If x= t, y+1= 1 t , then e^ -x d² x d y² + d x d y =

Step-by-step solution

Given x= t and y+1= 1 t d x d t = 1 t and d y d t = -1 t² d x d y = 1 t (-t² )=-t d² x d y² = d d t ( d x d y ) d t d y = d d t (-t) 1 ( d y d t ) = (-1) ( -1 t² ) =t²e^ -x =e^ - t =e^ (t)⁻¹ = 1 t Thus e^ -x d² y d x² + d x d y = ( 1 t ) (t² )+(-t)=t-t=0