If y= (x+ 1+x² )^ n , then (1+x² ) d² y d x² +x d y d x is

Step-by-step solution

y= (x+ 1+x² )^ n array c d y d x =n (x+ 1+x² )^ n-1 (1+ 1 2 (1+x² )^ -1 / 2 2 x ) array c d y d x =n (x+ 1+x² )^ n-1 ( 1+x² +x ) 1+x² = n ( 1+x² +x )^ n 1+x² array array or 1+x² d y d x =n y or 1+x² y₁=n y (y₁= d y d x ) Squaring, (1+x² ) y₁²=n² y² Differentiating, (1+x² ) 2 y₁ y₂+y₁² 2 x=n² 2 y y₁ or (1+x² ) y₂+x y₁=n² y