Given that
[tex]g(x) = x^3 + x[/tex]
the inverse [tex]g^{-1}(x)[/tex] is such that
[tex]g\left(g^{-1}(x)\right) = g^{-1}(x)^3 + g^{-1}(x) = x[/tex]
or
[tex]g\left(f(x)\right) = f(x)^3 + f(x) = x[/tex]
Differentiating both sides using the chain rule gives
[tex]3f(x)^2f'(x) + f'(x) = 1 \\\\ f'(x) \left(3f(x)^2+1\right) = 1 \\\\ f'(x) = \dfrac1{3f(x)^2+1}[/tex]
Then the derivative of f at 2 is
[tex]f'(2) = \dfrac1{3f(2)^2+1} = \boxed{\dfrac14}[/tex]
