Mathematics · Ch 7 — Limits and Derivatives

Algebra of Derivatives

Exactly as the algebra of limits (Section 2) allows the limit of a

combination of functions to be built from the limits of the simpler functions inside it, the

algebra of derivatives allows the derivative of a sum, difference, product or quotient of

two functions to be built from the derivatives of those two functions -- derived here directly from the first-principles definition of Section 8, not merely stated.

Throughout, let $u(x)$ and $v(x)$ be two functions, both differentiable at $x$ , with derivatives

$u'(x)$ and $v'(x)$ .

Constant multiple rule. For any constant $k$ ,

\frac{d}{dx}\big[k\,u(x)\big] = \lim_{h\to0}\frac{k\,u(x+h)-k\,u(x)}{h} = k\lim_{h\to0}\frac{u(x+h)-u(x)}{h} = k\,u'(x).

Theorem (sum and difference rule).

\frac{d}{dx}\big[u(x)+v(x)\big] = u'(x)+v'(x), \qquad \frac{d}{dx}\big[u(x)-v(x)\big] = u'(x)-v'(x).

Proof (sum case). By the first-principles definition,

\frac{d}{dx}\big[u(x)+v(x)\big] = \lim_{h\to0}\frac{\big[u(x+h)+v(x+h)\big]-\big[u(x)+v(x)\big]}{h} = \lim_{h\to0}\frac{\big[u(x+h)-u(x)\big]+\big[v(x+h)-v(x)\big]}{h}.

By the sum rule for limits (Section 2), this splits into two separate limits:

= \lim_{h\to0}\frac{u(x+h)-u(x)}{h} + \lim_{h\to0}\frac{v(x+h)-v(x)}{h} = u'(x)+v'(x).

The difference case is identical, using the limit difference rule in place of the sum rule.

$\blacksquare$

Theorem (product rule).

\frac{d}{dx}\big[u(x)v(x)\big] = u'(x)v(x) + u(x)v'(x).

Proof. By definition,

\frac{d}{dx}\big[u(x)v(x)\big] = \lim_{h\to0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}.

The key step is to subtract and add the term $u(x+h)v(x)$ in the numerator -- a quantity

that changes nothing, since it is added right back:

u(x+h)v(x+h)-u(x)v(x) = u(x+h)\big[v(x+h)-v(x)\big] + v(x)\big[u(x+h)-u(x)\big].

\frac{d}{dx}\big[uv\big] = \lim_{h\to0}\left[u(x+h)\cdot\frac{v(x+h)-v(x)}{h} + v(x)\cdot\frac{u(x+h)-u(x)}{h}\right].

As $h\to0$ : $u(x+h)\to u(x)$ (a differentiable function is necessarily continuous, so its own

value cannot jump as $h\to0$ ); $\dfrac{v(x+h)-v(x)}{h}\to v'(x)$ ; and

$\dfrac{u(x+h)-u(x)}{h}\to u'(x)$ , by definition. Hence, by the product and sum rules for

limits,

\frac{d}{dx}\big[uv\big] = u(x)\,v'(x) + v(x)\,u'(x). \qquad\blacksquare

Theorem (quotient rule). Provided $v(x)\neq0$ ,

\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x)-u(x)v'(x)}{\big[v(x)\big]^2}.

Proof. By definition,

\frac{d}{dx}\left[\frac{u}{v}\right] = \lim_{h\to0}\frac{1}{h}\left[\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}\right] = \lim_{h\to0}\frac{u(x+h)v(x)-u(x)v(x+h)}{h\,v(x+h)v(x)},

combining the two fractions over the common denominator $v(x+h)v(x)$ first. Now apply the same

subtract-and-add trick to the numerator, this time inserting $u(x)v(x)$ :

u(x+h)v(x)-u(x)v(x+h) = v(x)\big[u(x+h)-u(x)\big] - u(x)\big[v(x+h)-v(x)\big].

So …

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