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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

by Level 11 User (81.5k points)

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