x=(6-2y²)/(1+y)

Question

differentiate with respect to y

I saw this in a question here but..I don't get how the person go that answer.

Answer 1

Sorry, the other answer was incorrect.
The correct answer should have been all expressed in y instead of x, since the given expression is already in y:

You can differentiate by the product rule or the quotient rule, I prefer the former:
x=(6-2y²)/(1+y)
=(6-2y^2)(1+y)^(-1)
dx/dy
=-4y(1+y)^(-1) +(6-2y^2)(-1)(1)(x+1)^(-2)
=-4y/(y+1)-(6-2y^2)/(y+1)^2
as in the other post (with x changed for y)

You can do some algebraic simplifications if you wish.

Answer 2

=(1+y)d/dy(6-2y^2)-(6-2y^2)d/dy(1+y)/(1+y)^2 =(1+y)(-4y)-(6-2y^2)(1)/(1+y)^2 =-4y-4y^2-6-2y^2/(1+y)^2 =-2y^2-4y-6/(1+y)^2