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)
=
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