If log 1 + x 2 - x = y 1 + x 2 , then 1 + x 2 d y d x + x y =

Step-by-step solution

Given, log 1 + x 2 - x = y 1 + x 2 Differentiating, both sides w.r.t x , d d x log 1 + x 2 - x = d d x y 1 + x 2 ⇒ 1 1 + x 2 - x d d x 1 + x 2 - x = 1 + x 2 d y d x + y d 1 + x 2 d x ⇒ 1 1 + x 2 - x x 1 + x 2 - 1 = 1 + x 2 d y d x + x y 1 + x 2 Multiplying by 1 + x 2 both the sides, ⇒ 1 + x 2 × 1 1 + x 2 - x x 1 + x 2 - 1 = 1 + x 2 d y d x + x y 1 + x 2 d y d x + x y = 1 + x 2 × 1 1 + x 2 - x × 1 + x 2 - x 1 + x 2 = - 1