Learning Objectives
Interpret average and instantaneous rates of change.
Calculate a derivative at a point from its limit definition.
Use a derivative to write a tangent line and attach meaningful units.
Prerequisites: limits, algebraic simplification, slope, and function notation. A derivative turns the idea of a rate over an interval into a rate at an instant.
1. Begin with an Average Rate
The average rate of change of f over [a, a + h], with h ≠ 0, is [f(a + h) − f(a)]/h. Geometrically, this is the slope of a secant line through two points on the graph. The numerator measures output change and the denominator measures input change.
As h approaches zero, the second point approaches the first. If the secant slopes approach one finite number, that number is the derivative f′(a). Formally, f′(a) = lim(h → 0) [f(a + h) − f(a)]/h.
2. Simplify before Taking the Limit
Worked Example: f(x) = x² at x = 3
Use the definition: f′(3) = lim(h → 0) [(3 + h)² − 9]/h.
Expand the numerator: 9 + 6h + h² − 9 = 6h + h². For h ≠ 0, division by h gives 6 + h.
Now take the limit to obtain f′(3) = 6. Substituting h = 0 before simplifying would have produced the indeterminate form 0/0.
The same argument at a general input x gives [(x + h)² − x²]/h = 2x + h for h ≠ 0, so the derivative function is f′(x) = 2x. A derivative function assigns a local rate to each input where that rate exists.
3. A Derivative Gives a Tangent Slope
If f′(a) exists, the tangent line at (a, f(a)) has equation y − f(a) = f′(a)(x − a). The derivative supplies the slope, but the function supplies the point. These two values must not be confused.
Worked Example: A Tangent Line
For f(x) = x² at a = 3, the point is (3, 9) and the slope is 6.
The tangent line is y − 9 = 6(x − 3), or y = 6x − 9.
It passes through the correct point, since 6(3) − 9 = 9. Near x = 3, it gives a local straight-line approximation to the curve.
4. Interpret Units and Sign
If s(t) is position in metres and t is time in seconds, then s′(t) is velocity in metres per second. If C(q) is cost in dollars and q is output in kilograms, then C′(q) has units dollars per kilogram. The derivative's units are output units divided by input units.
A positive derivative indicates a positive instantaneous rate at that point; a negative derivative indicates a negative rate. A zero derivative gives a horizontal tangent when the graph is differentiable there. It does not by itself prove a maximum or minimum.
Common Mistakes
The numerator is f(a + h) − f(a), not f(a) + h − f(a).
A tangent-line equation needs both the slope and the point.
Keep h nonzero during algebraic cancellation, then take its limit.
5. Practice
1. Use the limit definition to find the derivative of f(x) = 4x − 1 at x = 2.
2. For f(x) = x², compare the average rate on [2, 3] with the instantaneous rate at x = 2.
3. Find the tangent line to y = x² at x = −1.
4. Water volume V(t) is measured in litres with t in minutes. Interpret V′(5) = −3.
Worked Solutions
1. [f(2 + h) − f(2)]/h = [7 + 4h − 7]/h = 4 for h ≠ 0. The derivative is 4.
2. The average rate is (9 − 4)/(3 − 2) = 5. The instantaneous rate is f′(2) = 2(2) = 4. They concern different input intervals.
3. The point is (−1, 1) and the slope is −2. Thus y − 1 = −2(x + 1), or y = −2x − 1.
4. At 5 minutes, the volume is decreasing at an instantaneous rate of 3 litres per minute. It does not say that the volume itself equals −3 litres.
6. Summary
- A derivative is a limit of average rates over shrinking intervals.
- Simplify the difference quotient before taking its limit.
- Use f(a) for the point and f′(a) for the tangent slope.
- Interpret the derivative with output-per-input units.