If _e y=3 ⁻¹ x , then (1-x^2 ) y^ -x y^ at x= 1 2 is equal to

Step-by-step solution

aligned & (y)=3 ⁻¹ x & 1 y y^ =3 ( 1 1-x^2 ) & y^ = 3 y 1-x^2 at x= 1 2 & y^ = 3 e^ 3 ( 6 ) 3 2 =2 3 e^ 2 & y^ =3 ( 1-x^2 y^ -y 1 2 1-x^2 (-2 x) (1-x^2 ) ) aligned aligned & (1-x^2 ) y^ =3 (3 y+ x y 1-x^2 ) & at x= 1 2 , y=e^ 3 ⁻¹ ( 1 2 ) = e ^ 3 ( 6 ) = e ^ 2 aligned aligned & . (1-x^2 ) y^ |_ at x= 1 2 =3 (3 e ^ 2 + 1 2 (e^ 2 ) 3 2 ) & =3 e ^ 2 (3+ 1 3 ) & (1-x^2 ) y^ - .x y^ |_ atx x= 1 2 & =3 e ^ 2 (3+ 1 3 )- 1 2 (2 3 e ^ 2 )=9 e ^ 2 aligned