Interpret a derived function and its notation
Connect a derivative to an instantaneous gradient or rate of change and read first- and second-derivative notation correctly.
Section 14 · Calculus
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which notation represents the first derivative of y with respect to x?
What you will learn
- Understand the derived function as the gradient of a curve at each input
- Interpret f′(x), f″(x), dy/dx and d²y/dx²
- Use the given derivative rules without attempting first-principles differentiation
A derivative describes a local rate of change
For a curve y = f(x), the derivative f′(x) gives the gradient of the tangent at input x. It can also represent the instantaneous rate at which y changes as x changes.
A nearby average gradient uses two points; the derivative is the limiting local gradient as the second point approaches the first. In 0606 only an informal idea of this limit is expected: differentiation from first principles is not required.
Several notations name the same derivative
For y = f(x), f′(x) and dy/dx both mean the first derivative. The second derivative is f″(x) or d²y/dx²; it describes how the gradient changes.
When a physical quantity depends on time, ds/dt or dv/dt makes that input explicit. The derivative's units are output units divided by input units, such as metres per second for displacement differentiated with respect to time.
Pause and explain
What does f″(x) describe?
Worked example
A curve has f′(2) = 5. Interpret this value and state what f″(2) would describe.
Show the worked solution
- At x = 2, the tangent gradient is 5.
- For a small change in x near 2, f changes at approximately 5 units of output per unit of x.
- f″(2) describes how the gradient itself is changing at x = 2.
Answer The tangent gradient at x = 2 is 5; f″(2) is the rate of change of the gradient there.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 s = s(t), explain the meaning and units of ds/dt if s is measured in metres and t in seconds.
Give me a hint
Divide the output unit by the input unit and interpret it as an instantaneous rate.
Compare my reasoning
- ds/dt is the derivative of displacement with respect to time.
- Its units are metres per second.
- It represents instantaneous velocity, including direction through its sign.
ds/dt is velocity, measured in m/s.
Look for these in your work
- Derivative variables identified
- Units divided in the correct order
- Physical meaning stated
A function has f′(3) = 0 and f″(3) = 8. State what each value tells you, without deciding the stationary point's type yet.
Give me a hint
The first derivative describes gradient; the second describes its change.
Compare my reasoning
- f′(3) = 0 means the tangent is horizontal at x = 3.
- f″(3) = 8 means the gradient is increasing locally at a rate of 8 gradient-units per x-unit.
- Classification needs a stationary-point test and nearby curve information.
Horizontal tangent at x = 3; the gradient's derivative there is 8. The point's type is not concluded here.
Look for these in your work
- First derivative interpreted
- Second derivative interpreted
- No unsupported classification made
A distance-time model is d = d(t). Which derivative represents speed and what are its units if d is in kilometres and t in hours?
Give me a hint
Differentiate the output quantity with respect to its input.
Compare my reasoning
- The derivative is dd/dt.
- Its units are kilometres per hour.
- It gives the instantaneous rate of distance change; signed displacement would instead give velocity.
dd/dt represents speed for distance d, in km/h.
Look for these in your work
- Correct derivative named
- Units follow the variables
- Distance distinguished from displacement
Common mistakes
- Confusing the value f(x) with its gradient f′(x)
- Reading f″(x) as a square of f′(x)
- Forgetting rate units depend on the input variable
- Trying to use first principles when the syllabus excludes it
What can you explain now?
What does dy/dx represent on a graph y = f(x)?
Compare with the explanation
The gradient of the curve at x
The derivative gives the instantaneous tangent gradient.
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Make it stick
What do first and second derivatives tell you, and which notation can represent each?
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.