Find dy/dx by implicit differentiation for the following
2x^2y+3y=-1
remember the product rule:
2(2x)y + 2(x^2)y' + 3y' = 0
y' (2x^2+3) = -4xy
y' = -4xy/(2x^2+3)
2x^2y+3y=-1
2(2x)y + 2(x^2)y' + 3y' = 0
y' (2x^2+3) = -4xy
y' = -4xy/(2x^2+3)