Differentiable Functions from Rn to Rm and Their Total Derivatives

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}