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7 November, 11:43

If x^3+2x^2y-4y=7, then when x=1, dy/dx=

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  1. 7 November, 12:00
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    dy/dx when x = 1 is (4y + 3) / 2.

    Step-by-step explanation:

    x^3 + 2x^2y - 4y=7

    Using implicit differentiation:

    3x^2 + 4xy + 2x^2. dy/dx - 4 dy/dx = 0

    2x^2. dy/dx - 4dy/dx = - 3x^2 - 4xy

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

    When x = 1:

    dy/dx = (-3 - 4y) / - 2

    = - 1 (4y + 3) / - 2

    (4y + 3) / 2 (answer).
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