Total Differentiability (Real Vector Functions)#

Definition: Total Differentiability of Real Vector Functions

Let \(f: \mathcal{D} \subseteq \mathbb{R}^m\to \mathbb{R}^n\) be a real vector function and let \(\boldsymbol{p}\) be an interior point of \(\mathcal{D}\).

We say that \(f\) is totally differentiable at \(\boldsymbol{p}\) if there exists a linear real vector function \(T_{\boldsymbol{p}}: \mathbb{R}^m \to \mathbb{R}^n\) such that \(||f(\boldsymbol{p}+\boldsymbol{h}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{h})||\) is little o of \(||\boldsymbol{h}||\) for \(||\boldsymbol{h}|| \to 0\):

\[||f(\boldsymbol{p}+\boldsymbol{h}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{h})|| = o(||\boldsymbol{h}||) \qquad \text{for} \qquad ||\boldsymbol{h}|| \to 0\]

In this case, \(T_{\boldsymbol{p}}\) is known as the total derivative or total differential of \(f\) at \(\boldsymbol{p}\).

Notation

We usually denote \(T_{\boldsymbol{p}}\) by \(\mathrm{d}f_{\boldsymbol{p}}\). When \(\boldsymbol{p}\) is clear from context, we can also write just \(\mathrm{d}f\).

If \(f\) is differentiable at every \(\boldsymbol{p}\) in some \(S \subseteq \mathbb{R}^m\), then we say that \(f\) is totally differentiable on \(S\). If \(S = \mathcal{D}\), we can also just say that \(f\) is totally differentiable.

Theorem: Uniqueness of the Total Derivative

Let \(f: \mathcal{D} \subseteq \mathbb{R}^m \to \mathbb{R}^n\) be a real vector function and let \(\boldsymbol{p}\) be an interior point of \(\mathcal{D}\).

If \(f\) is totally differentiable at \(\boldsymbol{p}\), then its total derivative there is unique.

Proof

TODO

Theorem: Total Differentiability via Limits

Let \(f: \mathcal{D} \subseteq \mathbb{R}^m \to \mathbb{R}^n\) be a real vector function and let \(\boldsymbol{p}\) be an interior point of \(\mathcal{D}\).

Then \(f\) is totally differentiable at \(\boldsymbol{p}\) if and only if there exists a linear real vector function \(T_{\boldsymbol{p}}: \mathbb{R}^m \to \mathbb{R}^n\) such that the following limit is zero:

\[\lim_{\boldsymbol{x} \to \boldsymbol{p}} \frac{||f(\boldsymbol{x}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{x} - \boldsymbol{p})||}{||\boldsymbol{x} - \boldsymbol{p}||} = 0\]

In this case, \(T_{\boldsymbol{p}}\) is the total derivative of \(f\) at \(\boldsymbol{p}\).

Proof

With the substitution \(\boldsymbol{h} = \boldsymbol{x} - \boldsymbol{p}\), we get:

\[\lim_{\boldsymbol{x} \to \boldsymbol{p}} \frac{||f(\boldsymbol{x}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{x} - \boldsymbol{p})||}{||\boldsymbol{x} - \boldsymbol{p}||} = \lim_{\boldsymbol{h} \to \boldsymbol{0}} \frac{||f(\boldsymbol{x} + \boldsymbol{h}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{h})||}{||\boldsymbol{h}||}\]

Since \(||\boldsymbol{h}|| \ne 0\) for \(\boldsymbol{h} \ne \boldsymbol{0}\), this implies that \(||f(\boldsymbol{p}+\boldsymbol{h}) - f(\boldsymbol{p}) - T_{\boldsymbol{p}}(\boldsymbol{h})||\) is little o of \(||\boldsymbol{h}||\) for \(||\boldsymbol{h}|| \to 0\):

Theorem: Total Differentiability via Component Functions

A real vector function \(f: \mathcal{D} \subseteq \mathbb{R}^m \to \mathbb{R}^n\) is totally differentiable at \(\boldsymbol{p} \in \operatorname{int} \mathcal{D}\) if and only if all of its component functions are totally differentiable there.

Proof

TODO

Theorem: Continuous Partial Differentiability \(\implies\) Total Differentiability

Let \(f: \mathcal{D} \subseteq \mathbb{R}^m \to \mathbb{R}^n\) be a real vector function, let \(\boldsymbol{p}\) be an interior point of \(\mathcal{D}\).

If \(f\) is partially differentiable on an open neighborhood of \(\boldsymbol{p}\) and is continuously partially differentiable at \(\boldsymbol{p}\), then \(f\) is totally differentiable at \(\boldsymbol{p}\).

Proof

TODO

Theorem: Differentiability \(\implies\) Continuity

If a real vector function \(f: \mathcal{D} \subseteq \mathbb{R}^m \to \mathbb{R}^n\) is totally differentiable at \(\boldsymbol{p} \in \operatorname{int} \mathcal{D}\), then it is also continuous there.

Proof

TODO

Theorem: Linearity of Differentiation

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^m \to \mathbb{R}^n\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^m \to \mathbb{R}^n\) be real vector functions.

If \(f\) and \(g\) are totally differentiable at \(\boldsymbol{p} \in \operatorname{int}(\mathcal{D}_f \cap \mathcal{D}_g)\), then so is \(\lambda f + \mu g\) for all \(\lambda, \mu \in \mathbb{R}\):

\[\mathop{\mathrm{d}(\lambda f + \mu g)_{\boldsymbol{p}}} = \lambda \mathop{\mathrm{d}f_{\boldsymbol{p}}} + \mu \mathop{\mathrm{d}g_{\boldsymbol{p}}}\]

Proof

TODO

Theorem: Chain Rule

Let \(g: \mathcal{D}_g \subseteq \mathbb{R}^m \to \mathbb{R}^n\) and \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}^p\) be real vector functions.

If \(g\) is totally differentiable at \(\boldsymbol{p} \in \operatorname{int} \mathcal{D}_g\) and \(f\) is totally differentiable at \(g(\boldsymbol{p}) \in \operatorname{int} \mathcal{D}_f\), then \(f \circ g\) is totally differentiable at \(\boldsymbol{p}\):

\[\mathrm{d}(f\circ g)_{\boldsymbol{p}} = \mathrm{d}f_{g(\boldsymbol{p})} \circ \mathrm{d}g_{\boldsymbol{p}}\]

Proof

TODO