MathematicsEasy63×since 2002Q3697d2xdy2{{{d^2}x} \over {d{y^2}}}dy2d2x equals:A−(d2ydx2)−1(dydx)−3- {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y)−1(dxdy)−3B(d2ydx2)(dydx)−2{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}(dx2d2y)(dxdy)−2C−(d2ydx2)(dydx)−3- \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y)(dxdy)−3D(d2ydx2)−1{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}(dx2d2y)−1Check answerSkip