If x= _e ( y 2 - y 2 y 2 + y 2 ), y 2 = 1-t 1+t . Then (y₁ )_ t= 1 2 has the value

Step-by-step solution

aligned & x= _ e ( y 2 - y 2 y 2 + y 2 ) & e ^x= 1- y 2 1+ y 2 aligned e ^x= ( 4 - y 2 ) ....(i) [ 4 =1 ] Differentiating w.r.t. x , we get e ^x= ^2 ( 4 - y 2 ) ( -1 2 ) d y ~d x d y ~d x =-2 e ^x ^2 ( 4 - y 2 ) aligned & When t = 1 2 , & y 2 = 1- 1 2 1+ 1 2 & y 2 = 1 3 & y 2 = 6 aligned Substituting y 2 = 6 in (i), we get e ^x= 12 =2- 3 ( d y ~d x )_ t = 1 2 =-2(2- 3 ) ^2 12 aligned & =-2(2- 3 ) ( 3 +1 2 2 )^2 & = -1 2 (2- 3 )(2+ 3 ) & =- 1 2 aligned