Then, suppose that the first derivative f ′ is invertible and let the inverse be g = ( f ′ ) − − 1 . Then for each p , the point g ( p ) is the unique critical point x ¯ ¯ of the function x ↦ ↦ p x − − f ( x ) (i.e., x ¯ ¯ = g ( p ) ) because f ′ ( g ( p ) ) = p and the function's first derivative with respect to x at g ( p ) is p − − f ′ ( g ( p ) ) = 0 . Hence we have f ∗ ∗ ( p ) = p ⋅ ⋅ g ( p ) − − f ( g ( p ) ) for each p . By differentiating with respect to p , we find ( f ∗ ∗ ) ′ ( p ) = g ( p ) + p ⋅ ⋅ g ′ ( p ) − − f ′ ( g ( p ) ) ⋅ ⋅ g ′ ( p ) . Since f ′ ( g ( p ) ) = p this simplifies to ( f ∗ ∗ ) ′ ( p ) = g ( p ) = ( f ′ ) − − 1 ( p ) . In other words, ( f ∗ ∗ ) ′ and f ′ are inverses to each other.