Question

How do you implicitly differentiate `-4=(x^2+y)^2-y^2x`?

Answer 1

`(y^2 - 4x^3 - 4 x y ) /( 2 ( x^2 + y - xy) )`

Explanation:

`d/(dx) { -4=(x^2+y)^2-y^2x}`

`\implies 0 = \color{red}{2(x^2 + y) ( 2x + y')} - \color{blue}{ 2y y' x - y^2}`

red bit is chain rule, blue bit is product rule with a bit of chain...

`\implies 0 = 4x(x^2 + y) + 2 y' ( x^2 + y - xy) - y^2`

`\implies y' = ( y^2 - 4x(x^2 + y) ) /( 2 ( x^2 + y - xy) )`

`= (y^2 - 4x^3 - 4 x y ) /( 2 ( x^2 + y - xy) )`