How do you find #(dy)/(dx)# given #x^3+y^3=2xy#?

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Answer 1 · 2016-08-18T14:42:33

#dy/dx=(2y-3x^2)/(3y^2-2x)#.

#x^3+y^3=2xy#

#:.d/dx(x^3+y^3)=d/dx(2xy)#.

#:.d/dxx^3+d/dxy^3=2d/dx(xy)#.

Here, by the Chain Rule, #d/dx(y^3)=d/dy(y^3)*dy/dx=3y^2*dy/dx#,

&, by, the Product Rule, #d/dx(xy)=x*d/dx(y)+y*d/dx(x)=xdy/dx+y*1#.

Therefore, #3x^2+3y^2dy/dx=2(xdy/dx+y)#.

#:. (3y^2-2x)dy/dx=2y-3x^2#

#:. dy/dx=(2y-3x^2)/(3y^2-2x)#.