Question

How do you find `dy/dx` by implicit differentiation of `x^3-3x^2y+2xy^2=12`?

Answer 1

`dy/dx=(6xy-3x^2-2y^2)/(4xy-3x^2)`

Explanation:

To differentiate `-3x^2y" and " 2xy^2` we use the `color(blue)"product rule"`

`rArr3x^2-(3x^2.dy/dx+6xy)+(2x.2ydy/dx+2y^2)=0`

`rArr3x^2-3x^2dy/dx-6xy+4xydy/dx+2y^2=0`

`rArrdy/dx(4xy-3x^2)=6xy-3x^2-2y^2`

`rArrdy/dx=(6xy-3x^2-2y^2)/(4xy-3x^2)`