[proofplan] Part 1 (linearity) is direct: add the [differentiability](/page/Derivative) expansions for $f$ and $g$ and observe the error terms sum to a vanishing remainder. Part 2 (product rule) expands the product $\phi(a + h)f(a + h)$ using both differentiability expansions, identifies the terms linear in $h$ as the candidate derivative, and verifies every remaining term is $o(|h|)$ by using the operator norm bound and the vanishing of the error functions. [/proofplan] [step:Prove linearity of the derivative by adding the differentiability expansions] By [differentiability](/page/Derivative) of $f$ and $g$ at $a$: \begin{align*} (f + g)(a + h) &= f(a) + Df_a(h) + |h|\varepsilon_1(h) + g(a) + Dg_a(h) + |h|\varepsilon_2(h) \\ &= (f + g)(a) + \bigl(Df_a + Dg_a\bigr)(h) + |h|\bigl(\varepsilon_1(h) + \varepsilon_2(h)\bigr). \end{align*} Since $\varepsilon_1(h) + \varepsilon_2(h) \to \mathbf{0}$ as $h \to \mathbf{0}$, the sum $f + g$ is differentiable at $a$ with derivative $Df_a + Dg_a$. [/step] [step:Expand the product $\phi f$ and identify the linear term] By [differentiability](/page/Derivative) of $\phi$ and $f$: \begin{align*} \phi(a + h) &= \phi(a) + D\phi_a(h) + |h|\delta(h), \\ f(a + h) &= f(a) + Df_a(h) + |h|\varepsilon(h), \end{align*} with $\delta(h) \to 0$ and $\varepsilon(h) \to \mathbf{0}$. Multiplying: \begin{align*} (\phi f)(a + h) = \phi(a)f(a) + \phi(a)Df_a(h) + D\phi_a(h)f(a) + R(h), \end{align*} where $R(h)$ collects all remaining terms from the expansion of the product. [/step] [step:Verify every term in the remainder $R(h)$ is $o(|h|)$] The remainder consists of: \begin{align*} R(h) = D\phi_a(h)Df_a(h) + \phi(a)|h|\varepsilon(h) + |h|\delta(h)f(a) + \text{(higher-order cross terms)}. \end{align*} For the first term, using the [Lipschitz bound](/theorems/321) on both $D\phi_a$ and $Df_a$: \begin{align*} |D\phi_a(h)Df_a(h)| \leq \|D\phi_a\| \cdot |h| \cdot \|Df_a\| \cdot |h| = O(|h|^2) = o(|h|). \end{align*} The second term satisfies $|\phi(a)|h|\varepsilon(h)| = |\phi(a)| \cdot |h| \cdot |\varepsilon(h)| = |h| \cdot o(1)$. The third satisfies $||h|\delta(h)f(a)| = |\delta(h)| \cdot |f(a)| \cdot |h| = |h| \cdot o(1)$. All remaining cross terms contain at least two vanishing factors and are likewise $o(|h|)$. Therefore $R(h) = |h|\varepsilon_3(h)$ with $\varepsilon_3 \to \mathbf{0}$, and $D(\phi f)_a(h) = \phi(a)Df_a(h) + D\phi_a(h)f(a)$. [guided] The product rule computation follows the same pattern as in single-variable calculus, but we need to be careful about the bookkeeping. We expand the product of two expressions, each of the form "value + linear term + error", and the candidate derivative picks up the cross terms that are linear in $h$: one from $\phi$ constant times $f$ linear, and one from $\phi$ linear times $f$ constant. Everything else involves at least two factors that are $O(|h|)$ or vanishing, making them $o(|h|)$. The critical estimates: - $D\phi_a(h) \cdot Df_a(h)$: both factors are $O(|h|)$ by the [Lipschitz bound](/theorems/321), so the product is $O(|h|^2) = o(|h|)$. - $\phi(a) \cdot |h|\varepsilon(h)$: the scalar $\phi(a)$ is constant, and $|h|\varepsilon(h) = o(|h|)$ by definition. - $|h|\delta(h) \cdot f(a)$: the vector $f(a)$ is constant, and $|h|\delta(h) = o(|h|)$ since $\delta \to 0$. Combining, $R(h) = o(|h|)$, confirming the [derivative](/page/Derivative) formula $D(\phi f)_a(h) = \phi(a)Df_a(h) + D\phi_a(h)f(a)$. [/guided] [/step]