Loav · VCE Mathematical Methods · Units 3 & 4
D01 · Differential calculus · Current curriculum
Build the derivative from a question about change, then use limits and domain checks to decide when that derivative exists.
1. A rate compares two changes
If a quantity y depends on x, a rate of change tells us how much y changes per unit change in x. The average rate across an interval [a, b], where a ≠ b, is [f(b) − f(a)]/(b − a). Both differences must use the same order. Geometrically this is the gradient of the secant joining (a, f(a)) and (b, f(b)). It summarises the whole interval; it does not claim that the function changes at that rate at every intermediate point.
The units are output units divided by input units. A volume measured in litres against time measured in minutes has a rate in litres per minute. A negative average rate means a net decrease between the two endpoints; the quantity need not decrease throughout the interval. A negative instantaneous rate describes the direction of change at that input. It does not mean the quantity itself is negative. Similarly, a positive function can have a negative derivative. Always distinguish the level of a quantity, the change in that quantity, and the rate of its change.
A straight-line function f(x) = mx + c has average rate m on every non-zero interval. Curved graphs generally have different average rates on different intervals. The expression f(b)/b is a rate from the origin only when that origin is genuinely one of the relevant points. It is not a substitute for the difference quotient. Reading the two endpoints carefully is often more important than performing the arithmetic quickly.
Worked example 1: one model, two different rates
A tank model is V(t) = 240 + 18t − 2t² litres for 0 ≤ t ≤ 4 minutes. Find the average rate from t = 1 to t = 3, and the instantaneous rate at t = 3.
V(1) = 256 and V(3) = 276, so the average rate is (276 − 256)/(3 − 1) = 10 L/min. For the instantaneous rate, V′(t) = 18 − 4t, giving V′(3) = 6 L/min. The tank is still gaining water at t = 3, but more slowly than its average gain across the preceding two-minute interval. The final volume, 276 L, is neither of these rates.
As a check, the derivative falls from 14 to 6 L/min between t = 1 and t = 3. An average rate of 10 lies between them, which is consistent with the graph. This is a useful reasonableness check, not a replacement for the secant calculation.
2. Moving from a secant to a tangent
Fix a point x = a and move a second point to x = a + h. Their secant gradient is [f(a + h) − f(a)]/h, with h ≠ 0. Positive h approaches from the right and negative h approaches from the left. If these gradients approach the same finite value as h approaches zero, that limiting value is f′(a), the derivative at a. The corresponding tangent has that gradient.
The notation h → 0 means that h becomes arbitrarily close to zero; it does not mean that the original quotient is evaluated at h = 0. Direct substitution initially produces 0/0, which is not a number. Algebra often reveals an equivalent expression for every non-zero h near zero. Its limiting value can then be identified. This distinction explains both the power and the care required in first-principles differentiation.
Worked example 2: differentiate without a memorised rule
Let f(x) = x² − 3x + 4. Start with f(x + h) = x² + 2xh + h² − 3x − 3h + 4. Subtract the complete expression f(x), including every sign, to obtain 2xh + h² − 3h.
For h ≠ 0, [f(x + h) − f(x)]/h = h(2x + h − 3)/h = 2x + h − 3. Taking the limit gives f′(x) = 2x − 3. At x = 2, the tangent gradient is 1. The derivative is a function of x because each input can have a different tangent gradient.
The cancellation is valid for h ≠ 0, exactly the condition under which the secant was defined. We only take the limit after simplifying. A common mistake is subtracting x² − 3x + 4 without brackets, which changes the signs of the wrong terms and prevents the required cancellation.
3. A limit describes nearby behaviour
The limit of f(x) as x approaches a concerns values near a, rather than necessarily the value at a. For example, (x² − 9)/(x − 3) equals x + 3 when x ≠ 3. Its limit as x approaches 3 is 6, although its original formula is undefined at 3. The graph has a missing point. Cancelling a factor helps calculate the limit; it does not restore an excluded input to the original domain.
For a two-sided finite limit to exist at an interior point, the left-hand and right-hand limiting values must agree. A jump has different limiting values on the two sides. An expression whose values grow without bound does not have a finite limit there. An unsuccessful algebraic simplification does not prove that a limit fails to exist: use the function’s structure, graph and appropriate reasoning. Keep this informal limit interpretation focused on rates and continuity.
4. Continuity is a value-and-limit condition
A function is continuous at an interior point a when f(a) is defined, the two-sided limit exists, and that limit equals f(a). All three requirements matter. A hole fails the first requirement unless a value is assigned there. Assigning the wrong value fills the hole at the wrong height and still fails continuity. A jump cannot be repaired merely by choosing a different value at the joining input.
At an endpoint of a restricted interval, continuity is considered using the side of the domain that is available. State the domain before testing endpoints. If a question concerns differentiability in the usual two-sided sense, distinguish interior points from boundaries instead of silently extending the formula beyond the model’s interval. A contextual model might be defined only for non-negative time even though its algebraic expression makes sense elsewhere.
5. Differentiability asks for a consistent local slope
Differentiability at an interior point implies continuity there. The reverse implication is false. For example, define p(x) = x for x ≤ 1 and p(x) = 2x − 1 for x > 1. Both branches meet at (1, 1), but their slopes are 1 on the left and 2 on the right. It has a corner, so no single finite derivative exists there. A vertical tangent also lacks a finite ordinary derivative. Neither a connected-looking drawing nor a defined function value is enough to establish differentiability.
For a piecewise function, inspect the join in two stages. First match the function values and limiting heights. Then match the one-sided derivative values, provided the pieces themselves are differentiable on their respective sides. Derivatives cannot repair a discontinuity. Two formulas can have equal derivatives at the joining input but still meet at different heights, so both stages are essential.
Worked example 3: make a smooth join
Define f(x) = x² for x ≤ 2, and f(x) = ax + b for x > 2. Find a and b so that the function is differentiable at x = 2.
Continuity requires the right-hand limiting value 2a + b to equal f(2) = 4. Thus 2a + b = 4. On the left the derivative is 2x, approaching 4; on the right the derivative is a. Matching slopes gives a = 4. Substitution into the continuity equation gives b = −4.
The right branch is therefore 4x − 4. Both branches approach height 4 and slope 4 at the join. Checking only a = 4 would leave infinitely many discontinuous choices of b. Checking only continuity would leave many corners. The pair (a, b) = (4, −4) satisfies both requirements.
6. Choose a method and communicate the result
For an average rate, use two values of the original function. For an instantaneous rate, find the derivative and then evaluate it. When explicitly asked for first principles, show the difference quotient, the expansion or factorisation, and the limit. Quoting the power rule alone does not demonstrate the requested method. When interpreting a graph, connect secants to intervals and tangents to individual inputs.
Keep exact fractions or surds during working. If a numerical approximation is requested, round at the end and retain the required units. A calculator’s numerical derivative can be a helpful check, but an apparent slope near a corner can depend on how the numerical routine samples either side. For piecewise formulas, establish the mathematical conditions rather than trusting a zoomed graph or a single decimal output.
A sound answer states what changes, with respect to what, at which input or over which interval. Finish by checking that the input belongs to the stated domain. A derivative formula may exist algebraically outside a contextual interval without having a meaningful interpretation there. These checks turn an otherwise correct calculation into a complete modelling answer.
Practice: seven original questions
- Find the average rate of change of f(x) = 3x² + 2 from x = −1 to x = 2.
- Use first principles to differentiate g(x) = x² + 5x. Show why cancellation by h is allowed.
- For q(x) = (x² − 9)/(x − 3), give the domain, the limit as x approaches 3, and the value that would make an extended function continuous there.
- Let p(x) = x for x ≤ 1 and p(x) = 2x − 1 for x > 1. Explain why p is continuous but not differentiable at x = 1.
- A piecewise function is x² for x ≤ 1 and ax + b for x > 1. Find a and b for differentiability at the join.
- W(t) = 100 + 24t − 3t² litres for 0 ≤ t ≤ 4 minutes. Calculate the average rate over the whole interval and the instantaneous rate at its final time. Interpret both.
- For s(x) = √(x − 2), state the real domain, the interval where the usual finite derivative exists, and what happens to right-hand secant gradients at x = 2.
Worked answers and reasoning
- f(−1) = 5 and f(2) = 14. The rate is (14 − 5)/(2 − (−1)) = 9/3 = 3 units of output per unit of input. The interval length is 3, not 1; subtracting a negative lower endpoint matters.
- g(x + h) − g(x) = 2xh + h² + 5h. For h ≠ 0, divide by h to get 2x + h + 5. Its limit as h approaches zero is 2x + 5. Cancellation is performed on non-zero secant increments; zero is approached afterwards, so no division by zero occurs.
- The original domain is ℝ excluding 3. Factoring gives q(x) = x + 3 for x ≠ 3, so the two-sided limit is 6. Defining an extended function with value 6 at x = 3 makes it continuous. The original function itself remains undefined there.
- Both branches approach 1 and p(1) = 1, establishing continuity. For x < 1 the expression is x with derivative 1; for x > 1 it is 2x − 1 with derivative 2. The unequal one-sided slopes prevent a derivative at the corner.
- Continuity gives a + b = 1. The derivative on the left approaches 2, while the right derivative is a, so a = 2. Therefore b = −1. Heights and slopes both match, which is stronger than satisfying either equation separately.
- W(0) = 100 and W(4) = 148. The average rate is 48/4 = 12 L/min. W′(t) = 24 − 6t, so the endpoint rate interpreted from within the model is 0 L/min. Water has accumulated over the interval even though the model’s rate has fallen to zero at its end.
- The real domain is [2, ∞). For x > 2, s′(x) = 1/[2√(x − 2)]. At x = 2, a right-hand increment h gives a secant gradient √h/h = 1/√h for h > 0. This grows without bound as h approaches zero. The endpoint is continuous from the right, but there is no finite derivative there.
Curriculum and further study
Aligned to the current VCAA VCE Mathematics Study Design, Mathematical Methods Units 3 and 4, Calculus area of study and relevant outcome knowledge and skills. The VCAA Mathematical Methods FAQ, updated February 2026, clarifies inflection points and strict increasing or decreasing behaviour. Curriculum checked 15 October 2026. Unit 3 is the suggested teaching sequence; the official design specifies combined Units 3 and 4 coverage.
Further study: Evans, Lipson and Greenwood, Cambridge Senior Mathematics VCE: Mathematical Methods Units 3 & 4, second edition (2023), sections 9A, 9L–9M and 10B. Independent Loav teaching material, not a VCAA or Cambridge publication. Explanations, worked examples, practice questions, answers and diagrams are original.
