If u = ( x-1 - x+1 ) and v = x+1 + x-1 then du dv = .

Step-by-step solution

The function v is given by v = x+1 + x-1 , with the condition that x 1 for real-valued square roots. Its product with the conjugate expression gives: v( x+1 - x-1 ) = ( x+1 )^2 - ( x-1 )^2 = (x+1) - (x-1) = 2 Solving for the conjugate, x+1 - x-1 = 2 v . The original expression for u , u = ( x-1 - x+1 ) , simplifies as: x-1 - x+1 = -( x+1 - x-1 ) = - 2 v Assuming real-valued functions, the logarithm of a negative number is undefined. Given that this is a derivative problem, we proceed with the absolute value or cons