How do you find #dy/dx# by implicit differentiation of #x^2y+y^2x=-2#?

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Answer 1 · 2016-12-06T06:07:13

The answer is #dy/dx=-(2xy+y^2)/(x^2+2xy)#

We use the product rule for differentiation

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

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

#(y^2x)'=y^2+2xydy/dx#

#(-2)'=0#

Putting it all together

#2xy+x^2dy/dx+y^2+2xydy/dx=0#

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

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