Question

if x^3+y^3=3axy so prove that d^2y/dx^2 = 2a^3xy/(ax-y^2)^3

Answer 1

${x}^{3} + {y}^{3} = 3axy \\ \frac{ {d}^{2} y}{d {x}^{2} } =( 2 {a}^{3} xy) \div {(ax - {y}^{2}) }^{3} \\ \frac{d}{dx} ( {x}^{3} + {y}^{3} = 3axy) \\ 3 {x}^{2} + 3 {y}^{2} \frac{dy}{dx} = 3a(y + x \frac{dy}{dx} ) \\ \frac{dy}{dx} (3 {y}^{2} - 3ax) = 3ay - 3 {x}^{2} \\ \frac{dy}{dx} = \frac{ay - {x}^{2} }{ {y}^{2} - ax } \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{( {y}^{2} - ax)(a \frac{dy}{dx} - 2x) - (ay - {x}^{2} )(2y \frac{dy}{dx} - a)}{ {( {y}^{2} - ax) }^{2} } \\ = \frac{ \frac{dy}{dx}(a {y}^{2} - {a}^{2} x - 2a {y}^{2} + 2 {x}^{2} y) - 2x {y}^{2} + 2a {x}^{2} + {a}^{2}y - a {x}^{2} }{{( {y}^{2} - ax)}^{2} } \\ = \frac{(ay - {x}^{2} )(a {y}^{2} - {a}^{2} x - 2a {y}^{2} + 2 {x}^{2}y) - ( {y}^{2} - ax)(2x {y}^{2} - 2a {x}^{2} - {a}^{2}y + a {x}^{2}) }{ {( {y}^{2} - ax) }^{3} } \\ = \frac{ - 2 {a}^{3}xy }{ { ({y}^{2} - ax) }^{3} } \\ = \frac{2 {a}^{3}xy }{ {(ax - {y}^{2}) }^{3} }$