How do I find dy/dx of the equation 4x^3-3xy^2+y^3=28 at the point (3,4)?

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asked Apr 19, 2017 116k views

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Zoom · Answer 1 · 2017-04-24T23:18:27+0000

To get the dy/ dx of the equation 4x^3-3xy^2+y^3=28 at the point (3,4), get the first derivative then substitute the given value. Y = 4x^3-3xy^2+y^3=28 Dy / dx = (2x / y) + 1 Substitute the given Dy /dx = ( (2*3)/ 4) + 1 Dy/dx = 5/2

How do I find dy/dx of the equation 4x^3-3xy^2+y^3=28 at the point (3,4)?

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