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Jaweria

  • one year ago

I need help with 11 questions of Derivatives problems in Calculus 1. Can anyone help me with that please let me know, I will really appreciate it.

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  1. AravindG
    • one year ago
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    we cant help with all 11 .maybe we can help in 1 or 2 with fond hope that you will understand how to do these and finish the rest yourself.

  2. Jaweria
    • one year ago
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    hmm thanks ok i ll need your help with atleast 2 of them.

  3. AravindG
    • one year ago
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    bring it on !! and "WELCOME TO OPENSTUDY!!"

  4. Jaweria
    • one year ago
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    one of them is : |dw:1359270057247:dw|

  5. Jaweria
    • one year ago
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    thank you :)

  6. Jaweria
    • one year ago
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    and we are finding dx/dy

  7. AravindG
    • one year ago
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    ok diff both sides with respect to y

  8. AravindG
    • one year ago
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    use product rule

  9. Jaweria
    • one year ago
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    ok

  10. Jaweria
    • one year ago
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    Like this:|dw:1359270336745:dw|

  11. Jaweria
    • one year ago
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    not sure though

  12. Jaweria
    • one year ago
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    can anyone help me here with my questions?

  13. nitz
    • one year ago
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    its \[x ^{2}y ^{2}=1 \] or \[xy ^{2}=1\]?

  14. Jaweria
    • one year ago
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    its x\[its x ^{2}y ^{2}=1\]

  15. nitz
    • one year ago
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    OK....differentiate both sides wrt to y using product rule:: \[x ^{2}(2y)+((y ^{2})2x (dx/dy))=0\]

  16. Jaweria
    • one year ago
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    ahan ok

  17. AravindG
    • one year ago
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    sorry i gt disconnected :O

  18. nitz
    • one year ago
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    so tell me from above equation what would be dx/dy?

  19. Jaweria
    • one year ago
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    oh its ok AravindG

  20. Jaweria
    • one year ago
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    ok i will tell you

  21. AravindG
    • one year ago
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    just solve for dx/dy after you applied product rule which will be .....

  22. Jaweria
    • one year ago
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    I need little bit of more discription for that if its possible.

  23. Jaweria
    • one year ago
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    everybody left me here :(

  24. Jaweria
    • one year ago
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    no one wants to help me :(

  25. Azteck
    • one year ago
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    \[y^2=\frac{ 1 }{ x^2 }\]

  26. Azteck
    • one year ago
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    \[y^2=x^{-2}\]

  27. Jaweria
    • one year ago
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    how did you get \[X ^{-2}\]

  28. Azteck
    • one year ago
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    Do you know how to differentiate explicitly?

  29. Azteck
    • one year ago
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    You don't know that ???? \[x^-2=\frac{1}{x^2}\]

  30. Azteck
    • one year ago
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    \[x^-2\]

  31. Jaweria
    • one year ago
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    oh yeah i do sorry

  32. Azteck
    • one year ago
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    \[x^{-2}\]

  33. Azteck
    • one year ago
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    So now, do you know how to differentiate explicitly?

  34. Azteck
    • one year ago
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    That means you differentiate boths sides at the same time.

  35. Jaweria
    • one year ago
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    yup

  36. Azteck
    • one year ago
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    Okay. Give it a shot and tell me what you get. Remember, when you differentiate the y part, you also have to attach a dy/dx.

  37. Jaweria
    • one year ago
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    ahan ok

  38. Jaweria
    • one year ago
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    is it \[-\frac{ y }{ x}\]

  39. Azteck
    • one year ago
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    Nope. Show me what you did as your first step.

  40. Azteck
    • one year ago
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    I gave you this to differentiate. \[y^2=x^{-2}\]

  41. Jaweria
    • one year ago
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    I did like this:|dw:1359271872658:dw|

  42. Azteck
    • one year ago
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    Then what's your next step.

  43. Azteck
    • one year ago
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    ?*

  44. Jaweria
    • one year ago
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    then I canceled \[Y ^{2} which became Y and Then I can canceled X ^{-2} and \it became -x\]

  45. Jaweria
    • one year ago
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    oppss sorry i did bad

  46. nitz
    • one year ago
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    u have any doubt in my way......should i explain?

  47. Jaweria
    • one year ago
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    lolz right now I dont have any doubt at anyone I just need help very badly :(

  48. Azteck
    • one year ago
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    When you differentiate y^2, you get 2y?

  49. Jaweria
    • one year ago
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    yes

  50. Azteck
    • one year ago
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    Why did you get y?

  51. Azteck
    • one year ago
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    and when you differentiate x^{-2}, you get -2x^{-3}

  52. Azteck
    • one year ago
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    why did you get -x?

  53. Jaweria
    • one year ago
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    Yes I did a mistake for -X though but I couldnt understand Y one

  54. Azteck
    • one year ago
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    differentiate x^2

  55. Azteck
    • one year ago
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    That's exactly the same as differentiating y^2. What are you doing?

  56. Azteck
    • one year ago
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    \[2y\times \frac{ dy }{ dx }=-\frac{ 2 }{ x^3 }\]

  57. Azteck
    • one year ago
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    Make dy/dx the subject.

  58. Jaweria
    • one year ago
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    sorry one question that why there is

  59. Jaweria
    • one year ago
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    Sorry but i have a question that how you got X^3

  60. nitz
    • one year ago
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    \[(2x ^{2}y)=-(2xy ^{2})(dx/dy)\] \[dx/dy=-(2x ^{2}y)/(2xy ^{2})\] dx/dy=-x/y

  61. Azteck
    • one year ago
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    http://www.wolframalpha.com/input/?i=differentiate+y%5E2%3Dx%5E-2

  62. Jaweria
    • one year ago
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    thanks Nitz for this :)

  63. Jaweria
    • one year ago
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    Aztec actually the answer I am having from the professor is matching with Nitz

  64. Azteck
    • one year ago
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    @Jaweria, do you have an answer sheet or something?

  65. Jaweria
    • one year ago
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    yes I do

  66. Azteck
    • one year ago
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    Ah okay. I can see that because you differentiate straight away. I moved the x^2 to the other side. Somehow that affected it.

  67. Jaweria
    • one year ago
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    oh ok

  68. Azteck
    • one year ago
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    nitz just differentiate straight away using the product rule.

  69. Jaweria
    • one year ago
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    but as Nitz just explained I am little bit confused there that how he is getting 2\[2X ^{2}Y\] isnt that when you get 2X then you dont need \[2X ^{2}\]

  70. Azteck
    • one year ago
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    Don't you know how to do the product rule?

  71. Jaweria
    • one year ago
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    I have it on my notes and I am still trying to learn this is my first homework of this Calculus class

  72. Jaweria
    • one year ago
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    I found it hard

  73. Azteck
    • one year ago
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    What's the answer say in your sheet, Does it say -x/y or -y/x?

  74. Jaweria
    • one year ago
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    it says \[-\frac{ y }{ x }\]

  75. Azteck
    • one year ago
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    Nitz is wrong then...

  76. Azteck
    • one year ago
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    <Look at his answer.

  77. Azteck
    • one year ago
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    He did dx/dy. You want dy/dx

  78. Jaweria
    • one year ago
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    yeah but I dont think so that really matters you can do either way

  79. Jaweria
    • one year ago
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    oh yeah

  80. Azteck
    • one year ago
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    ...

  81. Jaweria
    • one year ago
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    so its just we are flipping dy/dx instead of dx/dy

  82. Azteck
    • one year ago
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    here's how you get -y/x

  83. Azteck
    • one year ago
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    \[u=x^2\] \[u'=2x\] \[v=y^2\] \[v'=2y\] \[\frac{ dy }{ dx }=(u'\times v)+(u\times v')\]

  84. Azteck
    • one year ago
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    Sorry about that. \[0=2xy^2+2x^2y(\frac{dy}{dx})\]

  85. Azteck
    • one year ago
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    When you differentiate y^2 you would write dy/dx next to it.

  86. Jaweria
    • one year ago
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    yup

  87. Azteck
    • one year ago
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    Then you make dy/dx as the subject.

  88. Azteck
    • one year ago
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    And you're done.

  89. Jaweria
    • one year ago
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    ahan cool thanks.

  90. Azteck
    • one year ago
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    No worries mate.

  91. Jaweria
    • one year ago
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    one last question plzz

  92. Azteck
    • one year ago
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    Okay. last one. And you can try and finish the others yourself using what you got from here.

  93. Jaweria
    • one year ago
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    alright.

  94. Jaweria
    • one year ago
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    \[y ^{2}\tan(x)=x\]

  95. AravindG
    • one year ago
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    to find dx/dy ?

  96. Jaweria
    • one year ago
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    yup

  97. AravindG
    • one year ago
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    product rule on left

  98. AravindG
    • one year ago
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    try using draw button i will check :)

  99. Jaweria
    • one year ago
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    hmm ok

  100. Jaweria
    • one year ago
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    I know I ll get stuck again :( but I am trying

  101. AravindG
    • one year ago
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    try ! try you will succeed ! :)

  102. Jaweria
    • one year ago
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    alright :)

  103. Jaweria
    • one year ago
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    ok first step so far: |dw:1359274476783:dw|

  104. AravindG
    • one year ago
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    can you write thhird term again ? i cant see it clearly

  105. AravindG
    • one year ago
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    *third

  106. Jaweria
    • one year ago
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    ok sure

  107. Jaweria
    • one year ago
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    |dw:1359274789970:dw|

  108. AravindG
    • one year ago
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    i mean third term fully from y^2 sec^2 x

  109. Jaweria
    • one year ago
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    oh ok sorry

  110. AravindG
    • one year ago
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    and why do you have 2 equal signs ?

  111. Jaweria
    • one year ago
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    |dw:1359274989073:dw|

  112. Jaweria
    • one year ago
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    I put X after equal sign because thats also in the question

  113. AravindG
    • one year ago
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    for x diff with respect to y it will be simply dx/dy isnt it?

  114. Jaweria
    • one year ago
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    yeah thats what I pot in paranthesis next to the X

  115. Jaweria
    • one year ago
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    put

  116. AravindG
    • one year ago
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    you wrote dy/dx there?

  117. Jaweria
    • one year ago
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    yes next to it

  118. Jaweria
    • one year ago
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    |dw:1359275398794:dw|

  119. AravindG
    • one year ago
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    what was the question ?

  120. AravindG
    • one year ago
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    what i think is your question is \[y^2 \tan(x)=x \]

  121. Jaweria
    • one year ago
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    \[Y ^{2}\tan (x)=x\]

  122. AravindG
    • one year ago
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    |dw:1359275616596:dw|

  123. AravindG
    • one year ago
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    this is what i get

  124. Jaweria
    • one year ago
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    sorry what is equal to dy/dx?

  125. AravindG
    • one year ago
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    nothing !!

  126. AravindG
    • one year ago
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    we just diff with respect to dy the whole equation

  127. Jaweria
    • one year ago
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    oh ok

  128. AravindG
    • one year ago
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    so why do you think dy/dx will appear?

  129. Jaweria
    • one year ago
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    so the way i did is that wrong?

  130. AravindG
    • one year ago
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    if you wanted dx/dy I believe my method is enough ...

  131. Jaweria
    • one year ago
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    ahan ok

  132. AravindG
    • one year ago
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    i will be glad if someone verifies this bcoz i have not done differentiation for a while :P

  133. Jaweria
    • one year ago
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    is Aztec here?

  134. AravindG
    • one year ago
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    dunno , maybe he is away

  135. AravindG
    • one year ago
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    have patience :) in the meantime try to work on other questions .BEST OF LUCK !

  136. Jaweria
    • one year ago
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    i have an answer sheet though

  137. Jaweria
    • one year ago
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    alright then i ll work on them :)

  138. AravindG
    • one year ago
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    ok whats the answer you have for this question ?

  139. Jaweria
    • one year ago
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    its \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]

  140. AravindG
    • one year ago
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    in btw i frgt to add tan x here |dw:1359276176708:dw|

  141. Jaweria
    • one year ago
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    ahan thats what I was thinking

  142. AravindG
    • one year ago
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    i get dx/dy as reciprocal of your answer...are you sure the question asks for dx/dy ?

  143. Jaweria
    • one year ago
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    nope its asking for dy/dx

  144. AravindG
    • one year ago
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    then my working is right !! :)

  145. Jaweria
    • one year ago
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    ok :)

  146. AravindG
    • one year ago
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    solve for dx/dy from my working...dy/dx=1/dx/dy

  147. AravindG
    • one year ago
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    understood?

  148. Jaweria
    • one year ago
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    ok

  149. Jaweria
    • one year ago
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    yup

  150. Jaweria
    • one year ago
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    thanks :)

  151. AravindG
    • one year ago
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    Ok bye I gtg

  152. Jaweria
    • one year ago
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    ok bye

  153. AravindG
    • one year ago
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    yw :)

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