The Total Derivative of a Function From Rn To Rm As A Linear

The Total Derivative of a Function From Rn To Rm as a Linear Combination of Its Partial Derivatives

Recall from the Differentiable Functions from Rn to Rm and Their Total Derivatives page that if $S \subseteq \mathbb{R}^n$, $\mathbf{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$ then if $\mathbf{f}$ is differentiable at $\mathbf{c}$ then all of the directional derivatives of $\mathbf{f}$ at $\mathbf{c}$ exist and for all $\mathbf{u} \in \mathbb{R}^n$ we have that:

\begin{align} \quad \mathbf{T}_{\mathbf{c}} (\mathbf{u}) = \mathbf{f}'(\mathbf{c}, \mathbf{u}) \end{align}

In particular, if $\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n$ are the standard basis vectors for $\mathbb{R}^n$ then we know that the partial derivatives of $\mathbf{f}$ at $\mathbf{c}$ exist. The following theorem will tell us that for any $\mathbf{v}$, the total derivative evaluated at $\mathbf{v}$ is a linear combination of the partial derivatives of $\mathbf{f}$.

Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is differentiable at $\mathbf{c}$ and if $\mathbf{v} = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + ... + v_n\mathbf{e}_n$ where $\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n$ are the standard basis vectors for $\mathbb{R}^n$ then $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = v_1\mathbf{f}'(\mathbf{c}, \mathbf{e}_1) + v_2\mathbf{f}'(\mathbf{c}, \mathbf{e}_2) + ... + v_n\mathbf{f}'(\mathbf{c}, \mathbf{e}_n)$.

Recall the notation $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = \mathbf{f}'(\mathbf{c})(\mathbf{v})$ and $\mathbf{f}'(\mathbf{c}, \mathbf{e}_k) = D_k \mathbf{f}(\mathbf{c})$. Thus the conclusion of the Theorem above can be rewritten as:

\begin{align} \quad \mathbf{f}'(\mathbf{c})(\mathbf{v}) = v_1 D_1 \mathbf{f}(\mathbf{c}) + v_2 D_2 \mathbf{f}(\mathbf{c}) + ... + v_n D_n \mathbf{f}(\mathbf{c}) \end{align}
  • Proof: Let $\mathbf{f}$ be differentiable at $\mathbf{c}$. For each $\mathbf{v} \in S$ let $\mathbf{v} = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + ... + v_n\mathbf{e}_n$. Then since the total derivative $\mathbf{T}_{\mathbf{c}}$ is a linear function we have that:
\begin{align} \quad \mathbf{T}_{\mathbf{c}} (\mathbf{v}) &= \mathbf{T}_{\mathbf{c}} (v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + ... + v_n\mathbf{e}_n ) \\ &= v_1 \mathbf{T}_{\mathbf{c}}(\mathbf{e}_1) + v_2 \mathbf{T}_{\mathbf{c}}(\mathbf{e}_2) + ... + v_n \mathbf{T}_{\mathbf{c}}(\mathbf{e}_n) \end{align}
  • But by the Theorem referenced above, for each $k \in \{ 1, 2, ..., n \}$ we have that $\mathbf{T}_{\mathbf{c}} (\mathbf{e}_k) = \mathbf{f}'(\mathbf{c}, \mathbf{u})$ and so:
\begin{align} \quad \mathbf{T}_{\mathbf{c}}(\mathbf{v}) = v_1\mathbf{f}'(\mathbf{c}, \mathbf{e}_1) + v_2\mathbf{f}'(\mathbf{c}, \mathbf{e}_2) + ... + v_n\mathbf{f}'(\mathbf{c}, \mathbf{e}_n) \quad \blacksquare \end{align}
Corollary 1: If $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in \mathbb{R}^n$, and $f : S \to \mathbb{R}$ is an $n$-variable real-valued function then $\mathbf{T}_{\mathbf{c}}(\mathbf{v}) = \nabla f(\mathbf{c}) \cdot \mathbf{v}$.
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