Question

Find y' by implicit differentiation: xy + 2x + 3x^2 = 4

Answer 1

(xy)' + (2x)' + (3x^2)' = (4)'

y + xy' + 2 + 6x = 0

xy' = -y  -2 -6x

y' = [-y -2 -6x] / x

Now solve y from the original equation and substitue

xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x

y' =  [(-2x - 3x^2 +4) / x - 2 - 6x ] / x

y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =

= [-9x^2 - 4x + 4] / x^2