Differentiability of $\Psi(x) = \frac{1}{\phi(x)}$, $\phi: G \rightarrow \mathbb{R}$, $G \subset \mathbb{R}^n$?

Question

How to infer the differentiability of

$$\Psi(x) = \frac{1}{\phi(x)}$$

where $\phi: G \rightarrow \mathbb{R}$, $G \subset \mathbb{R}^n$.

Can this be done with the one-dimensional definition of derivative, i.e.

$$\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$

since $\phi (x) \in \mathbb{R}$?

Answer 1

Let $\theta: t \in \Bbb R \setminus \{0\} \mapsto \frac1t$. Then $\psi = \theta \circ \phi$, and provided $\phi$ is never zero, i.e. that $\phi(G) \subset \Bbb R \setminus \{0\}$, we get that $\theta$ is differentiable on $\phi(G)$, hence $\psi$ is differentiable on $G$ and by chain rule,

$$D \psi(x) = \theta'(\phi(x)) D \phi(x) = -\frac1{(\phi(x))^2} D \phi(x)$$