Definition:Differentiable Mapping

At a Point
Let $f$ be a real function defined on an open interval $I$.

Let $\xi \in I$ be a point in $I$.

Then $f$ is differentiable at the point $\xi$ iff the limit:
 * $\displaystyle \lim_{x \to \xi} \frac {f \left({x}\right) - f \left({\xi}\right)} {x - \xi}$

exists.

This limit, if it exists, is called the derivative of $f$ at $\xi$.

On an Open Interval
Let $f$ be a real function defined on an open interval $I$.

Let $f$ be differentiable at each point of $I$.

Then $f$ is differentiable on $I$.

On a Closed Interval
Let $f$ be differentiable on $I=(a..b)$ as defined above.

If the following limit from the right exists:


 * $\displaystyle \lim_{x \to a^+} \frac {f \left({x}\right) - f \left({a}\right)} {x - a}$

as well as this limit from the left:
 * $\displaystyle \lim_{x \to b^-} \frac {f \left({x}\right) - f \left({b}\right)} {x - b}$

then $f$ is differentiable on the closed interval $[a..b]$.

On the Real Number Line
In the definition of differentiable on an interval let that interval be the real number line $\R$.

Let $f$ be differentiable at each point of $\R$.

Then $f$ is differentiable everywhere (on $\R$).

In the Complex Plane
Let $f \left({z}\right): \C \to \C$ be a single-valued continuous complex function in a domain $D \subseteq \C$.

Let $z_0 \in D$ be a point in $D$.

Then $f \left({z}\right)$ is complex-differentiable at $z_0$ iff the limit:
 * $\displaystyle \lim_{h \to 0} \frac {f \left({z_0+h}\right) - f \left({z_0}\right)} h$

exists as a finite number and is independent of how the complex increment $h$ tends to $0$.

If such a limit exists, it is called the derivative of $f$ at $z_0$.

If $f \left({z}\right)$ is complex-differentiable at every point in $D$, it is differentiable in $D$. Such a function is called analytic.

At a Point
Let $f: \mathbb X \to \R^n$ be a real-valued function, where $\mathbb X \subseteq \R$.

Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb X$.

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

where $\Delta \mathbf x = \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \\ \vdots \\ \Delta x_n \end{bmatrix}$.

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

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

Here:


 * The limit being taken is the limit of a neighborhood
 * $\nabla$ is the gradient operator
 * $\bullet$ is the dot product
 * $\mathbf 0$ is the zero vector with $n$ entries
 * $\varepsilon_i$ is some real number

In a Region
Let $S \subseteq \mathbb X$.

We say that $f$ is differentiable in a region $S$ iff $f$ is differentiable at each $\mathbf x$ in $S$.

See Characterization of Differentiability for an explanation of these definitions.