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Jaweria

Please anyone help me with this question. Find Y prime: Y=sin^-1(2x^3).

  • one year ago
  • one year ago

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  1. Sujay
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    When you say y prime, do you mean the derivative?

    • one year ago
  2. Jaweria
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    yup

    • one year ago
  3. Jaweria
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    Please someone help me I m stuck and its due tomorrow

    • one year ago
  4. goformit100
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    Yo Yo, diffrentiate on both sides...

    • one year ago
  5. Jaweria
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    These are kind of new problems to me if you dont mind can you teach me and help me solving them?

    • one year ago
  6. mathsmind
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    the answer is:

    • one year ago
  7. Jaweria
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    can we go step by step?

    • one year ago
  8. Sujay
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    Ya sure, I'll set it up.

    • one year ago
  9. mathsmind
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    you want to go by steps sure will do

    • one year ago
  10. mathsmind
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    at the end i will show u how u can work such equation in ur head, without paper and pen

    • one year ago
  11. Jaweria
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    Thanks to both of you Sujay and Mathsmind :)

    • one year ago
  12. Sujay
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    Original Equation:\[y=\sin^2-1(2x-3)?\]

    • one year ago
  13. Sujay
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    with an x after the sin

    • one year ago
  14. Sujay
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    I want to make sure my original is correct

    • one year ago
  15. mathsmind
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    \[y=\sin^{-1}(2x^3)\]

    • one year ago
  16. mathsmind
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    please confirm is this what u r asking for?

    • one year ago
  17. mathsmind
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    are u there?

    • one year ago
  18. Jaweria
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    ok let me write down the equation here. Its \[y=\sin^{-1} (2x ^{3})\]

    • one year ago
  19. mathsmind
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    i did write it

    • one year ago
  20. Jaweria
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    yeah I m here sorry I was just writing that equation down

    • one year ago
  21. Jaweria
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    yeah you are right Mathmind

    • one year ago
  22. mathsmind
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    ok there are 2 main ways to solve this problem at ur level

    • one year ago
  23. Jaweria
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    is there any easy way bcoz I have 4 more to do after this one so I need to understand the whole thing my professor method was too hard

    • one year ago
  24. mathsmind
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    \[y = \arcsin(x)\]

    • one year ago
  25. mathsmind
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    \[\sin(y)=x\]

    • one year ago
  26. mathsmind
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    differentiate both sides

    • one year ago
  27. Sujay
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    Original:\[\sin^{-1} (2x^3)\] Use the chain rule: Derivative out the outside times the derivative of the inside. \[-\sin(2x^3)[\cos(2x^3)](6x^2) \] and there you have it, now simplify.

    • one year ago
  28. mathsmind
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    \[cosy \frac{ dy }{ dx }=1\]

    • one year ago
  29. mathsmind
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    \[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]

    • one year ago
  30. mathsmind
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    recall

    • one year ago
  31. Sujay
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    Apparently my answer is wrong, terribly sorry, I'll see where I went wrong :(.

    • one year ago
  32. mathsmind
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    \[\sin^2y+\cos^2y=1\]

    • one year ago
  33. Sujay
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    Oh yes, my sine needs to be raised to a negative 2

    • one year ago
  34. mathsmind
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    be patient both of u i will solve it in different way and u pick up the easiest method ok, all the best

    • one year ago
  35. Jaweria
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    oh ok

    • one year ago
  36. mathsmind
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    \[cosy=\pm \sqrt{1-\sin^2y}\]

    • one year ago
  37. Jaweria
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    thank u so much Mathsmind and its ok Sujay

    • one year ago
  38. mathsmind
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    \[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-\sin^2y} }\]

    • one year ago
  39. mathsmind
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    now remember our rearrangement\[\sin(y)=x\] so substitute that in the above equation

    • one year ago
  40. mathsmind
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    \[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-x^2} }\]

    • one year ago
  41. mathsmind
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    now this is the general derivative of arcsin or sin inverse ok

    • one year ago
  42. mathsmind
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    now if you follow the same steps for\[y = \sin^{-1}(2x^3)\]

    • one year ago
  43. mathsmind
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    \[\frac{ dy }{ dx } = \frac{ 6x^2 }{ \sqrt{1-4x^6} }\]

    • one year ago
  44. mathsmind
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    and that's is your final answer achieved using the chain rule

    • one year ago
  45. mathsmind
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    let me prove that for you

    • one year ago
  46. Jaweria
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    ok plzz if you can do that

    • one year ago
  47. mathsmind
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    step One:\[2x^3=siny\]

    • one year ago
  48. mathsmind
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    step two: differentiate both sides

    • one year ago
  49. mathsmind
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    \[6x^2=\frac{ dy }{ dx }cosy\]

    • one year ago
  50. mathsmind
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    step three : rearrange the differential equation

    • one year ago
  51. mathsmind
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    \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ cosy }\]

    • one year ago
  52. mathsmind
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    step four: recall that \[cosy=\pm \sqrt{1-\sin^2y}\]

    • one year ago
  53. mathsmind
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    step five: substitute cosy in the differential equation \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-\sin^2y} }\]

    • one year ago
  54. Jaweria
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    wow I am liking this method better than before

    • one year ago
  55. mathsmind
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    step six: recall\[siny=2x^3s\]

    • one year ago
  56. mathsmind
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    Step seven: substitute siny in the last deferential

    • one year ago
  57. Jaweria
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    what number is that under the power of 3?

    • one year ago
  58. mathsmind
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    that is a mistake an s

    • one year ago
  59. mathsmind
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    sorry i will do it again

    • one year ago
  60. Jaweria
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    oh ok

    • one year ago
  61. mathsmind
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    back to step six ok

    • one year ago
  62. mathsmind
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    \[siny=2x^3\]

    • one year ago
  63. Jaweria
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    ok

    • one year ago
  64. mathsmind
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    step seven : substitute step 6 in step 5

    • one year ago
  65. mathsmind
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    \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-(2x^3)^2} }\]

    • one year ago
  66. mathsmind
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    finally step eight : simplify

    • one year ago
  67. mathsmind
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    \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-4x^6} }\]

    • one year ago
  68. mathsmind
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    and this concludes our proof good luck

    • one year ago
  69. mathsmind
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    are you there? am sorry my computer is slow today

    • one year ago
  70. Jaweria
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    oh wow you explained really good thanks :)

    • one year ago
  71. mathsmind
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    so i deserve a medal then hehehe

    • one year ago
  72. Jaweria
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    Now I have to work on 4 more of these huh lets see how am I going to do it.

    • one year ago
  73. Jaweria
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    yup you did :)

    • one year ago
  74. mathsmind
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    ok if you do those 4 by yourself i will give you a medal, does that sound fair?

    • one year ago
  75. Jaweria
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    yup it does :)

    • one year ago
  76. mathsmind
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    you can use the triangle method as well to solve this

    • one year ago
  77. Jaweria
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    if you dont mind can I just step by step show you how to do that?

    • one year ago
  78. mathsmind
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    sure

    • one year ago
  79. mathsmind
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    how am i gonna give u a medal then ! hehehe

    • one year ago
  80. Jaweria
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    thanks

    • one year ago
  81. Jaweria
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    oh yeah sorry i forgot hehe

    • one year ago
  82. mathsmind
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    ur welcome!

    • one year ago
  83. mathsmind
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    u better not make one mistake

    • one year ago
  84. Jaweria
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    ohhh hmmm I am not sure about that but I m doing it

    • one year ago
  85. mathsmind
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    all the best

    • one year ago
  86. mathsmind
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    there is another method using Taylor's series

    • one year ago
  87. Jaweria
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    my professor also gave us the formula like for that equation the formula for \[\sin^{-1} is 1/\sqrt{1-u ^{2}}timesdu\]

    • one year ago
  88. mathsmind
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    yes that is the method of substitution which i see it as a lazy method

    • one year ago
  89. Jaweria
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    oh ok

    • one year ago
  90. mathsmind
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    but if u are used to that method then go for it, as soon as u get the right answer who cares, you know!

    • one year ago
  91. Jaweria
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    yeah

    • one year ago
  92. Jaweria
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    i m confuse

    • one year ago
  93. Jaweria
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    the answer that I m getting is weird

    • one year ago
  94. Jaweria
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    are u there?

    • one year ago
  95. mathsmind
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    yes

    • one year ago
  96. mathsmind
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    post the question

    • one year ago
  97. mathsmind
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    r u there?

    • one year ago
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