A Sum/Product Rule for Directional Derivatives of Functions from Rn to Rm

# A Sum and Product Rule for Directional Derivatives of Functions from Rn to Rm

Recall from the Directional Derivatives of Functions from Rn to Rm page that if $S \subseteq \mathbb{R}^n$ is open, $\mathbb{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ then the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ in the direction of $\mathbf{u} \in \mathbb{R}^n$ is defined as the following limit provided that it exists:

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

We will now look at two nice analogues of the sum and product rule for differentiation with respect to directional derivatives.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and let $\mathbf{f}, \mathbf{g} : S \to \mathbb{R}^n$. Let $\mathbb{u} \in \mathbb{R}^n$ and assume that the directional derivatives of $\mathbf{f}$ and $\mathbf{g}$ at $\mathbf{c}$ in the direction of $\mathbf{u}$ exists, i.e., $\mathbf{f}'(\mathbf{c}, \mathbf{u})$ and $\mathbf{g}'(\mathbf{c}, \mathbf{u})$ both exist. Then the directional derivative of $\mathbf{f} + \mathbf{g}$ exists and $(\mathbf{f} + \mathbf{g})'(\mathbf{c}, \mathbf{u}) = \mathbf{f}'(\mathbf{c}, \mathbf{u}) + \mathbf{g}'(\mathbf{c}, \mathbf{u})$.
• Proof: Using the definition of the directional derivative on $\mathbf{f} + \mathbf{g}$ gives us:
(2)
\begin{align} \quad (\mathbf{f} + \mathbf{g})'(\mathbf{c}, \mathbf{u}) &= \lim_{h \to 0} \frac{(\mathbf{f} + \mathbf{g})(\mathbf{c} + h\mathbf{u}) - (\mathbf{f} + \mathbf{g})(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) + \mathbf{g}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c}) - \mathbf{g}(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c})}{h} + \lim_{h \to 0} \frac{\mathbf{g}(\mathbf{c} + h\mathbf{u}) - \mathbf{g}(\mathbf{c})}{h} \\ &= \mathbf{f}'(\mathbf{c}, \mathbf{u}) + \mathbf{g}'(\mathbf{c}, \mathbf{u}) \end{align}
 Theorem 2: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and let $\mathbf{f}, \mathbf{g} : S \to \mathbb{R}^n$. Let $\mathbb{u} \in \mathbb{R}^n$ and assume that the directional derivatives of $\mathbf{f}$ and $\mathbf{g}$ at $\mathbf{c}$ in the direction of $\mathbf{u}$ exists, i.e., $\mathbf{f}'(\mathbf{c}, \mathbf{u})$ and $\mathbf{g}'(\mathbf{c}, \mathbf{u})$ both exist. Then the directional derivative of $\mathbf{f} \cdot \mathbf{g}$ exists and $(\mathbf{f} \cdot \mathbf{g})'(\mathbf{c}, \mathbf{u}) = \mathbf{f}(\mathbf{c}) \cdot \mathbf{g}'(\mathbf{c}, \mathbf{u}) + \mathbf{g}(\mathbf{c}) \cdot \mathbf{f}'(\mathbf{c}, \mathbf{u})$.
• Proof: Using the definition of the directional derivative on $\mathbf{f} \cdot \mathbf{g}$ gives us:
(3)
\begin{align} \quad (\mathbf{f} \cdot \mathbf{g})'(\mathbf{c}, \mathbf{u}) &= \lim_{h \to 0} \frac{(\mathbf{f} \cdot \mathbf{g})(\mathbf{c} + h\mathbf{u}) - (\mathbf{f} \cdot \mathbf{g})(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) \cdot \mathbf{g}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c}) \cdot \mathbf{g}(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) \cdot \mathbf{g}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c} + h\mathbf{u}) \cdot \mathbf{g}(\mathbf{c}) + \mathbf{f}(\mathbf{c} + h\mathbf{u}) \cdot \mathbf{g}(\mathbf{c}) - \mathbf{f}(\mathbf{c}) \cdot \mathbf{g}(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \mathbf{f}(\mathbf{c} + h\mathbf{u}) \cdot \frac{\mathbf{g}(\mathbf{c} + h\mathbf{u}) - \mathbf{g}(\mathbf{c})}{h} + \lim_{h \to 0} \mathbf{g}(\mathbf{c}) \cdot \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c})}{h} \\ &= \mathbf{f}(\mathbf{c}) \cdot \mathbf{g}'(\mathbf{c}, \mathbf{u}) + \mathbf{g}(\mathbf{c}) \cdot \mathbf{f}'(\mathbf{c}, \mathbf{u}) \quad \blacksquare \end{align}