How do you use implicit differentiation to find dy/dx given #xy^2+x^2y=x#?

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Answer 1 · 2016-11-05T20:00:19

#dy/dx = (1 - 2xy - y^2)/(2xy + x^2)#

The differentiation of the given expression is determined by using differentiation of the sum and the product differentiation.

Differentiation of the sum:

#color(blue)(d/dx(u+v)=(du)/dx+(dv)/dx)#

Product differentiation:

#color(brown)((uv)'=u'v+v'u)#

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

#rArrd/dx(xy^2+x^2y)=(dx)/dx#

#rArrcolor(blue)((d(xy^2))/dx+(d(x^2y))/dx)=1#

#rArrcolor(brown)((y^2xxdx/dx+x xx(dy^2)/dx))+color(brown)((yxxdx^2/dx+x^2xxdy/dx)=1#

#rArry^2 + 2xydy/dx +2xy + x^2dy/dx = 1#

#rArr2xydy/dx + x^2dy/dx +2xy +y^2 = 1#

#rArrdy/dx(2xy + x^2) +2xy +y^2 = 1#

#rArrdy/dx(2xy + x^2) = 1 - 2xy - y^2#

#rArrdy/dx = (1 - 2xy - y^2)/(2xy + x^2)#