Not sure how to do this (calculus)?

Question

Implicitly differentiate 3x^2=(2-y)/(y+2)

help please!

Answer 1

3x^2 = (2 - y)/(y + 2)

All you have to keep in mind when doing implicit differentiation is that whenever you take the derivative of y, it is equal to dy/dx.

Differentiating, we get

6x = [ (dy/dx)(y + 2) - (2 - y)(dy/dx) ] / (y + 2)^2

To isolate dy/dx, multiply both sides by (y + 2)^2.

6x(y + 2)^2 = (dy/dx)(y + 2) - (2 - y)(dy/dx)

By symmetry,

(dy/dx)(y + 2) - (2 - y)(dy/dx) = 6x(y + 2)^2

Factor dy/dx,

[dy/dx] (y + 2 - (2 - y)) = 6x(y + 2)^2

[dy/dx] (y + 2 - 2 + y) = 6x(y + 2)^2

[dy/dx] (2y) = 6x(y + 2)^2

Divide both sides by 2y,

dy/dx = [ 6x(y + 2)^2 ] / [ 2y ]

dy/dx = [ 3x(y + 2)^2 ] / y

Answer 2

6x= 1/(y+2)^2 *[(y+2)(-y´)-(2-y)*y´)
6x= (-4y´)/(y+2)^2