Differentiability and the Total Derivative of Functions from Rn to Rm

# Differentiability and the Total Derivative of Functions from Rn to Rm

 Definition: Let $S \subseteq \mathbb{R}^n$ be 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 $\mathbf{T}_c : \mathbb{R}^n \to \mathbb{R}^m$ called the Total Derivative of $\mathbf{f}$ such that $\mathbf{f}(\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + \mathbf{T}_{\mathbf{c}} (\mathbf{v}) + \| \mathbf{v} \| \mathbf{E}_{\mathbf{c}} (\mathbf{v})$ where we have that $\mathbf{E}_{\mathbf{c}} (\mathbf{v}) \to \mathbf{0}$ as $\mathbf{v} \to \mathbf{0}$.

The formula above is sometimes referred to as the "First order Taylor formula of $\mathbf{f}$ at $\mathbf{c}$".

The notation "$\mathbf{T}_{\mathbf{c}}(\mathbf{v})$" is sometimes replaced with the notation "$\mathbf{f}'(\mathbf{c})(\mathbf{v})$" where the notation "$\mathbf{f}'(\mathbf{c})$" in this context is denoting the linear function described above.

Recall that a function $T : A \to B$ is said to be Linear Function if for all $x, y \in A$ we have that $T(x + y) = T(x) + T(y)$ (the additivity property) and for all scalars $k$, $T(kx) = kT(x)$ (the homogeneity property).

For example, suppose that $f : S \to \mathbb{R}$ where $S \subseteq \mathbb{R}$ is open. If $f$ is differentiable at a point $c \in S$ then we have that the following limit exists:

(1)
\begin{align} \quad f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \end{align}

So let $E_c(h)$ be defined as follows:

(2)
\begin{align} \quad E_c(h) = \left\{\begin{matrix} \frac{f(c + h) - f(c)}{h} - f'(c) & \mathrm{if} \: h \neq 0\\ 0 & \mathrm{if} \: h = 0 \end{matrix}\right. \end{align}

Then for all $h$ we have that:

(3)
\begin{align} \quad hE_c(h) = f(c + h) - f(c) - hf'(c) \end{align}
(4)
\begin{align} \quad f(c + h) = f(c) + f'(c) h + hE_c(h) \end{align}

Notice that then the total derivative of $f$ is simply $T_c(h) = f'(c)h$, and since $f$ is differentiable we have that $E_c(h) \to 0$ as $h \to 0$ since:

(5)
\begin{align} \quad \lim_{h \to 0} E_c(h) = \lim_{h \to 0} \left [ \frac{f(c + h) - f(c)}{h} - f'(c) \right ] = \lim_{h \to 0} \left [ \frac{f(c + h) - f(c)}{h} \right ] - f'(c) = f'(c) - f'(c) = 0 \end{align}

Notice that $T_c(h)$ is indeed a linear function. To show this, note that for a defined $h_1$ and $h_2$ we have that:

(6)
\begin{align} \quad T_c(h_1 + h_2) = f'(c)(h_1 + h_2) = f'(c)h_1 + f'(c)h_2 = T_c(h_1) + T_c(h_2) \end{align}

So $T_c(h)$ satisfies the additivity property. Now for $k \in \mathbb{R}$ we have that:

(7)
\begin{align} \quad T_c(kh) = f'(c)kh = kf'(c)h = kT_c(h) \end{align}

So $T_c(h)$ satisfies the homogeneity property and thus $T_c : \mathbb{R} \to \mathbb{R}$ is indeed a linear map. So the notion of differentiability of a single variable real-valued function is consistent with the definition of differentiability made above.