Answer:
Step-by-step explanation:
Ok
(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
No. when plugging 4 into the equation, 0 is not the final result. instead it is 6.
square root of 2(4)+1 = square root of 9 = 3
3+3 = 6
and 6 is not 0 so x cannot = 4
