f(x)=x^3-(4/x)
find [f^(-1)(x)]' at x=6
If g = f^-1,
f(g) = 1
f'(g)g' = 1
g' = 1/f'(g)
so, (f^-1(x))' = 1/(3x^2+4/x^2)
now evaluate at x=6
find [f^(-1)(x)]' at x=6
f(g) = 1
f'(g)g' = 1
g' = 1/f'(g)
so, (f^-1(x))' = 1/(3x^2+4/x^2)
now evaluate at x=6