The From Average to Instantaneous Rate at a point is important enough to get its own name and symbol. Computing it means evaluating a $\frac{0}{0}$ limit, so the methods from Limits by Factoring and Canceling will do the work.
Key idea
The derivative of $f$ at $a$ is $$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},$$ when this limit exists. It is the slope of the tangent line to $y = f(x)$ at $(a, f(a))$ and the instantaneous rate of change of $f$ at $a$. An equivalent form is $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$.
Why it works
At $h = 0$ the difference quotient is $\frac{0}{0}$: there is only one point, and one point does not determine a line. So we never set $h = 0$ first. Instead we simplify the quotient algebraically. For polynomials, $f(a + h) - f(a)$ always has a factor of $h$, because it is 0 when $h = 0$. Canceling that $h$ leaves an expression we can evaluate at $h = 0$.
The method in four steps
- Write $f(a + h)$ by replacing every $x$ with $a + h$, and expand.
- Subtract $f(a)$; the constant terms cancel.
- Divide by $h$ (every remaining term has a factor of $h$).
- Only now let $h \to 0$.
Worked example: A quadratic
Use the definition to find $f'(2)$ for $f(x) = x^2 + 3x$.
- $$f(2 + h) = (2 + h)^2 + 3(2 + h) = 4 + 4h + h^2 + 6 + 3h = 10 + 7h + h^2$$.
- $f(2) = 10$, so $f(2 + h) - f(2) = 7h + h^2$.
- Divide by $h$: $7 + h$.
- Let $h \to 0$: $f'(2) = 7$.
Answer: $f'(2) = 7$
Worked example: A reciprocal
Find $f'(2)$ for $f(x) = \frac{1}{x}$.
- $\frac{f(2 + h) - f(2)}{h} = \frac{1}{h}\left(\frac{1}{2 + h} - \frac{1}{2}\right)$.
- Combine: $\frac{1}{2 + h} - \frac{1}{2} = \frac{2 - (2 + h)}{2(2 + h)} = \frac{-h}{2(2 + h)}$.
- Divide by $h$: $\frac{-1}{2(2 + h)}$.
- Let $h \to 0$: $-\frac{1}{4}$.
Answer: $f'(2) = -\frac{1}{4}$: the curve falls a quarter unit per unit there.
Worked example: A square root
Find $f'(4)$ for $f(x) = \sqrt{x}$.
- The quotient is $\frac{\sqrt{4 + h} - 2}{h}$.
- Multiply by the conjugate: $\frac{(4 + h) - 4}{h(\sqrt{4 + h} + 2)} = \frac{1}{\sqrt{4 + h} + 2}$.
- Let $h \to 0$: $\frac{1}{4}$.
Answer: $f'(4) = \frac{1}{4}$
The second form, $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$, is sometimes quicker. For $f(x) = x^2$ at $a = 3$: $\frac{x^2 - 9}{x - 3} = x + 3 \to 6$. Both forms describe the same secant slopes; one names the second point $a + h$, the other names it $x$.
Common mistake
Two classic slips. First, writing $f(a + h)$ as $f(a) + h$. For $f(x) = x^2$, $f(3 + h) = (3 + h)^2 = 9 + 6h + h^2$, not $9 + h$. Second, plugging in $h = 0$ before simplifying, which gives $\frac{0}{0}$ and nothing else. Check: after you divide by $h$, no $h$ should be left in a denominator for polynomials.