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December 21, 2014

December 21, 2014

Posted by **amy** on Wednesday, November 22, 2006 at 7:21am.

1.y^3=4(x^2+y^2)

2.y^2-3x+2y=0

help please

You consider y to be a function of x, but you don't explicitely solve for it. Then you formally differentiate w.r.t. x using the chain rule:

y^3(x) = 4(x^2+y^2(x)) --->

3y^2dy/dx = 8x + 8y dy/dx

y^2-3x+2y=0 --->

2ydy/dx - 3+2dy/dx = 0

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