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Jaweria

  • 3 years ago

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

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  1. Sujay
    • 3 years ago
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    When you say y prime, do you mean the derivative?

  2. Jaweria
    • 3 years ago
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    yup

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

  4. goformit100
    • 3 years ago
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    Yo Yo, diffrentiate on both sides...

  5. Jaweria
    • 3 years ago
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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?

  6. mathsmind
    • 3 years ago
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    the answer is:

  7. Jaweria
    • 3 years ago
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    can we go step by step?

  8. Sujay
    • 3 years ago
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    Ya sure, I'll set it up.

  9. mathsmind
    • 3 years ago
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    you want to go by steps sure will do

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

  11. Jaweria
    • 3 years ago
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    Thanks to both of you Sujay and Mathsmind :)

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

  13. Sujay
    • 3 years ago
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    with an x after the sin

  14. Sujay
    • 3 years ago
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    I want to make sure my original is correct

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

  16. mathsmind
    • 3 years ago
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    please confirm is this what u r asking for?

  17. mathsmind
    • 3 years ago
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    are u there?

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

  19. mathsmind
    • 3 years ago
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    i did write it

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

  21. Jaweria
    • 3 years ago
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    yeah you are right Mathmind

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

  23. Jaweria
    • 3 years ago
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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

  24. mathsmind
    • 3 years ago
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    \[y = \arcsin(x)\]

  25. mathsmind
    • 3 years ago
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    \[\sin(y)=x\]

  26. mathsmind
    • 3 years ago
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    differentiate both sides

  27. Sujay
    • 3 years ago
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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.

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

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

  30. mathsmind
    • 3 years ago
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    recall

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

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

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

  34. mathsmind
    • 3 years ago
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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

  35. Jaweria
    • 3 years ago
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    oh ok

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

  37. Jaweria
    • 3 years ago
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    thank u so much Mathsmind and its ok Sujay

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

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

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

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

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

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

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

  45. mathsmind
    • 3 years ago
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    let me prove that for you

  46. Jaweria
    • 3 years ago
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    ok plzz if you can do that

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

  48. mathsmind
    • 3 years ago
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    step two: differentiate both sides

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

  50. mathsmind
    • 3 years ago
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    step three : rearrange the differential equation

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

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

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

  54. Jaweria
    • 3 years ago
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    wow I am liking this method better than before

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

  56. mathsmind
    • 3 years ago
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    Step seven: substitute siny in the last deferential

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

  58. mathsmind
    • 3 years ago
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    that is a mistake an s

  59. mathsmind
    • 3 years ago
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    sorry i will do it again

  60. Jaweria
    • 3 years ago
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    oh ok

  61. mathsmind
    • 3 years ago
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    back to step six ok

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

  63. Jaweria
    • 3 years ago
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    ok

  64. mathsmind
    • 3 years ago
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    step seven : substitute step 6 in step 5

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

  66. mathsmind
    • 3 years ago
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    finally step eight : simplify

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

  68. mathsmind
    • 3 years ago
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    and this concludes our proof good luck

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

  70. Jaweria
    • 3 years ago
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    oh wow you explained really good thanks :)

  71. mathsmind
    • 3 years ago
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    so i deserve a medal then hehehe

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

  73. Jaweria
    • 3 years ago
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    yup you did :)

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

  75. Jaweria
    • 3 years ago
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    yup it does :)

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

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

  78. mathsmind
    • 3 years ago
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    sure

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

  80. Jaweria
    • 3 years ago
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    thanks

  81. Jaweria
    • 3 years ago
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    oh yeah sorry i forgot hehe

  82. mathsmind
    • 3 years ago
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    ur welcome!

  83. mathsmind
    • 3 years ago
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    u better not make one mistake

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

  85. mathsmind
    • 3 years ago
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    all the best

  86. mathsmind
    • 3 years ago
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    there is another method using Taylor's series

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

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

  89. Jaweria
    • 3 years ago
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    oh ok

  90. mathsmind
    • 3 years ago
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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!

  91. Jaweria
    • 3 years ago
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    yeah

  92. Jaweria
    • 3 years ago
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    i m confuse

  93. Jaweria
    • 3 years ago
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    the answer that I m getting is weird

  94. Jaweria
    • 3 years ago
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    are u there?

  95. mathsmind
    • 3 years ago
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    yes

  96. mathsmind
    • 3 years ago
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    post the question

  97. mathsmind
    • 3 years ago
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    r u there?

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