Let y = ⁻¹ ( 1-x^2 1+x^2 ) - ⁻¹ ( 2x 1+x^2 ) . The set of all values of x for which dy dx = 0 is

Step-by-step solution

We know the principal value branches of the given inverse trigonometric functions: ⁻¹ ( 1-x^2 1+x^2 ) = cases 2 ⁻¹x, & x 0 -2 ⁻¹x, & x ⁻¹ ( 2x 1+x^2 ) = cases - - 2 ⁻¹x, & x 1 cases Now, we evaluate y = ⁻¹ ( 1-x^2 1+x^2 ) - ⁻¹ ( 2x 1+x^2 ) in different intervals: Case 1: x (- , -1) y = (-2 ⁻¹x) - (- - 2 ⁻¹x) = Since y is a constant, dy dx = 0 . Case 2: x (-1, 0) y = (-2 ⁻¹x) - (2 ⁻¹x) = -4 ⁻¹x Here, dy dx = -4 1+x^2 0 . Case 3: x (0, 1) y = (2 ⁻¹x) - (2 ⁻¹x) = 0 Since y is a constant, dy dx = 0 . Case 4: x (1, ) y