Understand a curve gradient as a limit of chord gradients
Compare an average gradient with the tangent gradient at one point and interpret first- and second-derivative notation.
Section 1.7 · Differentiation
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Calculate a chord gradient from two coordinate pairs.
- Expand a square or cube and distinguish a small non-zero change from zero.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the chord gradient through (1, 2) and (3, 10)?
What you will learn
- Explain how chord gradients approach the tangent gradient at a point.
- Interpret f′(x), dy/dx, f″(x) and d²y/dx², distinguishing a rule from its value at one input.
Two-point averages approach a one-point gradient
A chord through two distinct curve points has gradient Δy/Δx, an average change over the input interval. Hold one point fixed and move the other towards it. If the chord gradients approach a single finite value from the permitted directions, that value is the tangent gradient at the fixed point. A small non-zero interval approximates the derivative; it is not the same as substituting a zero denominator.
For y = x² + 1, fix P(2, 5) and take Q at x = 2 + h with h ≠ 0. The chord gradient is [(2 + h)² + 1 − 5]/h = (4h + h²)/h = 4 + h. For h = 0.8, 0.3 and 0.03 the gradients are 4.8, 4.3 and 4.03; negative h gives values approaching 4 from below. The tangent gradient at x = 2 is 4. This is an informal example of a limit, rather than a requirement to derive every rule from first principles.
A derivative is a function; evaluation gives one gradient
For y = f(x), the notations f′(x) and dy/dx mean the first derivative: the gradient rule as x varies. f′(2) is its value at input 2. For f(x) = x² + 1, f′(x) = 2x and f′(2) = 4. The second derivative f″(x), also written d²y/dx², is the derivative of that gradient rule; here it is 2, not the square of f′(x).
If an output is distance in metres and the input is time in seconds, its first derivative has units m/s and its second derivative m/s². The first is an instantaneous rate, while a chord gives an average over an interval. Always state which variable is changing and at which input you evaluate a requested rate; an average and an instantaneous rate can be different even over a short interval.
Pause and explain
For a chord gradient 4 + h with h ≠ 0, what happens as h tends to zero?
Worked example
For y = x² + 1, find the chord gradient from x = 2 to x = 2 + h, and describe its limit as h approaches zero.
Show the worked solution
- The output difference is (2 + h)² + 1 − 5 = 4h + h².
- For h ≠ 0, divide by h to obtain chord gradient 4 + h.
- As h approaches zero, this approaches 4; y′ = 2x also gives tangent gradient 4 at x = 2.
Answer Chord gradient 4 + h for h ≠ 0; limiting tangent gradient 4.
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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 = x², calculate the chord gradient from x = 3 to x = 3 + h for h = 0.2 and −0.2, and find its limiting value.
Give me a hint
Expand (3 + h)² and simplify the quotient while h is non-zero.
Compare my reasoning
- The quotient is [(3 + h)² − 9]/h = (6h + h²)/h = 6 + h.
- For h = 0.2 it is 6.2, and for h = −0.2 it is 5.8.
- Both approach 6 as h tends to zero, giving the tangent gradient at x = 3.
Chord gradients 6.2 and 5.8; limiting tangent gradient 6.
Look for these in your work
- Use the signed input change in the quotient.
- Distinguish the two average gradients from their shared limit.
For y = x³, fix x = 1 and take a second point at x = 1 + h. Find the chord gradient, its values at h = ±0.1 and the limiting tangent gradient.
Give me a hint
Expand the cube, cancel h only for h ≠ 0 and then consider small signed changes.
Compare my reasoning
- The quotient is [(1 + h)³ − 1]/h = 3 + 3h + h² for h ≠ 0.
- At h = 0.1 it is 3.31; at h = −0.1 it is 2.71.
- The limiting tangent gradient is 3, consistent with y′ = 3x² at x = 1.
Chord gradient 3 + 3h + h²; values 3.31 and 2.71; limiting gradient 3.
Look for these in your work
- Retain every term when expanding and dividing the cube difference.
- Check the limit from both signed directions.
A supplied distance model is s(t) = t² + 2t metres. Compare the average speed from t = 2 to 2.1 seconds with the instantaneous speed and its rate of change at t = 2.
Give me a hint
Calculate the two-point average separately from the evaluated first and second derivatives.
Compare my reasoning
- s(2) = 8 and s(2.1) = 8.61, so the interval average is (8.61 − 8)/0.1 = 6.1 m/s.
- The first derivative is s′(t) = 2t + 2, giving instantaneous speed 6 m/s at t = 2.
- The second derivative is s″(t) = 2, so the speed's rate of change is 2 m/s².
Average 6.1 m/s; instantaneous speed 6 m/s; rate of change of speed 2 m/s².
Look for these in your work
- Use the actual elapsed interval for the average.
- Evaluate the correct derivative and give its corresponding units.
Common mistakes
- An average chord gradient is different from a tangent gradient at one input.
- Keep h non-zero in the chord quotient; take a limit instead of dividing by zero.
- The second derivative is the derivative of the first, not its square.
What can you explain now?
For f(x) = x² + 1, give f′(x), f′(2) and f″(x).
Compare with the explanation
f′(x) = 2x; f′(2) = 4; f″(x) = 2.
The first derivative is a rule, its evaluated value is one gradient, and differentiating the rule again gives the second derivative.
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Make it stick
Explain how a chord with non-zero input change can approximate a tangent gradient. Distinguish a first-derivative rule, an evaluated gradient and a second derivative.
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.