Verified answer

Note that f(0) = 6 + 0^8 + tan(pi*0/2) = 6.

From a general formula relating the derivative of a function to the derivative of its inverse,

(f^(-1))'(a) = 1/f'(f^(-1)(a))

so in particular

(f^(-1))'(6) = 1/f'(f^(-1)(6)).

What is f^(-1)(6)? It's the value of x that makes f(x) = 6. If we just stare at the formula for f(x) it's clear that f(0) = 6 + 0^8 + tan(pi*0/2) = 6 + 0 + 0 = 6, so that f^(-1)(6) = 0. [They've deliberately made f(x) so complicated that you can't generally solve the equation y = f(x) for y, so we really have to depend on the fact that we can "just see" that f(0) = 6 here].

So the formula reads

(f^(-1))'(6) = 1/f'(0).

From standard derivative rules,

f'(x) = 8x^7 + sec^2(pi x/2) * pi/2

so that

f'(0) = 8*0 + sec^2(pi*0/2) * pi/2 = 0 + 1*pi/2 = pi/2

so

(f^(-1))'(6) = 1/f'(0) = 2/pi