If y= ⁻¹ ( 2 x 1+x^2 )+ ⁻¹ ( 1+x^2 1-x^2 ) then the value of dy d x at x= 3 is

Step-by-step solution

Given y = ⁻¹ ( 2x 1 + x^2 ) + ⁻¹ ( 1 + x^2 1 - x^2 ) , we aim to find dy dx at x = 3 . Let x = with = ⁻¹ x . For the first term, A = ⁻¹ ( 2x 1 + x^2 ) = ⁻¹( 2 ) . Since x > 1 implies > 4 , then 2 ( 2 , ) , so A = - 2 ⁻¹ x . For the second term, B = ⁻¹ ( 1 + x^2 1 - x^2 ) = ⁻¹ ( 1 - x^2 1 + x^2 ) = ⁻¹( 2 ) . For x 0 , 2 [0, ) , so B = 2 ⁻¹ x . Thus, at x = 3 (where x > 1 ), y = ( - 2 ⁻¹ x) + 2 ⁻¹ x = . Differentiating, dy dx = 0 since is constant. Solution: dy dx = 0 . 0