Let y = y (x) be the implicit function determined by the functional equation E ^ (x + y) = 2 + X + 2Y at points (1, - 1), and find y "| (1, - 1) and Let y = y (x) be the implicit function determined by the functional equation E ^ (x + y) = 2 + X + 2Y at points (1, - 1), and find the quadratic differential of Y "| (1, - 1) and dy

Let y = y (x) be the implicit function determined by the functional equation E ^ (x + y) = 2 + X + 2Y at points (1, - 1), and find y "| (1, - 1) and Let y = y (x) be the implicit function determined by the functional equation E ^ (x + y) = 2 + X + 2Y at points (1, - 1), and find the quadratic differential of Y "| (1, - 1) and dy

E ^ (x + y) = 2 + X + 2Y, both sides of X are derived at the same time
E ^ (x + y) * (1 + y ') = 1 + 2Y', take y (1) = - 1; get 1 + y '(1) = 1 + 2Y' (1); then y '(1) = 0
At the same time on both sides of the derivation
E ^ (x + y) * (1 + y ') ^ 2 + e ^ (x + y) * y' '= 2Y' ', y (1) = - 1, y' (1) = 0 is brought in
Then: 1 + y '' (1) = 2Y '' (1),
That is, y '' (1) = 1
The quadratic differential of Dy is equal to e ^ (x + y) * (1 + y ') ^ 2 + e ^ (x + y) * y' '= 2Y' 'and e ^ (x + y) * (1 + y') = 1 + 2Y 'and y' can be eliminated~