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Math Help - Implicit differentiation

  1. #1
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    Post Implicit differentiation

    Find dy/dx by implicit differentiation.

    (3xy^2 + 1)^4 = 2x - 3y

    4(3xy^2 + 1)^3 (3xy^2 + 1) = 2 -3

    (3xy^2 + 1) +3 = 2-4(3xy^2 +1)^3

    dy/dx = 2 - 4(3xy^2 +1)^3/(3xy^2 +1) +3

    Am I getting any closer to understanding this?
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  2. #2
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    You started out ok, but your second line is incorrect.

    This part is right 4(3xy^2+1)^3. Now you must multiply by the derivative of the inside function.

    So d/dx (3xy^2+1)= 6xy(dy/dx)+3y^2.

    Putting the LHS together it's 4(3xy^2+1)^3 * (6xy(dy/dx)+3y^2)

    The RHS of your work is correct.
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  3. #3
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    Hello, Becky!

    I don't see dy/dx or y' in your work.
    Then it suddenly appears in your last step.


    Find dy/dx by implicit differentiation: .(3xy + 1)^4 .= .2x - 3y

    . . . . . . 4(3xy + 1)(6xyy' + 3y) . = . 2 - 3y'

    . . . . . . 24xy(3xy + 1)3y' + 12y(3xy + 1) . = . 2 - 3y'

    . . . . . . 24xy(3xy + 1)3y' + 3y' . = . 2 - 12y(3xy + 1)

    Factor: .3[8xy(3xy(3xy + 1) + 1]y' . = . 2[1 - 6y(3xy + 1)]

    . . . . . . . . . . . . . . .2[1 - 6y(3xy + 1)]
    Therefore: . y' . = . ---------------------------
    . . . . . . . . . . . . . . 3[8xy(3xy + 1) + 1]

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  4. #4
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    Quote Originally Posted by becky View Post
    Find dy/dx by implicit differentiation.

    (3xy^2 + 1)^4 = 2x - 3y...
    Hi,

    I've attached an image to show you what I've calculated:
    Attached Thumbnails Attached Thumbnails Implicit differentiation-impl_ableitg1.gif  
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