# A Sufficient Condition for the Differentiability of Functions from Rn to Rm

So far we have stated many properties of functions $\mathbf{f} : S \to \mathbb{R}^m$ (where $S \subseteq \mathbb{R}^n$) that are differentiable at a point $\mathbb{c} \in S$. For example, on the Differentiable Functions from Rn to Rm are Continuous page we saw that $\mathbf{f}$ being differentiable at $\mathbf{c}$ implies $\mathbf{f}$ is continuous at $\mathbf{c}$.

We have not yet developed any conditions under which we can determine whether a function $\mathbf{f}$ is differentiable or not (apart from using the definition outright). The following theorem gives us a sufficient (though not necessary) condition for differentiability of a function $\mathbf{f}$ at a point $\mathbf{c}$.

Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and let $\mathbf{f} : S \to \mathbb{R}^m$ where $\mathbf{f} = (f_1, f_2, ..., f_m)$. If at least one of the partial derivatives $D_1 \mathbf{f}$, $D_2 \mathbf{f}$, …, $D_n \mathbf{f}$ exist at $\mathbf{c}$ and if the remaining $n - 1$ partial derivatives are continuous at $\mathbf{c}$ and exist on an open ball centered at $\mathbf{c}$ then $\mathbf{f}$ is differentiable at $\mathbf{c}$. |

Let's look at an example of applying this theorem. Consider the function $f : \mathbb{R}^3 \to \mathbb{R}$ defined for all $(x, y, z) \in \mathbb{R}^3$ by:

(1)We claim that $f$ is differentiable at every point $(x, y, z) \in \mathbb{R}^3$.

First $S = \mathbb{R}^3$ is an open subset of $\mathbb{R}^3$ and $\mathbf{c} = (x, y, z) \in \mathbb{R}^3 = S$.

The three partial derivatives of $f$ are:

(2)So at least one of the partial derivatives of $D_1 f$, $D_2 f$, $D_3 f$ exist at $\mathbf{c} = (x, y, z)$ (they all exist), and the remaining partial derivatives are certainly continuous on an open ball centered at $\mathbf{c} = (x, y, z)$ (in fact all of the partial derivatives above are continuous on every open ball centered at $\mathbf{c} = (x, y, z)$).

So by Theorem 1 we have that $f$ is differentiable at $\mathbf{c}$ for every $\mathbf{c} = (x, y, z) \in \mathbb{R}^3$. Since $\mathbf{c}$ was arbitrary, this means that $f$ is differentiable on all of $\mathbb{R}^3$.