The Total Derivative of a Linear Function from Rn to Rm

# The Total Derivative of a Linear Function from Rn to Rm

Recall from the Differentiability and the Total Derivative of Functions from Rn to Rm page that if $S \subseteq \mathbb{R}^n$, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ then $\mathbf{f}$ is said to be differentiable at $\mathbf{c}$ if there exists a linear function $\mathbf{T}_{\mathbf{c}} : \mathbb{R}^n \to \mathbb{R}^m$ called the total derivative of $\mathbf{f}$ at $\mathbf{c}$ such that for all $\mathbf{v} \in \mathbb{R}^n$ we have that:

(1)
\begin{align} \quad \mathbf{f}(\mathbf{c} +\mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}} (\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) \end{align}

Where $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.

Now suppose that $\mathbf{f}$ is a linear function. Then the total derivative of $\mathbf{f}$ at any defined point $\mathbf{c}$ is extremely easy to determine as we will see in the following proposition.

 Proposition 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is a linear function defined at $\mathbf{c}$ then $\mathbf{f}$ is differentiable at $\mathbf{c}$ and the total derivative of $\mathbf{f}$ at $\mathbf{c}$ is $\mathbf{f}$ itself. In other words, for all $\mathbf{v} \in \mathbb{R}^n$, $\mathbf{T}_{\mathbf{c}} (\mathbf{v}) = \mathbf{f}(\mathbf{v})$.
• Proof: Let $\mathbf{f}$ be a linear function. Then for any $\mathbf{v} \in \mathbb{R}^n$ we have that:
(2)
\begin{align} \quad \mathbf{f} (\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{f}(\mathbf{v}) \end{align}
• Set $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = \mathbf{f}(\mathbf{v})$ and $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) = \mathbf{0}$. Then we have that:
(3)
\begin{align} \quad \mathbf{f} (\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}}(\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v}) \end{align}
• Where $\mathbf{T}_{\mathbf{c}} = \mathbf{f}$ is a linear function and $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$. So $\mathbf{f}$ is differentiable at $\mathbf{c}$ and for all $\mathbf{v} \in \mathbb{R}^n$:
(4)