Year 12 · Mathematics · Algebra
Rates of change and the derivative
Move from the average rate of change between two points to the instantaneous rate of change, and use the power rule to differentiate simple polynomials.
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These lessons were written around Australian Curriculum strands and matched to New Zealand year levels by age (NZ Year 1 is the first year of school). They are not yet mapped to The New Zealand Curriculum, so check them against your own learning programme.
Learning goals
Learners will
- Calculate the average rate of change of a function over an interval.
- Understand the derivative as the gradient of a tangent, found as a limit.
- Use the power rule to differentiate polynomial functions.
Success looks like
- I can find the gradient of a secant between two points on a curve.
- I can explain how the gradient of a tangent is found from first principles for f(x) = x^2.
- I can differentiate expressions such as 3x^2 + 4x and evaluate the derivative at a point.
- I can interpret a derivative as a rate, such as velocity.
The big idea
For a straight line, the gradient is the same everywhere. For a curve, the steepness keeps changing. Calculus is the mathematics of describing how quickly things change at each point.
The average rate of change of f between x = a and x = b is (f(b) − f(a)) / (b − a). This is the gradient of the secant joining the two points on the curve. For f(x) = x^2 from x = 1 to x = 3, it is (9 − 1) / (3 − 1) = 4.
To find the steepness at a single point, imagine sliding the second point closer and closer to the first. The secants approach the tangent at that point. Writing the second point as x + h, the gradient of the secant is (f(x + h) − f(x)) / h, and we find the value it approaches as h approaches 0. This is called differentiation from first principles.
For f(x) = x^2: (f(x + h) − f(x)) / h = ((x + h)^2 − x^2) / h = (x^2 + 2xh + h^2 − x^2) / h = (2xh + h^2) / h = 2x + h. As h approaches 0, this approaches 2x. So the derivative of x^2 is 2x.
Repeating this for other powers reveals the power rule: the derivative of x^n is n × x^(n − 1). Bring the power down in front and reduce the power by 1. A number multiplying the term stays as a multiplier, the derivative of a sum is the sum of the derivatives, and the derivative of a constant is 0.
For example, if f(x) = 3x^2 + 4x − 7, then f'(x) = 6x + 4. The gradient of the curve at x = 1 is f'(1) = 6 + 4 = 10.
Derivatives describe rates in the real world. If d(t) is the distance travelled after t seconds, then d'(t) is the velocity at time t.
Worked examples
Average rate of change
Find the average rate of change of f(x) = x^2 between x = 1 and x = 3.
- f(1) = 1 and f(3) = 9.
- Average rate of change = (9 − 1) / (3 − 1) = 8 / 2 = 4.
Answer: 4
First principles for x^2
Use first principles to find the derivative of f(x) = x^2.
- Write the secant gradient: ((x + h)^2 − x^2) / h.
- Expand: (x^2 + 2xh + h^2 − x^2) / h = (2xh + h^2) / h.
- Divide each term by h (h is not zero): 2x + h.
- As h approaches 0, 2x + h approaches 2x.
Answer: f'(x) = 2x
Gradient of a tangent with the power rule
Find the gradient of the tangent to y = x^3 at x = 2.
- Differentiate with the power rule: dy/dx = 3x^2.
- Substitute x = 2: 3 × 2^2 = 3 × 4 = 12.
Answer: 12
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Find the average rate of change of f(x) = x^2 between x = 0 and x = 4.
4 — (f(4) − f(0)) / (4 − 0) = (16 − 0) / 4 = 4.
What is the derivative of x^5?
5x^4 — Power rule: bring down the 5 and reduce the power by 1, giving 5x^4.
What is the derivative of the constant function f(x) = 7?
0 — The graph of y = 7 is a horizontal line, which has gradient 0 everywhere.
A car's distance travelled, in metres, after t seconds is d(t) = t^2. What is its velocity at t = 3 seconds?
6 m/s — Velocity is the derivative: d'(t) = 2t, so d'(3) = 6 m/s. (9 m is the distance travelled, not the velocity.)
As the second point on a secant moves closer to the first point (h approaches 0), what does the gradient of the secant approach?
The gradient of the tangent — This limiting process is exactly how the gradient of the tangent, the derivative, is defined.
If f(x) = 3x^2 + 4x, find f'(1).
10 — f'(x) = 6x + 4, so f'(1) = 6 + 4 = 10.
Find the average rate of change of f(x) = 2x + 1 between x = 1 and x = 5.
2 — f(1) = 3 and f(5) = 11, so (11 − 3) / (5 − 1) = 8 / 4 = 2. For a straight line, this is just its gradient.
Find the gradient of the tangent to y = x^2 at x = −3.
−6 — dy/dx = 2x, so at x = −3 the gradient is 2 × (−3) = −6.
Watch out for
- Substituting into the original function instead of the derivative when asked for a gradient. The gradient comes from f'(x), not f(x).
- Reducing the power but forgetting to multiply by it, such as differentiating x^5 to x^4.
- Thinking the derivative of a constant is the constant itself. A constant does not change, so its rate of change is 0.
- Believing a tangent can only touch a curve once anywhere. A tangent touches at the point of interest, but it may cross the curve elsewhere.
Tips for parents
- Ask your learner to sketch the curve and a tangent line, and estimate whether the gradient should be positive, negative or zero before calculating.
- Use a car trip to discuss the difference between average speed for the whole trip and the speed shown on the speedometer at one moment.
- Calculus builds on algebra. If expanding or simplifying slows your learner down, a short review of those skills will help.
Go further
- Calculate the gradient of the secant to y = x^2 from x = 2 to x = 2.1, then to x = 2.01 and x = 2.001. What value do the answers approach?
- Use first principles to show that the derivative of x^3 is 3x^2. (Hint: (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3.)
- Roll a toy car down a ramp, record its position at regular times, and estimate its speed at different moments.