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A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.7 · ~9 min read

Differentiation

The gradient as the limit of chord gradients, differentiating xⁿ for any rational n, the chain rule, tangents and normals, increasing and decreasing functions, stationary points and connected rates of change.

Gradient as a limitPower ruleChain ruleTangents & normalsIncreasing & decreasingStationary pointsSecond derivativeOptimisation
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • The derivative dy/dx is the gradient of the curve at a point: the limit of the gradients of chords as they shrink towards that point.
  • Power rule: d/dx(xⁿ) = nxⁿ⁻¹ for any rational n. Rewrite roots and fractions as powers first.
  • Chain rule: d/dx[(ax + b)ⁿ] = an(ax + b)ⁿ⁻¹; in general dy/dx = dy/du × du/dx.
  • Stationary points are where dy/dx = 0. If d²y/dx² < 0 it is a maximum; if > 0, a minimum.
  • Connected rates: dy/dt = dy/dx × dx/dt. Write the chain down before substituting numbers.

By the end you can

  • Explain the gradient of a curve as the limit of chord gradients.
  • Differentiate sums of powers of x and composite functions with the chain rule.
  • Find equations of tangents and normals.
  • Decide where a function is increasing or decreasing, and prove a function is always increasing.
  • Find stationary points, determine their nature, and solve optimisation problems.
  • Solve connected rates of change problems.

1What a derivative is

Chords become a tangent

xy0.511.522.512340P(1, 1)chord to x = 2: gradient 3to x = 1.5: gradient 2.5to x = 1.2: gradient 2.2to x = 1 + h: gradient 2 + hred tangent: gradient 2As Q slides towards P, the chord gradient 2 + h tends to 2: the gradient of the curve at P.
The chord PQ on y = x² from x = 1 to x = 1 + h has gradient 2 + h. As h → 0 this tends to 2.
  • Gradient of the chord from (1, 1) to (1 + h, (1 + h)²): [(1 + h)² − 1]/h = (2h + h²)/h = 2 + h. As h → 0, it tends to 2.
  • Notation: dy/dx or f'(x) is the gradient function; d²y/dx² or f''(x) is the derivative of the derivative (the rate of change of the gradient).
  • Paper 1 only needs this informal idea of a limit. You won't be asked to differentiate from first principles.

2Differentiating powers

The power rule

d/dx(xⁿ) = nxⁿ⁻¹ FORMULA LIST (MF19)
for any rational n; d/dx(k) = 0 for a constant k
  • Constant multiples and sums: d/dx[af(x) + bg(x)] = af'(x) + bg'(x).
  • Rewrite first: √x = x1/2, 1/x³ = x⁻³, 5/(2x) = (5/2)x⁻¹. The 2024 report saw 1/(2x) differentiated as 2x⁻¹: it is ½x⁻¹, with derivative −½x⁻².
Worked example

Find dy/dx for y = 4√x − 3/x² + 5.

  1. y = 4x1/2 − 3x⁻² + 5 M1
  2. dy/dx = 2x−1/2 + 6x⁻³ = 2/√x + 6/x³ A1 A1

The chain rule

dy/dx = dy/du × du/dx
for y = f(u) where u = g(x)
d/dx[(ax + b)ⁿ] = an(ax + b)ⁿ⁻¹
the most common case in Paper 1
  • Why: if u changes by a small δu and y by δy, then δy/δx = (δy/δu)(δu/δx). Letting δx → 0 gives the chain rule.
Worked example

Differentiate y = (3x − 1)⁵ and y = 6/√(2x + 5).

  1. dy/dx = 5(3x − 1)⁴ × 3 = 15(3x − 1)⁴ B1
  2. y = 6(2x + 5)−1/2 ⇒ dy/dx = 6 × (−½)(2x + 5)−3/2 × 2 M1
  3. = −6(2x + 5)−3/2 A1
Common mistakeForgetting the inner derivative (the × 3 or × 2). It is the most common chain-rule error.

3Tangents and normals

Method

  • 1 · Find the y-coordinate at the given x (if not given). 2 · Find dy/dx at that x: this is the tangent gradient m. 3 · The normal gradient is −1/m. 4 · Use y − y₁ = m(x − x₁).
Worked example

Find the equations of the tangent and the normal to y = x³ − 3x at the point where x = 2.

  1. y = 8 − 6 = 2; dy/dx = 3x² − 3 = 9 at x = 2 M1 A1
  2. tangent: y − 2 = 9(x − 2) ⇒ y = 9x − 16 A1
  3. normal gradient −1/9: y − 2 = −(1/9)(x − 2) ⇒ x + 9y = 20 M1 A1

4Increasing and decreasing functions

  • Increasing where f'(x) > 0; decreasing where f'(x) < 0.
  • "Show that f is increasing for all x" needs an argument for every x, such as completing the square on f'(x). Substituting a few values is not a proof; the 2024 report penalised exactly that.
Worked example

Find the set of values of x for which f(x) = x³ − 6x² + 9x + 1 is decreasing.

  1. f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) M1
  2. decreasing ⇒ f'(x) < 0 ⇒ 1 < x < 3 A1
Worked example

Show that f(x) = x³ + 3x² + 5x is an increasing function.

  1. f'(x) = 3x² + 6x + 5 B1
  2. = 3(x + 1)² + 2 M1
  3. (x + 1)² ≥ 0, so f'(x) ≥ 2 > 0 for all x, hence f is increasing A1

5Stationary points

Finding and classifying

  • Stationary point: dy/dx = 0. Solve for x, then find y.
  • Second derivative test: d²y/dx² < 0 ⇒ maximum; d²y/dx² > 0 ⇒ minimum. If d²y/dx² = 0, the test fails: check the sign of dy/dx on either side instead.
  • Points of inflexion are not in the 9709 syllabus, so you won't be asked to find them.
xy-2-1123-4-2240max (−1, 2)min (1, −2)P(2, 2)tangentm = 9y = x³ − 3xdy/dx = 3x² − 3d²y/dx² = 6xStationary points where dy/dx = 0. Sign of d²y/dx² decides: negative → max, positive → min.
y = x³ − 3x: a maximum at (−1, 2), a minimum at (1, −2), and the tangent at (2, 2).
Worked example

Find the stationary points of y = x³ − 3x and determine their nature.

  1. dy/dx = 3x² − 3 = 0 ⇒ x = ±1 M1 A1
  2. points (1, −2) and (−1, 2) A1
  3. d²y/dx² = 6x: at x = 1 it is 6 > 0 ⇒ minimum (1, −2); at x = −1 it is −6 < 0 ⇒ maximum (−1, 2) M1 A1
Common mistakeNot showing the sign of the second derivative. Write "d²y/dx² = 6 > 0, so minimum". Just stating "minimum" can lose the mark.

Optimisation

  • 1 · Write the quantity to maximise or minimise in terms of one variable, using any constraint given. 2 · Differentiate and set to zero. 3 · Check it is a max or min. 4 · Answer the question asked (often the value of the quantity, not x).
Worked example

A farmer has 60 m of fencing to make a rectangular pen against a long straight wall, so only three sides need fencing. Find the maximum area.

  1. let the sides perpendicular to the wall be y, the parallel side x: x + 2y = 60 B1
  2. A = xy = (60 − 2y)y = 60y − 2y² M1
  3. dA/dy = 60 − 4y = 0 ⇒ y = 15, x = 30 M1 A1
  4. d²A/dy² = −4 < 0 ⇒ maximum; A = 450 m² A1

6Rates of change

Connected rates

dy/dt = dy/dx × dx/dt
chain rule with time
  • Method: write the formula linking the two quantities, differentiate it, then multiply by the given rate. Label each rate with its meaning and units.
Worked example

The radius of a spherical balloon increases at 0.2 cm s⁻¹. Find the rate at which its volume increases when r = 5 cm.

  1. V = (4/3)πr³ ⇒ dV/dr = 4πr² = 100π when r = 5 M1
  2. dV/dt = dV/dr × dr/dt = 100π × 0.2 M1
  3. = 20π ≈ 62.8 cm³ s⁻¹ A1
Worked example

A point moves along y = √(x² + 9) so that its x-coordinate increases at 3 units per second. Find the rate of increase of y when x = 4.

  1. dy/dx = x(x² + 9)−1/2 = 4/5 at x = 4 M1 A1
  2. dy/dt = (4/5) × 3 = 2.4 units per second A1
Linking questionPaper 3 adds e^x, ln x, trig functions, products, quotients and implicit equations in Further differentiation; integration reverses differentiation in Integration; and Mechanics uses derivatives for velocity and acceleration in Kinematics in a straight line.

Test yourself — cover the answers first

Quick check

  1. Differentiate y = 3x⁴ − 2x + 7.
  2. Differentiate y = 1/√x.
  3. Differentiate y = (5 − 2x)³.
  4. What is the gradient of the normal if the tangent gradient is −4?
  5. At a stationary point d²y/dx² = −3. Is it a max or a min?

Exam-style

  1. The curve y = 2x + 8/x has two stationary points. Find them and determine their nature.
  2. A curve has equation y = (2x − 3)³ − 6x. (a) Find dy/dx and d²y/dx². (b) Find the x-coordinates of the stationary points and their nature.
  3. The volume of water in a tank is V = 20h² + 30h cm³ when the depth is h cm. Water is added at 70 cm³ s⁻¹. Find the rate at which h is increasing when h = 4.
Show answers
  1. 12x³ − 2
  2. −½x−3/2
  3. −6(5 − 2x)²
  4. ¼
  5. maximum
  6. dy/dx = 2 − 8/x² = 0 ⇒ x = ±2; (2, 8) minimum (d²y/dx² = 16/x³ = 2 > 0); (−2, −8) maximum
  7. (a) 6(2x − 3)² − 6; 24(2x − 3)
    (b) (2x − 3)² = 1 ⇒ x = 1 (max, d²y/dx² = −24) and x = 2 (min, +24)
  8. dV/dh = 40h + 30 = 190; dh/dt = 70/190 = 0.368 cm s⁻¹
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.