Find the 1st and 2nd derivative of 3xy=4x+y^2
Differentiate both sides of 3xy=4x+y^2 implicitly.
3y + 3xy' = 4 + 2yy' ----- (1)
y'(3x - 2y) = 4 - 3y
y' = (4 - 3y)/(3x - 2y) ---- 1st derivative
To get the 2nd derivative, it would probably be simpler to implicitly differentiate eqn (1).
3y' + 3y' + 3xy'' = 2(y')^2 + 2yy''
y''(3x - 2y) = 2(y')^2 - 6y' = 2y'(y' - 3)
substituting for y' into the above expression.
y''(3x - 2y) = 2{(4 - 3y)/(3x - 2y)}*{(4 - 3y)/(3x - 2y) - 3}
y''(3x - 2y) = 2{(4 - 3y)/(3x - 2y)}*{[(4 - 3y) - 3(3x - 2y)]/(3x - 2y)}
y''(3x - 2y) = 2{(4 - 3y)}*{[4 - 3y - 9x + 6y]}/(3x - 2y)^2
y'' = 2{(4 - 3y)}*{4 + 3y - 9x}/(3x - 2y)^3 --- 2nd derivative