Differentiation, tangents and stationary points
Calculate a polynomial's local gradient and use its sign to identify maxima and minima.
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use algebraic indices and solve linear or quadratic equations.
- Interpret a graph's gradient and coordinate pairs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a negative gradient describe locally?
What you will learn
- Differentiate simple sums of polynomial power terms.
- Calculate the gradient at a given input.
- Find and classify stationary points.
The derivative is a local gradient rule
A tangent gives a curve's gradient at one point. Its slope can be estimated from a drawing, but a derivative gives a formula for the gradient. For y = x², the derivative dy/dx = 2x, so at x = 3 the gradient is six and at x = −1 it is −2. The derivative is not the curve's y-coordinate.
For a term axⁿ with non-negative integer n, differentiating gives anxⁿ⁻¹ when n ≥ 1. A constant term has derivative zero. For y = 3x³ − 4x + 7, dy/dx = 9x² − 4. Differentiate each term separately and then combine. This syllabus's differentiation work uses simple polynomial sums; reciprocal and fractional-power differentiation are beyond this lesson's scope.
Use an input for gradient and for height separately
For y = 2x³ − 3x² + 5, dy/dx = 6x² − 6x. At x = 2 the gradient is 12, while the curve's height is 16 − 12 + 5 = 9. Thus the point is (2, 9) and its tangent slope is 12. Substituting into the original and derivative answers different questions.
An estimated tangent gradient is found from two points on the tangent, not from two arbitrary points on the curve. Compare an estimate with a derivative where one is available. The units of a derivative are output units per input unit, so a distance differentiated with respect to time has speed units.
A zero gradient needs classification
Stationary points satisfy dy/dx = 0. For y = x³ − 3x, the derivative is 3x² − 3 and vanishes at x = ±1. Substituting into y gives (−1, 2) and (1, −2). The derivative is positive to the left of −1, negative between the two and positive to the right of one.
A change from positive to negative gradient identifies a local maximum; negative to positive identifies a local minimum. Here (−1, 2) is a maximum and (1, −2) is a minimum. A second derivative can also classify these when its value is non-zero, but a sign check works directly. A zero gradient alone does not guarantee either kind; a maximum over a restricted interval also requires checking its endpoints.
Pause and explain
What is the derivative of 5x³ − 2x + 9?
Worked example
Find and classify the stationary point of y = 2x² − 8x + 5.
Show the worked solution
- Differentiate: dy/dx = 4x − 8. Set it to zero to obtain x = 2.
- Substitute in the original: y = 8 − 16 + 5 = −3.
- The derivative is negative for x < 2 and positive for x > 2, so (2, −3) is a minimum.
Answer Minimum at (2, −3)
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For y = 4x² − 3x + 2, find dy/dx and its value at x = 2.
Give me a hint
Apply the power rule one term at a time.
Compare my reasoning
- The derivatives of the three terms are 8x, −3 and 0.
- Thus dy/dx = 8x − 3.
- At x = 2 the gradient is 16 − 3 = 13.
dy/dx = 8x − 3; gradient 13
Look for these in your work
- Multiplied by the exponent and reduced its power.
- Removed the constant term.
Find and classify the stationary points of y = x³ − 3x.
Give me a hint
Solve the derivative equation, then inspect its sign on the three intervals.
Compare my reasoning
- dy/dx = 3x² − 3, so x = −1 or 1.
- The points are (−1, 2) and (1, −2).
- Gradient changes + to − at −1 and − to + at 1, giving a maximum and minimum respectively.
Maximum (−1, 2); minimum (1, −2)
Look for these in your work
- Found coordinates using the original function.
- Classified by gradient sign changes.
A model gives area A = x(12 − x) m² for 0 ≤ x ≤ 12. Find the greatest area using differentiation.
Give me a hint
Expand the expression, find the zero derivative and compare endpoints.
Compare my reasoning
- A = 12x − x², so dA/dx = 12 − 2x.
- The derivative is zero at x = 6 and changes from positive to negative.
- A(6) = 36 m²; both endpoints have area zero, so this is the greatest area on the domain.
36 m² at x = 6 m
Look for these in your work
- Used the derivative to locate the maximum.
- Checked the restricted domain's endpoints.
Common mistakes
- Reporting the derivative value as the curve's height.
- Keeping a constant when differentiating.
- Calling every zero-gradient point a maximum without classification.
What can you explain now?
For y = x³ + 2x, find the gradient at x = −2.
Compare with the explanation
14
dy/dx = 3x² + 2, so the gradient is 3 × 4 + 2 = 14.
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Make it stick
Differentiate a polynomial of up to three terms. Evaluate its height and gradient at the same x, then explain why these are different quantities.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.