Question

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

Answer 1

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