Basic Rules of Differentiation

Once you can compute a derivative from the limit definition, the goal is to avoid repeating that calculation for every new function. The three rules below — linearity, the product rule, and the quotient rule — let you differentiate any polynomial or rational function by inspection.

Linearity

Theorem. If $f$ and $g$ are differentiable at $x$, and $c \in \mathbb{R}$, then

$$(cf + g)'(x) \;=\; c\,f'(x) + g'(x).$$

Proof. By definition,

$$\lim_{h\to 0}\frac{(cf+g)(x+h)-(cf+g)(x)}{h} \;=\; \lim_{h\to 0}\left[c\cdot\frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}\right].$$

Both difference quotients converge, so the sum of limits equals the limit of the sum: $c f'(x) + g'(x)$. $\square$

Product rule

Theorem. If $f$ and $g$ are differentiable at $x$, then

$$(fg)'(x) \;=\; f'(x)\,g(x) + f(x)\,g'(x).$$

Proof. Add and subtract $f(x)g(x+h)$ in the numerator:

$$\frac{f(x+h)g(x+h) - f(x)g(x)}{h} \;=\; \frac{f(x+h)-f(x)}{h}\cdot g(x+h) \;+\; f(x)\cdot\frac{g(x+h)-g(x)}{h}.$$

As $h \to 0$: the first term converges to $f'(x) \cdot g(x)$ (using continuity of $g$ at $x$, which follows from differentiability), and the second to $f(x) \cdot g'(x)$. $\square$

Quotient rule

Theorem. If $f$ and $g$ are differentiable at $x$ and $g(x) \neq 0$, then

$$\left(\frac{f}{g}\right)'(x) \;=\; \frac{f'(x)\,g(x) - f(x)\,g'(x)}{g(x)^2}.$$

Proof. First derive the reciprocal rule for $1/g$. With $\delta$ in place of $h$:

$$\frac{\tfrac{1}{g(x+\delta)}-\tfrac{1}{g(x)}}{\delta} \;=\; -\frac{g(x+\delta)-g(x)}{\delta}\cdot\frac{1}{g(x+\delta)\,g(x)}.$$

As $\delta \to 0$, this converges to $-g'(x)/g(x)^2$ (since $g(x+\delta) \to g(x) \neq 0$). Then apply the product rule to $f \cdot (1/g)$:

$$\left(\frac{f}{g}\right)' = f' \cdot \frac{1}{g} + f \cdot \left(-\frac{g'}{g^2}\right) = \frac{f'g - fg'}{g^2}. \;\square$$

Applications

Polynomials

From linearity and $(x^n)' = nx^{n-1}$:

$$\left(a_n x^n + \cdots + a_1 x + a_0\right)' \;=\; n a_n x^{n-1} + \cdots + a_1.$$

Rational functions

$$\frac{d}{dx}\left(\frac{x^2+1}{x-1}\right) \;=\; \frac{2x(x-1)-(x^2+1)}{(x-1)^2} \;=\; \frac{x^2-2x-1}{(x-1)^2}.$$

Summary

• Linearity: $(cf+g)' = cf' + g'$.
• Product rule: $(fg)' = f'g + fg'$.
• Quotient rule: $(f/g)' = (f'g - fg')/g^2$.
• All three follow from the limit definition via elementary limit arithmetic.
• Every polynomial and rational function is differentiable wherever defined.