Proposition 4

Let E be a complete normed algebra with a unit element over $\mathbf{R}$ and let $\mathbf{f}$ be a function defined on an interval $I \subset \mathbf{R}$, taking values in E, and differentiable at the point $x_0 \in I$. If $y_0 = \mathbf{f}(x_0)$ is invertible $^2$ in E, then the function $x \mapsto (\mathbf{f}(x))^{-1}$ is defined on a neighbourhood of $x_0$ (relative to I), and has a derivative equal to $-(\mathbf{f}(x_0))^{-1} \mathbf{f}'(x_0) (\mathbf{f}(x_0))^{-1}$ at $x_0$.

Indeed, the set of invertible elements in E is an open set on which the function $y \mapsto y^{-1}$ is continuous (Gen. Top., IX, p. 178); since $\mathbf{f}$ is continuous (relative to I) at $x_0$, $(\mathbf{f}(x))^{-1}$ is defined on a neighbourhood of $x_0$, and we have

The proposition thus follows from the continuity of $y^{-1}$ on a neighbourhood of $y_0$ and the continuity of $xy$ on $E \times E$.

$^2$ Recall from (Alg., I, p. 15) that an element $z \in E$ is said to be invertible if there exists an element of $E$, denoted by $z^{-1}$, such that $zz^{-1} = z^{-1}z = e$ (e being the unit element of $E$).