To solve this problem, the following theorem needs to be used:

 

If f is a differentiable function on an interval, and if f has an inverse function g , then  g is differentiable at any x for which f’(g(x) 0 and g’(x)=.

 

Here f(x)= (2x-π)3 +2x-cosx and it can be readily seen that at x=π/2, f(x)= π

Or,     f(π/2)= π

 

Or,    f-1(π)= π/2

 

Applying the theorem,

 

                                                                                =  (x=π/2)

 

                                                                               = 

 

                                                                     ****