If x^3 + y^3 = 6 and d^2y dx^2 d^2x dy^2 = m (xy)^n , where m, n R then m n =

Step-by-step solution

Given x^3 + y^3 = 6 . Differentiating with respect to x , we get: 3x^2 + 3y^2 dy dx = 0 dy dx = - x^2 y^2 Differentiating again with respect to x : d^2y dx^2 = - d dx ( x^2 y^2 ) = - y^2(2x) - x^2 (2y dy dx ) y^4 Substituting dy dx = - x^2 y^2 : d^2y dx^2 = - 2xy^2 - 2x^2y (- x^2 y^2 ) y^4 = - 2xy^2 + 2x^4 y y^4 = - 2x(y^3 + x^3) y^5 Since x^3 + y^3 = 6 : d^2y dx^2 = - 12x y^5 By symmetry, differentiating x^3 + y^3 = 6 with respect to y twice gives: d^2x dy^2 = - 12y x^5 Multiplying the two second derivatives: d^2y