If y is a function of x and (x+y)=2 x y , then the value of y^ (0) is

Step-by-step solution

(x+y)=2 x y Differentiating w.r.t. x , we get aligned & 1 x+y (1+ d y ~d x )=2 x d y ~d x +2 y & 1 x+y + 1 (x+y) d y ~d x =2 x d y ~d x +2 y & ( 1 x+y -2 x ) d y ~d x =2 y- 1 x+y & d y ~d x ( 1 x+y -2 x )=2 y- 1 x+y & d y ~d x = (2 y- 1 x+y ) ( 1 x+y -2 x ) & For x=0, (y)=0 & y=1 & . d y ~d x |_ (0,1) = (2- 1 0+1 ) ( 1 0+1 -0 ) =1 aligned