Differentiable Functions from Rn to Rm and Their Total Derivatives

Recall from the Differentiability and the Total Derivative of Functions from Rn to Rm page that if $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in S$, 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 $T_c(h) : \mathbb{R}^m \to \mathbb{R}^n$ called the total derivative of $\mathbf{f}$ at $\mathbf{c}$ such 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 we have that $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.

We will now look at a nice theorem which tells us that if $\mathbf{f}$ is differentiable at $\mathbf{c}$ then all of the directional derivatives of $\mathbf{f}$ at $\mathbf{c}$ exist and the value of the total derivative at $\mathbf{u}$ will be that directional derivative.

Theorem 1: Let $S \subseteq \mathbb{R}^n$, $\mathbf{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is differentiable at $\mathbf{c}$ then all directional derivatives of $\mathbf{f}$ at $\mathbf{c}$ exist, that is, $\mathbf{f}'(\mathbf{c}, \mathbf{u})$ exists for all $\mathbf{u} \in \mathbb{R}^n$ and if $T_c : \mathbb{R}^m \to \mathbb{R}^n$ is the total derivative of $\mathbf{f}$ then $T_c(\mathbf{u}) = \mathbf{f}'(\mathbf{c}, \mathbf{u})$.

• Proof: Suppose that $\mathbf{f}$ is differentiable at $\mathbf{c}$ and let $T_c : \mathbb{R}^m \to \mathbb{R}^n$ be the total derivative of $\mathbf{f}$ at $c$. Then we have that:

(2)

\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 we have that $\mathbf{E}_{\mathbf{c}}(\mathbf{v}) \to 0$ as $\mathbf{v} \to 0$. Let $\mathbf{v} = h\mathbf{u}$. Since $\mathbf{T}_{\mathbf{c}}$ is a linear function we have that $\mathbf{T}_{\mathbf{c}}(h\mathbf{u}) = h \mathbf{T}_{\mathbf{c}}(\mathbf{u})$. Also, note that $\| h\mathbf{u} \| = \mid h \mid \| \mathbf{u} \|$. So:

(3)

\begin{align} \quad \mathbf{f}(\mathbf{c} + h\mathbf{u}) & = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}} (h\mathbf{u}) + \| h\mathbf{u} \| \mathbf{E}_{\mathbf{c}}(h\mathbf{u}) \\ \quad \mathbf{f}(\mathbf{c} + h\mathbf{u}) & = \mathbf{f}(\mathbf{c}) + h\mathbf{T}_{\mathbf{c}} (\mathbf{u}) + \mid h \mid \| \mathbf{u} \| \mathbf{E}_{\mathbf{c}}(h\mathbf{u}) \end{align}

• We rearrange the terms in this equation to get:

(4)

\begin{align} \quad \mathbf{T}_{\mathbf{c}} (\mathbf{u}) = \frac{\mathbf{f}(\mathbf{c} + h \mathbf{u}) - \mathbf{f} (\mathbf{c})}{h} - \frac{\mid h \mid}{h} \| \mathbf{u} \| \mathbf{E}_{\mathbf{c}}(h\mathbf{u}) \end{align}

• This holds for all $h$. So we take the limit as $h \to 0$ of both sides noting that $\frac{\mid h \mid}{h}$ is bounded as as $h \to 0$, $\| u \|$ is fixed, $h\mathbf{u} \to 0$ so $\mathbf{E}_{\mathbf{c}} (h\mathbf{u}) \to 0$ and $\displaystyle{\frac{\mid h \mid}{h} \| \mathbf{u} \| \mathbf{E}_{\mathbf{c}}(h\mathbf{u}) \to 0}$, i.e.:

(5)

\begin{align} \quad \mathbf{T}_{\mathbf{c}} (\mathbf{u}) &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h \mathbf{u}) - \mathbf{f} (\mathbf{c})}{h} \\ &= \mathbf{f}'(\mathbf{c}, \mathbf{u}) \quad \blacksquare \end{align}