Question 1034949
{{{f(x) = x^3 +2x }}} ==> {{{f(f^-1(x)) = x = (f^-1(x))^3 + 2f^-1(x) }}} after plugging in the inverse {{{f^-1(x)}}} into the original equation.

Now differentiate implicitly wrt x.  

==> {{{1 = 3*(f^-1(x))^2*(df^-1(x)/dx) + 2(df^-1(x)/dx) =  (3*(f^-1(x))^2 + 2)(df^-1(x)/dx)  }}} 

Now {{{f(x) = x^3 +2x }}} ==> {{{df(x)/dx = 3x^2 +2 > 0 }}} for all real x hence f(x) increasing ==> f(x) is one-to-one ==> f(x) has an inverse function.  Incidentally, f(1) = 3, and so {{{f^-1(3) = 1}}}.


==> {{{1 =  (3*(f^-1(3))^2 + 2)(df^-1(3)/dx)  }}} 

==> {{{1 = (3*1^2 + 2)(df^-1(3)/dx) = 5(df^-1(3)/dx)}}}

Therefore, {{{df^-1(3)/dx = 1/5}}}