f(x)=3+x2+tan(½πx).
If g(x)=f-1(x), then f(g(x))=g(f(x))=x.
Assuming that we need g'(f(x)) (inverse prime) then differentiate g(f(x)) wrt x:
(d/dx)(g(f(x)))=g'(f(x))f'(x).
But g(f(x))=x, so g'(f(x))f'(x)=1 and:
g'(f(x))=1/f'(x).
f'(x)=2x+½πsec2(½πx), so g'(f(x))=1/(2x+½πsec2(½πx)).
g'(f(a))=1/(2a+½πsec2(½πa)). When a=3:
g'(f(3))=1/(6+½πsec2(3π/2)). cos(3π/2)=0 so sec(3π/2)→∞.
Therefore inverse prime=0.