If `e^(y) +xy = e`, the ordered pair ` ((dy)/(dx),(d^(2)y)/(dx^(2)))` at x = 0 is equal to

Options

A. `(1/e,1/e^(2))`
B. `(-1/e,1/e^(2))`
C. `(1/e,1/e^(2))`
D. `(-1/e, -1/e^(2))`

Answer

B

Explanation

Key Idea Differentiating the given equation twice w.r.t. 'x'.
Given equation is
`e^(y)+xy=e` …(i)
On differentiating both sides w.r.t.x, we get
`" "e^(y)(dy)/(dx) + x (dy)/(dx) + y= 0` ....(ii)
`rArr" " (dy)/(dx) =- (y/(e^(y)+x)) ` ...(iii)
Again differentiating Eq. (ii) w.r.t.t. 'x', we get
`e^(y)(d^(2)y)/(dx^(2))+e^(y)((dy)/(dx))^(2) + x(d^(2)y)/(dx^(2)) + (dy)/(dx) + (dy)/(dx) = 0 ` ...(iv)
Now, on putting x = 0 in Eq. (i), we get
` " "e^(y) = e^(1) `
` rArr" " y = 1`
On putting x = 0, y = 1 in Eq. (iii) , we get
`(dy)/(dx) =- 1/(e+0) =- 1/e`
Now, on putting x = 0, y=1 and `(dy)/(dx) =- 1/e ` in Eq. (iv), we get
`e^(1) (d^(2)y)/(dx^(2))+e^(1)(-1/e)^(2)+0((d^(2)y)/(dx^(2)))+(-1/e)+(-1/e) =0`
` rArr" "(d^(2)y)/(dx^(2))underset("(0,1)")(|)=1/e^(2)`
So, `" "((dy)/(dx),(d^(2)y)/(dx^(2)))" at " (0, 1)" is "(-1/e,1/e^(2))`