Characterization of Differentiability

Theorem
Let $f: \R^n \to \R, \mathbf x \mapsto f\left({\mathbf x}\right)$ be a real-valued function.

Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \R^n$.

Let $\Delta f\left({\mathbf x}\right) = f\left({\mathbf x + \Delta \mathbf x}\right) - f \left({\mathbf x}\right)$

Let $\dfrac {\partial f}{\partial x_j}$ is the partial derivative of $f$ WRT $x_j$.

Then $f$ is differentiable iff there is some $\Delta f\left({\mathbf x}\right)$ such that:

where $\forall i: 1 \le i \le n: \varepsilon_i \to 0$ as $\Delta x_i \to 0$.

Proof
Suppose $f: \R^1 \to \R$.

Define:


 * $f\left({x}\right) = y$
 * $\Delta y = f\left({x + \Delta x}\right) - \Delta x$

From the definition of the derivative of a real function, we can say that $f$ is differentiable iff:

Clearly, this is equivalent to saying that $f$ is differentiable iff:

where $\varepsilon$ is some real number.

Solving this equation for $\Delta y$:

That is, iff the real function is differentiable, $\varepsilon \to 0$ as $\Delta x \to 0$.

Now consider $f:\R^n \to \R$, $n > 1$.

From the definition of differentiability of a real-valued function, $f$ is differentiable iff:

such that $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ as $\Delta \mathbf x \to \mathbf 0$.

Observe that:

where $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \begin{bmatrix} \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$ as $\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \to \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$.