 3 years ago
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    (uv)' = uv'+u'v

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

  29. DLS
    • 3 years ago
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    isnt it :/..

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

  31. DLS
    • 3 years ago
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    I guess I studied too much :P

  32. DLS
    • 3 years ago
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    thats what I did >.<

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

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

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

  36. DLS
    • 3 years ago
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    stil confused :|

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

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

  39. DLS
    • 3 years ago
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    let me write

  40. hartnn
    • 3 years ago
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    don't write in latex, takes time unnecessarily...

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

  42. DLS
    • 3 years ago
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    i get

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

  44. DLS
    • 3 years ago
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    -cosx cosec^2 x -sinxcotx :///////

  45. DLS
    • 3 years ago
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    on solving cosxcotx by product rule thats it

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

  47. DLS
    • 3 years ago
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    I dont think so loll :/

  48. hartnn
    • 3 years ago
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    log u = cos x log sin x this ?

  49. DLS
    • 3 years ago
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    yes

  50. DLS
    • 3 years ago
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    cos x (1/sinx * cosx)

  51. DLS
    • 3 years ago
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    log dissapeared

  52. hartnn
    • 3 years ago
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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 ?

  53. hartnn
    • 3 years ago
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    just simple product rule.

  54. DLS
    • 3 years ago
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    okay

  55. hartnn
    • 3 years ago
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    want to continue ? or should i ?

  56. DLS
    • 3 years ago
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    i will

  57. DLS
    • 3 years ago
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    logsinx(-sinx)+cosx(logsinx) okay?

  58. DLS
    • 3 years ago
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    DUH

  59. hartnn
    • 3 years ago
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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 ? ^

  60. hartnn
    • 3 years ago
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    use of chain rule.

  61. DLS
    • 3 years ago
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    1minute and its over!

  62. DLS
    • 3 years ago
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    give me

  63. DLS
    • 3 years ago
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    lol been 30 minutes:/

  64. hartnn
    • 3 years ago
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    let me do it...

  65. hartnn
    • 3 years ago
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    u just see.

  66. DLS
    • 3 years ago
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    ivedone

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

  68. hartnn
    • 3 years ago
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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 ?

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

  70. DLS
    • 3 years ago
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    x

  71. DLS
    • 3 years ago
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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

  72. hartnn
    • 3 years ago
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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.

  73. hartnn
    • 3 years ago
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    and to get final answer, just add them

  74. DLS
    • 3 years ago
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    YES! correct!

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

  76. hartnn
    • 3 years ago
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    but easy.

  77. DLS
    • 3 years ago
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    for u!

  78. hartnn
    • 3 years ago
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    sum of 2 eaSY Q's

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

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