Algebraic Rules of Differentiation

Technique: direct manipulation of the difference quotient, then the sum and scalar-multiple laws for limits. By definition, for $h \neq 0$ small enough that $a+h$ lies in the domain of both $f$ and $g$, form the difference quotient of $\alpha f + \beta g$:

Regroup the numerator by which function each term came from — this is legitimate because addition of real numbers is commutative and associative, so the terms may be reordered without changing the value:

This last equality is just distributing division by $h$ over a sum, valid since $h \neq 0$. Now take $h \to 0$. By hypothesis, $f$ is differentiable at $a$, so

exists, and likewise $\lim_{h\to 0} \frac{g(a+h)-g(a)}{h} = g'(a)$ exists. By the constant-multiple law for limits (if $\lim_{h\to0} \varphi(h) = L$ exists, then $\lim_{h\to 0}\alpha \varphi(h) = \alpha L$), each of the two terms above has a limit as $h \to 0$, namely $\alpha f'(a)$ and $\beta g'(a)$ respectively. By the sum law for limits (if two individual limits exist, the limit of the sum exists and equals the sum of the limits), the limit of the whole difference quotient exists and equals

Since the difference quotient of $\alpha f + \beta g$ converges to this value as $h \to 0$, the function $\alpha f + \beta g$ is by definition differentiable at $a$ with derivative $\alpha f'(a) + \beta g'(a)$. $\blacksquare$

Technique: add and subtract a well-chosen cross term inside the numerator, then split the limit using <a class="wiki-link" href="/wiki/math-analysis-differentiability-implies-continuity" data-track="wiki_article.body.wiki_link">Differentiability Implies Continuity</a>. Fix $h \neq 0$ small enough that $a + h$ is in the common domain of $f$ and $g$. The difference quotient of $fg$ at $a$ is

The obstruction to splitting this immediately is that the numerator mixes $f(a+h), f(a)$ with $g(a+h), g(a)$ inseparably. The standard resolution is to insert the term $f(a+h)g(a)$, which is subtracted immediately after being added, so the numerator's value is unchanged:

Factor each bracket: the first factors $f(a+h)$ out of both terms, the second factors $g(a)$ out of both terms:

Divide by $h$ (legitimate since $h \neq 0$), distributing over the sum:

Now let $h \to 0$ and examine each of the two terms. Second term. Since $f$ is differentiable at $a$, $\dfrac{f(a+h)-f(a)}{h} \to f'(a)$ as $h \to 0$. Since $g(a)$ is a fixed constant (it does not depend on $h$), the constant-multiple limit law gives

First term. We need two facts: $\dfrac{g(a+h)-g(a)}{h} \to g'(a)$, which holds since $g$ is differentiable at $a$; and $f(a+h) \to f(a)$ as $h \to 0$. The latter is precisely the continuity of $f$ at $a$, which follows from the differentiability of $f$ at $a$ by Differentiability Implies Continuity (this is the one place that theorem is needed — differentiability alone does not immediately hand us $f(a+h) \to f(a)$ without invoking it). By the product law for limits (the limit of a product of two functions, each possessing a limit at $a$, is the product of the limits), we get

By the sum law for limits, since both individual limits above exist, the full difference quotient converges as $h\to 0$ to the sum:

Hence $fg$ is differentiable at $a$ with $(fg)'(a) = f'(a)g(a) + f(a)g'(a)$. $\blacksquare$

Technique: continuity to guarantee $g$ is nonzero nearby, then a single common-denominator algebraic manipulation. Since $g$ is differentiable at $a$, it is continuous at $a$ by Differentiability Implies Continuity, so $\lim_{x\to a} g(x) = g(a) \neq 0$. By the definition of a limit (applied with $\varepsilon = |g(a)|/2 > 0$), there is $\delta > 0$ such that for all $x$ with $0 < |x-a| < \delta$ (and also at $x=a$ itself, trivially), $|g(x) - g(a)| < |g(a)|/2$, which by the reverse triangle inequality forces $|g(x)| > |g(a)|/2 > 0$. So $g(x) \neq 0$ for all $x$ within $\delta$ of $a$, and $1/g$ is well-defined on this punctured neighborhood together with $a$. For $h \neq 0$ with $|h| < \delta$, form the difference quotient of $1/g$:

where the middle equality is a single common-denominator combination of the two fractions $\frac1{g(a+h)}$ and $\frac1{g(a)}$, and the last step just factors out a sign and separates the difference quotient of $g$. As $h \to 0$: $\dfrac{g(a+h)-g(a)}{h} \to g'(a)$ since $g$ is differentiable at $a$; and $g(a+h) \to g(a)$ by continuity of $g$ at $a$, so $g(a+h)g(a) \to g(a)^2 \neq 0$, whence $-\dfrac{1}{g(a+h)g(a)} \to -\dfrac{1}{g(a)^2}$ by the quotient law for limits (valid since the denominator's limit $g(a)^2$ is nonzero). By the product law for limits, the whole expression converges to

Hence $1/g$ is differentiable at $a$ with the claimed derivative. $\blacksquare$

Technique: reduce to the already-proved product rule and reciprocal rule (composition of previously established results, not the raw limit definition again). Write $f/g = f \cdot (1/g)$ as an equality of functions on the neighborhood of $a$ where $g$ is nonzero (which exists by continuity of $g$ at $a$, as established inside the proof of the Reciprocal Rule above). By the Reciprocal Rule, $1/g$ is differentiable at $a$ with $(1/g)'(a) = -g'(a)/g(a)^2$. Since $f$ is differentiable at $a$ by hypothesis, both factors in the product $f \cdot (1/g)$ are differentiable at $a$, so the Product Rule applies directly:

Combine the two fractions over the common denominator $g(a)^2$ (multiplying the first fraction's numerator and denominator by $g(a)$, which is nonzero, so this is a valid algebraic step):

This is precisely the claimed formula. $\blacksquare$ Note that this proof is legitimately non-circular: it invokes the Product Rule and the Reciprocal Rule, both already proved directly from the difference-quotient definition, and does not smuggle in the Quotient Rule itself or anything depending on it.