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Calcules

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find y" by implicit differentiation.
4x^3 + 3y^3 = 4
=====================================
4x^3 + 3y^3 = 4
12x + 9yy' = 0
24x + 9(2yy')y' + 9y^2(y") = 0

y" = -6[4x-3y(y')^2] answer. (it this right? )

  • Calcules -

    Your 1st deriv should have been
    12x^2 + 9y^2 y' = 0 ----> y' = -12x^2/(9y^2)

    2nd derivative:
    24x + 9y^2 (y'') + y' (18y) = 0
    replacing y'
    24x + 9y^2 y'' + (-12x^2)/(9y^2) (18y) = 0
    24x + 9y^2 y'' - 24x/y = 0
    times y
    24xy + 9y^3 - 24x = 0
    y'' = (24x - 24xy)/(9y^3) = (8x-8xy)/(3y^3)

    I don't think they wanted to see that y' in your answer.

    btw, how did you ever get you first derivative ?

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