If y = 10 ( ⁻¹ ( 1+x^2 -1 x ) )^2 , where x 0 , then the value of (1+x^2)^2 y'' + 2x(1+x^2)y' is

Step-by-step solution

Given y = 10 ( ⁻¹ ( 1+x^2 -1 x ) )^2 Put x = , so = ⁻¹x . The inner expression simplifies as: 1+ ^2 -1 = - 1 = 1 - Using half-angle formulas: 2 ^2( /2) 2 ( /2) ( /2) = ( 2 ) So, ⁻¹ ( ( 2 ) ) = 2 = 1 2 ⁻¹x Substitute this back into the equation for y : y = 10 ( 1 2 ⁻¹x )^2 = 10 4 ( ⁻¹x)^2 = 5 2 ( ⁻¹x)^2 Differentiating with respect to x : y' = 5 2 2 ⁻¹x 1 1+x^2 = 5 ⁻¹x 1+x^2 Rearranging gives: (1+x^2)y' = 5 ⁻¹x Squaring both sides to avoid the quotient rule: (1+x^2)^2 (y')^2 = 25( ⁻¹x)^2 Since y = 5 2 ( ⁻¹x)^2 , we