If `e^y+xy=e` then the value of `(d^2y)/(dx^2)` for `x=0` isA. `e^(-1)`B. `e^(-2)`C. eD. 1

If `e^y+xy=e` then the value of `(d^2y)/(dx^2)` for `x=0` isA. `e^(-1)

1.	If `e^y+xy=e` then the value of `(d^2y)/(dx^2)` for `x=0` isA. `e^(-1)`B. `e^(-2)`C. eD. 1
Answer» Correct Answer - B Given `e^(y)+xy=e` On differentiating w.r.t. x, we get `e^(y)(dy)/(dx)+y+x(dy)/(dx)=0" …(i)"` At x = 0 we get `e^(y)+0.y=e rArr e^(y)=e rArr y=1` `therefore` By putting y = 1 in equation (i) we get `e(dy)/(dx)+1+0=0` `rArr(dy)/(dx)=-(1)/(e)` Again differentiating Eq. (i), we get `e^(y)(d^(2)y)/(dx^(2))+e^(y)((dy)/(dx))^(2)+(dy)/(dx)+x(d^(2)y)/(dx^(2))+(dy)/(dx)=0` `rArr(d^(2)y)/(dx^(2))(e^(y)+x)+e^(y)((dy)/(dx))^(2)+(2dy)/(dx)=0` Now, At x = 0, y = 1 `(d^(2)y)/(dx^(2))(e+0)+e(-(1)/(e))^(2)+2(-(1)/(e))=0` `rArr e(d^(2)y)/(dx^(2))-(1)/(e)=0` `rArr (d^(2)y)/(dx^(2))=(1)/(e^(2))=e^(-2)`