Question

Question #bcb5d

Answer 1

The answer is `=-(y(2x+y))/(x(x+2y))`

Explanation:

We use

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

`x^2y+xy^2=6`

`(x^2y)'+(xy^2)'=(6)'`

`x^2dy/dx+2xy+x2ydy/dx+y^2=0`

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

`dy/dx=-(y(2x+y))/(x(x+2y))`