If f (x)= ^2 x 1+ x + ^2 x 1+ x , then the value of f ^ ( 6 ) is equal to

Step-by-step solution

Simplify the function: f (x) = ^2 x 1+ x + ^2 x 1+ x . Express x and x in terms of sine and cosine: f (x) = ^2 x 1 + x x + ^2 x 1 + x x = ^2 x x + x x + ^2 x x + x x = ^3 x x + x + ^3 x x + x . Combine terms: f (x) = ^3 x + ^3 x x + x . Apply the sum of cubes identity: f (x) = ( x + x)( ^2 x - x x + ^2 x) x + x . Cancel the common factor (valid for x + x 0 ): f (x) = ^2 x + ^2 x - x x = 1 - x x . Use the double angle identity (2x) = 2 x x : f (x) = 1 - 1 2 (2x) . Differentiate: f '(x) = d dx (1 - 1 2 (2x) ) = - 1 2