---
tag: F0BV90294A
kind: proof
name: Proof of the linearity of differentiation
msc: [26A24]
proves: 0M0TKFRX2A
uses: [0F1WR9C83K, 0A61B60B5R, 07EGCNHAVW, 000MXHEY22, 0RAM4E07NR, 0734RSVF2V, 09G6KHS6GY]
sources:
  - url: https://en.wikipedia.org/wiki/Linearity_of_differentiation
    locator: Linearity (directly)
---

## Proof

We use the [[sequential characterization of limits|000MXHEY22]] and the limit laws for sequences.

1. Let $H = \set{h \ne 0 : a + h \in I}$. Since $I$ is an open interval containing $a$, there is $r > 0$ with $(a - r, a + r) \subseteq I$. So $0$ is a [[limit point|07EGCNHAVW]] of $H$.
2. For a function $u \colon I \to \R$ and $h \in H$, write $q_u(h) = \frac{u(a + h) - u(a)}{h}$. By the definition of the [[derivative|0F1WR9C83K]], $u'(a) = L$ means $q_u(h) \to L$ as $h \to 0$.
3. Put $s = \alpha f + \beta g$. For $h \in H$, expanding the numerator gives $q_s(h) = \alpha \, q_f(h) + \beta \, q_g(h)$.
4. Let $(h_n)$ be a sequence in $H$ with $h_n \to 0$. By the sequential characterization, $q_f(h_n) \to f'(a)$ and $q_g(h_n) \to g'(a)$.
5. The constant sequence $\alpha, \alpha, \dots$ converges to $\alpha$, since $\abs{\alpha - \alpha} = 0 < \varepsilon$ for every $\varepsilon > 0$ (see the [[epsilon-N form|09G6KHS6GY]]). The same holds for $\beta$.
6. By the [[limit of a product of sequences|0734RSVF2V]], $\alpha \, q_f(h_n) \to \alpha f'(a)$ and $\beta \, q_g(h_n) \to \beta g'(a)$.
7. By the [[limit of a sum of sequences|0RAM4E07NR]] and step 3, $q_s(h_n) \to \alpha f'(a) + \beta g'(a)$.
8. This holds for every such sequence $(h_n)$. By the sequential characterization, $q_s(h) \to \alpha f'(a) + \beta g'(a)$ as $h \to 0$.
9. So $s = \alpha f + \beta g$ is [[differentiable|0A61B60B5R]] at $a$, and $(\alpha f + \beta g)'(a) = \alpha f'(a) + \beta g'(a)$.

## Idea

The difference quotient of $\alpha f + \beta g$ is the same combination of the difference quotients of $f$ and $g$. Limits respect sums and constant multiples. So the limit of the combination is the combination of the limits.