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Jaweria

  • 3 years 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. Curry
    • 3 years ago
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    just post, i got you.

  2. Jaweria
    • 3 years ago
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    ok first one is: (2x-3)^2+(4y-5)^2=10

  3. Jaweria
    • 3 years ago
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    and we need to find dy/dx for all of them.

  4. binarymimic
    • 3 years ago
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    you can either use implicit differentiation from the start, or you can begin by isolating the y variable.

  5. Jaweria
    • 3 years ago
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    second question is: x^2sin(x)+y^2cos(y)=1

  6. binarymimic
    • 3 years ago
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    likewise for the second

  7. Jaweria
    • 3 years ago
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    yeah we have to do implicit diffentiation but I am really stuck in all of them.

  8. binarymimic
    • 3 years ago
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    i'll get you started with (2x-3)^2+(4y-5)^2=10

  9. Jaweria
    • 3 years ago
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    ohh thank you so much

  10. binarymimic
    • 3 years ago
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    |dw:1359266294450:dw|

  11. binarymimic
    • 3 years ago
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    idk if you can read the edge of it, it should say = 0

  12. Jaweria
    • 3 years ago
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    yeah i can

  13. Jaweria
    • 3 years ago
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    but after the paranthesis is that a multiply by 2 or its a power of 2 for the first one

  14. binarymimic
    • 3 years ago
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    multiplication

  15. Jaweria
    • 3 years ago
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    ok

  16. Jaweria
    • 3 years ago
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    but then we have to simplify more right?

  17. binarymimic
    • 3 years ago
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    yeah you would have to isolate y'

  18. Jaweria
    • 3 years ago
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    ok can we go through this together till the end?

  19. binarymimic
    • 3 years ago
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    sure

  20. Jaweria
    • 3 years ago
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    thank u :)

  21. binarymimic
    • 3 years ago
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    |dw:1359266644304:dw|

  22. binarymimic
    • 3 years ago
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    |dw:1359266753423:dw|

  23. Jaweria
    • 3 years ago
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    how did u get -2?

  24. binarymimic
    • 3 years ago
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    oh my bad, that should be a 12

  25. binarymimic
    • 3 years ago
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    |dw:1359266876031:dw|

  26. Jaweria
    • 3 years ago
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    oh ok

  27. binarymimic
    • 3 years ago
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    |dw:1359266992506:dw|

  28. Jaweria
    • 3 years ago
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    and do we have to put y^1 before paranthesis?

  29. binarymimic
    • 3 years ago
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    where?

  30. Jaweria
    • 3 years ago
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    its ok I got it

  31. binarymimic
    • 3 years ago
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    ok. try the second one whenever you are comfortable

  32. Jaweria
    • 3 years ago
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    ok

  33. Jaweria
    • 3 years ago
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    ok I have a feeling that I ll again get stuck with this one:

  34. Jaweria
    • 3 years ago
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    x^2sin(x)+y^2cos(y)=1

  35. binarymimic
    • 3 years ago
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    just think of differentiating y like you were differentiating any other variable (except x). remember to write the "dy/dx" or " y' " after you differentiate

  36. Jaweria
    • 3 years ago
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    ok

  37. binarymimic
    • 3 years ago
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    like here's an example: if i differentiate y*sin y with respect to x: = y cos(y) * y' + sin(y) * y' = y' [y cos(y) + sin(y)]

  38. binarymimic
    • 3 years ago
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    it's the same as if you were differentiating 'u', as in differentiate: sin(x^2) let u = x^2 and differentiate sin(u) d/dx sin(u) = cos(u) * du = cos(x^2) * du Another way of saying that is: cos(x^2) * u' The only difference here is, we can write du in terms of x very easily. With implicit differentiation, we normally can't do that, but it's the same concept.

  39. Jaweria
    • 3 years ago
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    hmmm if you dont mind can we go over this problem because my brain is not working at all with all of these i am sorry but thats the truth all my 11 questions are due tomorrow online :(

  40. Jaweria
    • 3 years ago
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    I m still learning

  41. binarymimic
    • 3 years ago
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    x^2sin(x)+y^2cos(y)=1 let's start with the easy part. if we're going to differentiate both sides, the right side becomes 0. what does x^2 sin(x) become ?

  42. Jaweria
    • 3 years ago
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    2x cos(x)

  43. binarymimic
    • 3 years ago
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    are you sure?

  44. Jaweria
    • 3 years ago
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    not much

  45. binarymimic
    • 3 years ago
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    don't forget the product rule

  46. Jaweria
    • 3 years ago
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    i always mess up on product rule and chain rule

  47. binarymimic
    • 3 years ago
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    \[\frac{ d }{ dx } x^{2} \sin x = x^{2} \cos x + \sin x * 2x\]

  48. binarymimic
    • 3 years ago
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    "first times the derivative of the second, plus second times the derivative of the first"

  49. Jaweria
    • 3 years ago
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    ohh ok

  50. binarymimic
    • 3 years ago
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    now try differentiating y^2cos(y)

  51. Jaweria
    • 3 years ago
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    ok

  52. Jaweria
    • 3 years ago
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    not sure if I did it right or not

  53. Jaweria
    • 3 years ago
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    |dw:1359268366069:dw|

  54. binarymimic
    • 3 years ago
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    the left part is almost right, just need the negative sign (the derivative of cos is -sin the other part would become: cos(y) * 2y * dy/dx

  55. Jaweria
    • 3 years ago
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    oh ok

  56. binarymimic
    • 3 years ago
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    so i think you got that right, too, just needed the dy/dx ?

  57. Jaweria
    • 3 years ago
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    yes

  58. binarymimic
    • 3 years ago
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    \[x^{2} \cos x + 2x \sin x + \frac{ dy }{ dx } * -y^{2} \sin y + \frac{ dy }{ dx } * 2y \cos y\]

  59. Jaweria
    • 3 years ago
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    so that will be |dw:1359268690019:dw|

  60. binarymimic
    • 3 years ago
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    |dw:1359268763927:dw|

  61. Jaweria
    • 3 years ago
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    ok

  62. binarymimic
    • 3 years ago
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    you need to isolate dy/dx now

  63. Jaweria
    • 3 years ago
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    yeah ok

  64. Jaweria
    • 3 years ago
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    |dw:1359269043320:dw|

  65. Jaweria
    • 3 years ago
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    is that right?

  66. binarymimic
    • 3 years ago
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    yep perfect you can eliminate the minus sign on the outside if you wanted to by swapping the terms in the denominator to make it: y^2 sin y - 2y cos y but thats just a matter of preference

  67. Jaweria
    • 3 years ago
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    ok thanks

  68. binarymimic
    • 3 years ago
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    wait let me double check something

  69. Jaweria
    • 3 years ago
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    couple more please if you dont mind?

  70. Jaweria
    • 3 years ago
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    ok

  71. binarymimic
    • 3 years ago
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    ok. sorry. if you had more questions you might want to start a new post

  72. Jaweria
    • 3 years ago
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    oh ok

  73. binarymimic
    • 3 years ago
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    after a few of these, implicit differentiation will be easy for you

  74. Jaweria
    • 3 years ago
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    this is the first time ever I am using this site can you suggest me that how to start new post?

  75. Jaweria
    • 3 years ago
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    Yeah I am trying my best.

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