If e ^ y +x y = e , then the ordered pair ( dy d x , ~d ^2 y d x^2 ) at x=0 is equal to

Step-by-step solution

Given the equation e^y + xy = e , we determine the ordered pair ( dy dx , d^2y dx^2 ) at x=0 . Substituting x=0 into the equation yields e^y = e , so y = 1 . Implicit differentiation gives e^y dy dx + y + x dy dx = 0 , so dy dx = - y e^y + x . At (x, y) = (0, 1) , dy dx = - 1 e . Differentiating again using the quotient rule, d^2y dx^2 = dy dx (e^y + x) + y(e^y dy dx + 1) (e^y + x)^2 . Substituting values yields d^2y dx^2 = -1 e^2 . The ordered pair is (- 1 e , - 1 e^2 ) . The provided options are single numerical