If $f: (a, b) \to \mathbb{R}^{1}$ and if $f'(x_{0})$ exists for $x_{0} \in (a, b)$, i.e., the limit $$ \lim_{ x \to x_{0} } \frac{f(x) - f(x_{0})}{x - x_{0}} $$ exists, then let $h = x - x_{0}$, we have the following $$ \lim_{ h \to 0 } \frac{f(x_{0} + h) - f(x_{0}) - hf'(x_{0})}{h} = 0. $$ For a scalar-valued single variable function, the existence of the derivative is equivalent to say that $$ \frac{f(x_{0} + h) - \left( f(x_{0}) + hf'(x_{0}) \right)}{h} \to 0 $$ with respect to $h$.

Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, and $\mathbf{x} \in E$. If there exists a linear transformation $A$ of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ such that $$ \lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0, $$ then we say $\mathbf{f}$ is differentiable at $\mathbf{x}$, and we write $$ \mathbf{f}'(\mathbf{x}) = A. $$ If $\mathbf{f}$ is differentiable at every $\mathbf{x} \in E$, we say that $\mathbf{f}$ is differentiable in $E$.

Suppose $E$ and $\mathbf{f}$ are as in Definition 911, $\mathbf{x} \in E$, and $$ \lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0 $$ holds with $A = A_{1}$ and with $A = A_{2}$. Then $A_{1} = A_{2}$.

When $\vec{h} \to 0$, $$ \begin{align*} \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) \rVert &= \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) - A \vec{h} + A \vec{h} \rVert \\ &\leq \lVert \vec{F}(\vec{x} + \vec{h}) - \vec{F}(\vec{x}) - A \vec{h} \rVert + \lVert A \vec{h} \rVert \\ &\leq \lVert \vec{h} \rVert + \lVert A \rVert \cdot \lVert \vec{h} \rVert \\ &= (\lVert A \rVert + 1)\lVert \vec{h} \rVert. \end{align*} $$ Hence $\vec{F}$ is continuous at $\vec{x}$, provided that $\vec{F}$ is differentiable at $\vec{x}$.

Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable at $\mathbf{x}_{0} \in E$, $\mathbf{g}$ maps an open set containing $\mathbf{f}(E)$ into $\mathbb{R}^{k}$, and $\mathbf{g}$ is differentiable at $\mathbf{f}(\mathbf{x}_{0})$. Then the mapping $\mathbf{F}$ of $E$ into $\mathbb{R}^{k}$ defined by $$ \mathbf{F}(\mathbf{x}) = \mathbf{g}\left( \mathbf{f}(\mathbf{x}) \right) $$ is differentiable at $\mathbf{x}_{0}$, and $$ \mathbf{F}'(\mathbf{x}_{0}) = \mathbf{g}'\left( \mathbf{f}(\mathbf{x}_{0}) \right) \mathbf{f}'(\mathbf{x}_{0}). $$

Consider $\mathbf{f}: E \to \mathbb{R}^{m}$ where $E \subset \mathbb{R}^{n}$, Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ and $\{ \mathbf{u}_{1}, \dots, \mathbf{u}_{m} \}$ be the standard bases of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$. The components of $\mathbf{f}$ are the real functions $f_{1}, \dots, f_{m}$ defined by $$ \mathbf{f}(\mathbf{x}) = \sum_{i=1}^{m} f_{i}(\mathbf{x}) \mathbf{u}_{i} \qquad (\mathbf{x} \in E), $$ or equivalently, by $f_{i}(\mathbf{x}) = \mathbf{f}(\mathbf{x}) \cdot \mathbf{u}_{i}, 1 \leq i \leq m$.

For, $\mathbf{x} \in E$, $1 \leq i \leq m, 1 \leq j \leq n$, we define $$ (D_{j}f_{i})(\mathbf{x}) = \lim_{ t \to 0 } \frac{f_{i}(\mathbf{x} + t \mathbf{e}_{j}) - f_{i}(\mathbf{x}) }{t}, $$ provided the limit exists. Writing $f_{i}(x_{1}, \dots, x_{n})$ in place of $f_{i}(\mathbf{x})$, we see that $D_{j}f_{i}$ is the derivative of $f_{i}$ with respect to $x_{j}$, keeping the other variables fixed. The notation $$ \frac{\partial f_{i}}{\partial x_{j}} $$ is therefore used in place of $D_{j}f_{i}$, and $D_{j}f_{i}$ is called a partial derivative.

Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, and $\mathbf{f}$ is differentiable at a point $\mathbf{x} \in E$. Then the partial derivative $(D_{j}f_{i})(\mathbf{x})$ exist, and $$ \mathbf{f}'(\mathbf{x})\mathbf{e}_{j} = \sum_{i=1}^{m}(D_{j}f_{i})(\mathbf{x})\mathbf{u}_{i} \qquad (1 \leq j \leq n). $$

If $\mathbf{F}(\mathbf{x})$ is a scalar-valued function, then $$D \mathbf{F}(\mathbf{x}) = \begin{bmatrix}D_{1}\mathbf{F} & D_{2}\mathbf{F} & \cdots & D_{n}\mathbf{F}\end{bmatrix}$$ It’s also called the gradient of $\mathbf{F}$.

Suppose $\mathbf{f}$ maps a convex open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable in $E$, and there is a real number $M$ such that $$ \lVert \mathbf{f}'(\mathbf{x}) \rVert \leq M $$ for every $\mathbf{x} \in E$. Then $$ \lvert \mathbf{f}(\mathbf{b}) - \mathbf{f}(\mathbf{a}) \rvert \leq M \lvert \mathbf{b} - \mathbf{a} \rvert $$ for all $\mathbf{a} \in E, \mathbf{b} \in E$.

A differentiable mapping $\mathbf{f}$ of an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ is said to be continuously differentiable in $E$ if $\mathbf{f}'$ is a continuous mapping of $E$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.

More explicitly, it is required that to every $\mathbf{x} \in E$ and to every $\epsilon > 0$ corresponds a $\delta > 0$ such that $$ \lVert \mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x}) \rVert < \epsilon $$ if $\mathbf{y} \in E$ and $\lvert \mathbf{x} - \mathbf{y} \rvert < \delta$.

If this is so, we also say that $\mathbf{f}$ is a $\mathscr{C}'$-mapping, or that $\mathbf{f} \in \mathscr{C}'(E)$.

Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$. Then $\mathbf{f} \in \mathscr{C}'(E)$ if and only if the partial derivatives $D_{j}f_{i}$ exist and are continuous on $E$ for $1 \leq i \leq m, 1 \leq j \leq n$.