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x^3y^3-y=x ... by implicit differentiation find dy/dx
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use product rule d(dx) (u*v) = udv+vdu x^3*3y^2 *dy/dx + y^3*3x^2 - dy/dx = 1 x^3*3y^2 *dy/dx - dy/dx = 1 - y^3*3x^2 (x^3*3y^2 - 1) * dy/dx = 1 - y^3*3x^2 (1 - y^3*3x^2) / (x^3*3y^2 -1) = dy/dx
thank you so much!!!!!!!!!
you're welcome. make sure to check the work and understand the problem.
yes, that's what im doing now
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