IGCSE Maths · Algebra and Graphs · Extended only

Differentiation: Finding the Exact Gradient of a Curve Extended only

This entire note is Extended-only. Differentiation replaces the “draw a tangent and estimate” method from earlier notes with an exact formula for the gradient at any point on a curve — and it’s what makes finding turning points precise instead of approximate.

Why this matters Earlier, estimating the speed at an instant meant drawing a tangent by eye and hoping it was accurate. Differentiation gives you the gradient formula directly — no drawing, no estimating, exact every time.

1. Recap: Estimating a Gradient with a Tangent

As covered in Graphs in Practical Situations, a tangent is a straight line touching a curve at one point — its gradient estimates the curve’s steepness right at that point. Differentiation is the algebraic tool that finds this gradient exactly, for any point, without drawing anything.

2. The Power Rule for Differentiation

To differentiate a term of the form axⁿ, multiply by the power, then reduce the power by 1. The result is called the derivative, written dy/dx.

d/dx (axⁿ) = anxⁿ⁻¹
Worked examples Differentiate y = x³.
dy/dx = 3x²  (multiply by the power 3, reduce the power to 2)

Differentiate y = 5x⁴.
dy/dx = 20x³  (5 × 4 = 20, power reduces from 4 to 3)

Differentiate y = x.
dy/dx = 1  (since x = x¹, the rule gives 1 × x⁰ = 1)

Differentiate y = 7 (a constant).
dy/dx = 0  (a constant is 7x⁰; its rate of change is always zero — a flat line has zero gradient everywhere)
Try it — Power Rule
  1. Differentiate y = x⁵.
  2. Differentiate y = 4x².
  3. Differentiate y = 6x.
  4. Differentiate y = −3x³.
Show answers
  1. dy/dx = 5x⁴
  2. dy/dx = 8x
  3. dy/dx = 6
  4. dy/dx = −9x²

3. Differentiating Sums of Terms

Differentiate each term separately, then add the results — this works for sums of up to three axⁿ terms.

Worked example Differentiate y = 3x³ − 5x² + 2x.
Term by term: d/dx(3x³) = 9x², d/dx(−5x²) = −10x, d/dx(2x) = 2
Answer: dy/dx = 9x² − 10x + 2
Worked example — including a constant Differentiate y = 4x² + 3x − 7.
d/dx(4x²) = 8x, d/dx(3x) = 3, d/dx(−7) = 0
Answer: dy/dx = 8x + 3
Try it — Differentiating Sums
  1. Differentiate y = 2x³ + 4x² − 5x.
  2. Differentiate y = x² − 6x + 9.
  3. Differentiate y = 5x³ − 2x + 8.
Show answers
  1. dy/dx = 6x² + 8x − 5
  2. dy/dx = 2x − 6
  3. dy/dx = 15x² − 2

4. Finding the Gradient at a Specific Point

Differentiate first, then substitute the given x-value into dy/dx.

Worked example Find the gradient of y = x³ − 4x at the point where x = 2.
Differentiate: dy/dx = 3x² − 4
Substitute x = 2: dy/dx = 3(4) − 4 = 12 − 4 = 8
Try it — Gradient at a Point
  1. Find the gradient of y = x² − 3x at the point where x = 4.
  2. Find the gradient of y = 2x³ − x² at the point where x = −1.
Show answers
  1. dy/dx = 2x − 3. At x=4: 8−3 = 5
  2. dy/dx = 6x² − 2x. At x=−1: 6(1) − 2(−1) = 6 + 2 = 8

5. Finding Stationary Points

A stationary point (turning point) is where the gradient is exactly zero — the curve is momentarily flat. Set dy/dx = 0 and solve for x, then substitute back into the original equation to find the matching y-value.

Worked example maximum minimum Find the stationary points of y = x³ − 3x² − 9x + 5.
Differentiate: dy/dx = 3x² − 6x − 9
Set equal to zero: 3x² − 6x − 9 = 0 → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1
Find y at each: at x = 3, y = 27 − 27 − 27 + 5 = −22. At x = −1, y = −1 − 3 + 9 + 5 = 10
Stationary points: (3, −22) and (−1, 10)
Try it — Stationary Points
  1. Find the stationary point(s) of y = x² − 6x + 5.
  2. Find the stationary points of y = x³ − 12x.
Show answers
  1. dy/dx=2x−6=0 → x=3. y=9−18+5=−4. Stationary point: (3, −4)
  2. dy/dx=3x²−12=0 → x²=4 → x=2 or x=−2. At x=2: y=8−24=−16. At x=−2: y=−8+24=16. Stationary points: (2,−16) and (−2,16)

6. Discriminating Between Maxima and Minima

Any of these three methods is acceptable: (1) an accurate sketch, (2) checking the sign of the second derivative, or (3) inspecting the gradient just before and after the point.

Method — second derivative Using y = x³ − 3x² − 9x + 5 from Section 5 (stationary points at x = 3 and x = −1):
Second derivative: differentiate dy/dx = 3x² − 6x − 9 again → d²y/dx² = 6x − 6
At x = 3: 6(3) − 6 = 12, which is positive → this is a minimum
At x = −1: 6(−1) − 6 = −12, which is negative → this is a maximum
Method — gradient either side Check the sign of dy/dx = 3x² − 6x − 9 just before and after x = 3.
At x = 2.5: 3(6.25) − 15 − 9 = 18.75 − 24 = −5.25 (negative — curve falling)
At x = 3.5: 3(12.25) − 21 − 9 = 36.75 − 30 = 6.75 (positive — curve rising)
Falling then rising means x = 3 is a minimum — consistent with the second derivative method.
Try it — Maxima and Minima
  1. For y = x² − 6x + 5 (stationary point at x = 3, from the previous practice set), use the second derivative to confirm whether it’s a maximum or minimum.
  2. For y = x³ − 12x (stationary points at x = 2 and x = −2), use the second derivative to classify each one.
Show answers
  1. dy/dx=2x−6, so d²y/dx²=2, which is always positive → minimum.
  2. d²y/dx² = 6x. At x=2: 12>0 → minimum. At x=−2: −12<0 → maximum.
Common mistakes to watch for
  • Forgetting to reduce the power by 1 after multiplying — differentiating x⁴ gives 4x³, not 4x⁴.
  • Differentiating a constant as though it were a variable term — any plain number differentiates to 0.
  • Stopping after finding the x-value of a stationary point, without substituting back to find the matching y-value.
  • Confusing the derivative dy/dx (used to find stationary points) with the second derivative d²y/dx² (used to classify them) — they answer different questions.
  • Assuming a positive second derivative means a maximum — it’s the opposite: positive means minimum, negative means maximum.

7. Quick-Fire Challenge Round

True or False?
  1. The derivative of x⁵ is 5x⁴.
  2. The derivative of a constant is the constant itself.
  3. At a stationary point, dy/dx = 0.
  4. A negative second derivative at a stationary point indicates a minimum.
Show answers
  1. True.
  2. False — the derivative of any constant is 0, not the constant itself.
  3. True.
  4. False — a negative second derivative indicates a maximum, not a minimum.
Spot the Error
  1. “Differentiate y = 3x⁴: dy/dx = 3x³.”
  2. “Differentiate y = 6x: dy/dx = 6x.”
  3. “Found a stationary point at x = 2, so the stationary point is (2, 0).”
Show answers
  1. The coefficient must also multiply by the original power: 3 × 4 = 12, so dy/dx = 12x³, not 3x³.
  2. y = 6x = 6x¹, so dy/dx = 6 × 1 × x⁰ = 6, a constant — not 6x. (Differentiating a linear term always gives just its gradient, a number.)
  3. Finding x = 2 only gives the x-coordinate — you must substitute x = 2 back into the original equation for y to find the matching y-coordinate; it is not automatically 0.

8. Mixed Practice — Bringing It All Together

  1. Differentiate y = 4x³ − 2x² + x.
  2. Find the gradient of y = x² − 5x + 2 at x = 3.
  3. Find the stationary point of y = x² + 4x − 1.
  4. Determine whether the stationary point in question 3 is a maximum or minimum, using the second derivative.
  5. Find the stationary points of y = x³ − 27x, and classify each using the second derivative.
Show answers
  1. dy/dx = 12x² − 4x + 1
  2. dy/dx = 2x−5. At x=3: 1
  3. dy/dx=2x+4=0 → x=−2. y=4−8−1=−5. Stationary point: (−2, −5)
  4. d²y/dx²=2 (always positive) → minimum
  5. dy/dx=3x²−27=0 → x²=9 → x=3 or x=−3. d²y/dx²=6x. At x=3: 18>0 → minimum. At x=−3: −18<0 → maximum.

Practice at Home

A large set again — take it steadily.

Part A — Power rule and sums
  1. Differentiate y = x⁶.
  2. Differentiate y = 7x².
  3. Differentiate y = 2x³ + 5x² − 4x + 9.
  4. Differentiate y = 10 (a constant on its own).
Part B — Gradient at a point
  1. Find the gradient of y = x³ − 2x at x = 1.
  2. Find the gradient of y = 3x² − 7x + 2 at x = 0.
  3. Find the gradient of y = x³ + x² − 5x at x = −1.
Part C — Stationary points
  1. Find the stationary point of y = x² − 10x + 21.
  2. Find the stationary points of y = x³ − 3x.
  3. Find the stationary points of y = x³ − 48x.
Part D — Maxima and minima
  1. Classify the stationary point of y = x² − 10x + 21 (from Part C) as a maximum or minimum.
  2. Classify each stationary point of y = x³ − 3x (from Part C) using the second derivative.
  3. Classify each stationary point of y = x³ − 48x (from Part C) using the second derivative.
Part E — Applied / challenge
  1. A ball’s height (in metres) after t seconds is h = 20t − 5t². Find dh/dt, and use it to find the time at which the ball reaches its maximum height. Then find that maximum height.
  2. A company’s profit (in thousands of dollars) is modelled by P = −2x² + 40x − 150, where x is the number of units sold (in hundreds). Find the stationary point, and explain what it represents in context.
  3. Explain why a cubic of the form y = ax³ + d (no x² or x term) never has a stationary point, using differentiation to justify your answer.

Key Vocabulary Recap

Derivative — the gradient function, written dy/dx
Power rule — d/dx(axⁿ) = anxⁿ⁻¹
Stationary point — a point where dy/dx = 0
Second derivative — the derivative of the derivative, written d²y/dx²
Maximum/minimum — a stationary point that is the highest/lowest locally

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section E2.12 (Extended subject content) — Differentiation: estimating gradients with tangents, the derivatives of axⁿ terms (sums of up to three), gradients at a point, stationary points, and discriminating between maxima and minima. There is no Core equivalent for this section. Points of inflection are not required at this level.

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