Lesson 6: Differentiation | Additional Mathematics

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1. Introduction — Gradient of a Curve

Derivative: The derivative dy/dx (or f'(x)) gives the gradient of the curve y = f(x) at any point. It is the instantaneous rate of change of y with respect to x — the gradient of the tangent to the curve at that point.

Differentiation from First Principles

Definition — First Principles

f'(x) = lim[h→0] [f(x+h) − f(x)] / h

This is the gradient of the chord from (x, f(x)) to (x+h, f(x+h)) as h → 0 (chord becomes tangent).

🔵 Proof — Differentiate f(x) = x² from First Principles

f(x+h) = (x+h)² = x²+2xh+h²

f(x+h)−f(x) = x²+2xh+h²−x² = 2xh+h²

[f(x+h)−f(x)]/h = (2xh+h²)/h = 2x+h

As h→0: f'(x) = 2x ✓

2. Standard Differentiation Rules

Basic Rules

Standard Results — Must Know All

d/dx (xⁿ) = nxⁿ⁻¹ (Power Rule)

d/dx (axⁿ) = anxⁿ⁻¹

d/dx (eˣ) = eˣ d/dx (eᵃˣ) = aeᵃˣ

d/dx (ln x) = 1/x d/dx (ln(ax)) = 1/x

d/dx (sin x) = cos x d/dx (sin ax) = a cos ax

d/dx (cos x) = −sin x d/dx (cos ax) = −a sin ax

d/dx (tan x) = sec²x d/dx (tan ax) = a sec²ax

All angles must be in radians when differentiating trig functions.

📐 Worked Example 1 — Basic Differentiation

Differentiate: (a) y = 4x³−5x²+3x−7 (b) y = 3√x − 2/x² (c) y = 2e³ˣ + sin 4x − ln 5x

(a) dy/dx = 12x² − 10x + 3

(b) Rewrite: y = 3x^(½) − 2x^(−2)
dy/dx = 3×½x^(−½) − 2×(−2)x^(−3) = 3/(2√x) + 4/x³

(c) dy/dx = 6e³ˣ + 4cos 4x − 1/x = 6e³ˣ + 4cos 4x − 1/x
Note: d/dx(ln 5x) = d/dx(ln 5 + ln x) = 1/x

The Chain Rule

Chain Rule — Composite Functions

dy/dx = (dy/du) × (du/dx)

If y = [f(x)]ⁿ then dy/dx = n[f(x)]ⁿ⁻¹ × f'(x)

Used whenever the function is a composition — "a function of a function." Differentiate the outside, keeping the inside unchanged, then multiply by the derivative of the inside.

📐 Worked Example 2 — Chain Rule

Differentiate: (a) y = (3x²−5)⁷ (b) y = √(4x+1) (c) y = e^(x²+3x) (d) y = sin³x

(a) Outside: (...)⁷ → 7(...)⁶. Inside: 3x²−5 → 6x
dy/dx = 7(3x²−5)⁶ × 6x = 42x(3x²−5)⁶

(b) y = (4x+1)^(½). Outside: ½(...)^(−½). Inside: 4x+1 → 4
dy/dx = ½(4x+1)^(−½) × 4 = 2/√(4x+1)

(c) Outside: e^(...) → e^(...). Inside: x²+3x → 2x+3
dy/dx = e^(x²+3x) × (2x+3) = (2x+3)e^(x²+3x)

(d) y = (sin x)³. Outside: 3(...)². Inside: sin x → cos x
dy/dx = 3sin²x × cos x = 3sin²x cos x

The Product Rule

Product Rule

If y = uv then dy/dx = u(dv/dx) + v(du/dx)

Used when two functions are multiplied together and neither can be simplified away. Memory aid: "first × diff of second + second × diff of first."

📐 Worked Example 3 — Product Rule

Differentiate: (a) y = x³ sin 2x (b) y = e²ˣ(3x−1)⁴

(a) u = x³, v = sin 2x. du/dx = 3x², dv/dx = 2cos 2x
dy/dx = x³(2cos 2x) + sin 2x(3x²) = 2x³cos 2x + 3x²sin 2x
Factorise: = x²(2x cos 2x + 3 sin 2x)

(b) u = e²ˣ, v = (3x−1)⁴. du/dx = 2e²ˣ, dv/dx = 4(3x−1)³×3 = 12(3x−1)³
dy/dx = e²ˣ × 12(3x−1)³ + (3x−1)⁴ × 2e²ˣ
= e²ˣ(3x−1)³[12 + 2(3x−1)]
= 2e²ˣ(3x−1)³(3x+5)

The Quotient Rule

Quotient Rule

If y = u/v then dy/dx = [v(du/dx) − u(dv/dx)] / v²

Used when one function is divided by another. Memory aid: "bottom × diff of top minus top × diff of bottom, all over bottom squared." (vdu − udv) / v²

📐 Worked Example 4 — Quotient Rule

Differentiate: (a) y = (2x+1)/(x²−3) (b) y = sin x / eˣ

(a) u=2x+1, v=x²−3. du/dx=2, dv/dx=2x
dy/dx = [(x²−3)(2) − (2x+1)(2x)] / (x²−3)²
= [2x²−6−4x²−2x] / (x²−3)²
= (−2x²−2x−6) / (x²−3)² = −2(x²+x+3)/(x²−3)²

(b) u=sin x, v=eˣ. du/dx=cos x, dv/dx=eˣ
dy/dx = [eˣ cos x − sin x × eˣ] / e²ˣ
= eˣ(cos x−sin x) / e²ˣ
= (cos x − sin x) / eˣ

Which Rule to Use:
• Single function of x with power → Power Rule
• Function inside another function → Chain Rule
• Two functions multiplied → Product Rule
• Two functions divided → Quotient Rule
• Often need Chain Rule inside Product or Quotient Rule — apply from outside in.

3. Tangents and Normals

Tangent: The straight line that touches the curve at a point, with gradient = dy/dx at that point.
Normal: The line perpendicular to the tangent at the same point. Its gradient = −1 / (dy/dx).

Finding the Tangent and Normal at (x₁, y₁):
1. Differentiate y = f(x) to find dy/dx.
2. Substitute x = x₁ to find the gradient m of the tangent.
3. Tangent equation: y − y₁ = m(x − x₁)
4. Normal gradient = −1/m. Normal equation: y − y₁ = (−1/m)(x − x₁)

📐 Worked Example 5 — Tangent and Normal

Find the equations of the tangent and normal to y = x³ − 4x + 2 at the point where x = 2.

Find y at x=2: y = 8−8+2 = 2. Point: (2, 2)

dy/dx = 3x²−4. At x=2: dy/dx = 12−4 = 8.
Tangent gradient m = 8.

Tangent: y−2 = 8(x−2) → y = 8x−14

Normal gradient: −1/8
Normal: y−2 = −⅛(x−2) → 8y−16 = −x+2 → x+8y = 18

📐 Worked Example 6 — Tangent Parallel to a Line

Find the point(s) on y = 2x³ − 9x² + 12x − 3 where the tangent is parallel to y = 3x − 1.

Parallel means equal gradient. Gradient of y=3x−1 is 3.
Set dy/dx = 3: dy/dx = 6x²−18x+12 = 3

6x²−18x+9 = 0 → 2x²−6x+3 = 0
x = (6±√(36−24))/4 = (6±√12)/4 = (3±√3)/2

x = (3+√3)/2 ≈ 2.37 or x = (3−√3)/2 ≈ 0.634
Substitute each back into y to find the coordinates.

4. Stationary Points and Curve Sketching

Stationary Point: A point where dy/dx = 0 — the tangent is horizontal. There are three types: local maximum, local minimum, and point of inflection.

Finding and Classifying Stationary Points:
Step 1: Differentiate → find dy/dx
Step 2: Set dy/dx = 0 → solve for x (stationary x-values)
Step 3: Find y by substituting x back into the equation
Step 4: Classify using second derivative d²y/dx²:
     d²y/dx² > 0 at point → Minimum (curve concave up ∪)
     d²y/dx² < 0 at point → Maximum (curve concave down ∩)
     d²y/dx² = 0 → inconclusive — use first derivative test (sign change)
OR: Step 4 alternative: First derivative test — check sign of dy/dx on either side of stationary point.

📐 Worked Example 7 — Stationary Points

Find and classify all stationary points of y = 2x³ − 3x² − 12x + 4.

dy/dx = 6x²−6x−12 = 6(x²−x−2) = 6(x−2)(x+1)
Set dy/dx=0: x=2 or x=−1

Find y-values:
x=2: y=16−12−24+4=−16 → point (2, −16)
x=−1: y=−2−3+12+4=11 → point (−1, 11)

Second derivative: d²y/dx² = 12x−6
At x=2: d²y/dx²=18 > 0 → minimum at (2, −16)
At x=−1: d²y/dx²=−18 < 0 → maximum at (−1, 11)

Increasing and Decreasing Functions

Increasing/Decreasing:
f is increasing on an interval if dy/dx > 0 for all x in that interval.
f is decreasing on an interval if dy/dx < 0 for all x in that interval.
Between stationary points, find the sign of dy/dx to determine whether the function is increasing or decreasing.

📐 Worked Example 8 — Increasing/Decreasing Intervals

For y = x³−6x²+9x+1, find the intervals where y is increasing and decreasing.

dy/dx = 3x²−12x+9 = 3(x²−4x+3) = 3(x−1)(x−3)

Sign of dy/dx:
x < 1: (+)(−)(−) = +ve → increasing
1 < x < 3: (+)(+)(−) = −ve → decreasing
x > 3: (+)(+)(+) = +ve → increasing

Increasing: x < 1 and x > 3. Decreasing: 1 < x < 3.

5. Optimisation Problems

Optimisation uses differentiation to find maximum or minimum values in real-world contexts. The method always follows the same structure.

Optimisation Method:
1. Define variables — assign letters to all unknown quantities.
2. Write a constraint equation (the condition given in the problem).
3. Write the objective function (the quantity to be maximised or minimised).
4. Use the constraint to express the objective in terms of one variable only.
5. Differentiate, set = 0, solve.
6. Verify maximum or minimum using second derivative or sign test.
7. Answer the question — state the maximum/minimum value.

📐 Worked Example 9 — Optimisation

A farmer has 120 m of fencing to enclose a rectangular field against a straight wall (the wall forms one side — no fencing needed on that side). Find the maximum area.

Let width = x m (two sides). Length along wall = y m (one side, no fence).
Constraint: 2x + y = 120 → y = 120−2x

Objective: Area A = xy = x(120−2x) = 120x−2x²

dA/dx = 120−4x = 0 → x = 30 m
y = 120−60 = 60 m

d²A/dx² = −4 < 0 → maximum ✓
Maximum area = 30×60 = 1800 m²

📐 Worked Example 10 — Optimisation with Volume

A closed cylinder has volume 500π cm³. Find the radius that minimises the total surface area.

Constraint: V = πr²h = 500π → h = 500/r²

Objective: SA = 2πr² + 2πrh = 2πr² + 2πr(500/r²)
= 2πr² + 1000π/r

dSA/dr = 4πr − 1000π/r² = 0
4πr = 1000π/r² → 4r³ = 1000 → r³ = 250
r = ∛250 = 5∛2 ≈ 6.30 cm

d²SA/dr² = 4π + 2000π/r³ > 0 → minimum ✓

6. Connected Rates of Change

Connected Rates of Change: Uses the chain rule to connect rates of change of different variables with respect to time. If y depends on x, and both depend on time t:
dy/dt = dy/dx × dx/dt

📐 Worked Example 11 — Connected Rates

The radius of a circular oil slick is increasing at 2 cm/s. Find the rate at which the area is increasing when the radius is 5 cm.

Given: dr/dt = 2. Find: dA/dt when r = 5.
A = πr² → dA/dr = 2πr

Chain rule: dA/dt = dA/dr × dr/dt = 2πr × 2 = 4πr
At r=5: dA/dt = 20π ≈ 62.8 cm²/s

📐 Worked Example 12 — Volume Rate of Change

Water is poured into a conical container (vertex down) of half-angle 30°. When the depth is 6 cm, the volume is increasing at 4 cm³/s. Find the rate at which the depth is increasing at this moment.

Half-angle 30°: radius r = h tan 30° = h/√3
V = ⅓πr²h = ⅓π(h/√3)²h = ⅓π(h²/3)h = πh³/9

dV/dh = πh²/3
At h=6: dV/dh = π(36)/3 = 12π

dV/dt = 4 cm³/s. Chain rule:
dV/dt = dV/dh × dh/dt → 4 = 12π × dh/dt
dh/dt = 4/(12π) = 1/(3π) ≈ 0.106 cm/s

Connected Rates — Sign Convention:
Increasing quantities have positive rates (dr/dt > 0 if r increasing).
Decreasing quantities have negative rates (e.g. if balloon deflating, dV/dt < 0).
Always state units in your final answer — cm²/s, cm³/s, m/s etc.

7. Kinematics — Motion in a Straight Line

Kinematics Relationships

v = ds/dt (velocity = rate of change of displacement)

a = dv/dt = d²s/dt² (acceleration = rate of change of velocity)

s = displacement (from fixed origin) | v = velocity | a = acceleration
Positive direction is usually to the right or upward.
v > 0: moving in positive direction | v < 0: moving in negative direction | v = 0: instantaneously at rest
a > 0: accelerating in positive direction | a < 0: decelerating (if v > 0)

📐 Worked Example 13 — Kinematics

A particle moves in a straight line. Its displacement from O at time t seconds is s = t³ − 6t² + 9t − 2 (t ≥ 0). Find: (a) when it is at rest (b) its acceleration at these times (c) whether it passes through O.

v = ds/dt = 3t²−12t+9 = 3(t²−4t+3) = 3(t−1)(t−3)

(a) At rest: v=0 → t=1 s and t=3 s

a = dv/dt = 6t−12
(b) At t=1: a=6−12=−6 m/s² (decelerating)
At t=3: a=18−12=6 m/s² (accelerating)
a = −6 m/s² at t=1, a = +6 m/s² at t=3

(c) At O: s=0 → t³−6t²+9t−2=0.
Test t=0: s=−2≠0. Try rational roots: t=2: 8−24+18−2=0 ✓
So t=2 is a root. Factor: (t−2)(t²−4t+1)=0
t = 2 or t=(4±√12)/2 = 2±√3
t = 2−√3 ≈ 0.27, t=2, t=2+√3 ≈ 3.73
Yes — passes through O three times.

8. The Second Derivative

Second Derivative d²y/dx²: The derivative of dy/dx — the rate of change of the gradient. Also written f''(x).
d²y/dx² > 0 → curve is concave upward (∪ shape) at that point
d²y/dx² < 0 → curve is concave downward (∩ shape) at that point
Point of inflection: Where d²y/dx² changes sign — the concavity changes.

📐 Worked Example 14 — Second Derivative Test

Find d²y/dx² for y = x⁴ − 8x² + 3. Find all stationary points and classify them. Find any points of inflection.

dy/dx = 4x³−16x = 4x(x²−4) = 4x(x−2)(x+2)
Stationary: x=0, x=2, x=−2

d²y/dx² = 12x²−16
x=0: d²y/dx²=−16 < 0 → maximum at (0, 3)
x=2: d²y/dx²=32 > 0 → minimum at (2, −13)
x=−2: d²y/dx²=32 > 0 → minimum at (−2, −13)

Points of inflection (d²y/dx²=0): 12x²−16=0 → x²=4/3 → x=±2/√3
Check sign change of d²y/dx² at these points → confirmed inflection points.

📝 Exam Practice Questions

Q1 [4 marks] — Differentiate with respect to x: (a) y = (5x²−3)⁸ (b) y = x²ln x (c) y = e^(3x)/(2x+1)

(a) Chain rule: dy/dx = 8(5x²−3)⁷×10x = 80x(5x²−3)⁷

(b) Product: u=x², v=ln x. du/dx=2x, dv/dx=1/x
dy/dx = x²(1/x)+ln x(2x) = x + 2x ln x = x(1+2ln x)

(c) Quotient: u=e^(3x), v=2x+1. du/dx=3e^(3x), dv/dx=2
dy/dx = [(2x+1)3e^(3x) − e^(3x)×2]/(2x+1)²
= e^(3x)[3(2x+1)−2]/(2x+1)² = e^(3x)(6x+1)/(2x+1)²

Q2 [4 marks] — Find the equation of the tangent and normal to y = x²eˣ at the point where x = 1.

At x=1: y=e. Point (1, e).
dy/dx (product): 2xeˣ + x²eˣ = eˣ(2x+x²)
At x=1: dy/dx = e(2+1) = 3e
Tangent: y−e = 3e(x−1) → y = 3ex − 2e
Normal gradient: −1/(3e)
y − e = −(1/3e)(x−1) → 3ey−3e² = −x+1 → x+3ey = 3e²+1

Q3 [5 marks] — A curve has equation y = 2x³ + 3x² − 36x + 5. Find all stationary points, classify each, and find the range of values of x for which y is decreasing.

dy/dx = 6x²+6x−36 = 6(x²+x−6) = 6(x+3)(x−2)
Stationary: x=−3 and x=2
y(−3) = −54+27+108+5 = 86 → point (−3, 86)
y(2) = 16+12−72+5 = −39 → point (2, −39)
d²y/dx² = 12x+6
x=−3: −36+6=−30<0 → maximum at (−3, 86)
x=2: 24+6=30>0 → minimum at (2, −39)
Decreasing (dy/dx<0): between the roots → −3 < x < 2

Q4 [4 marks] — A rectangle has perimeter 40 cm. Find the maximum possible area and prove it is a maximum.

Let length = x, width = y. 2x+2y=40 → y=20−x
Area A = xy = x(20−x) = 20x−x²
dA/dx = 20−2x = 0 → x = 10
y = 10. Maximum area = 100 cm² (square)
d²A/dx² = −2 < 0 → maximum confirmed ✓

Q5 [3 marks] — The volume of a sphere is increasing at 8π cm³/s. Find the rate of increase of the surface area when the radius is 3 cm.

V = (4/3)πr³ → dV/dr = 4πr²
SA = 4πr² → dSA/dr = 8πr
dV/dt = 8π → dV/dr × dr/dt = 8π → 4πr² × dr/dt = 8π
At r=3: 36π × dr/dt = 8π → dr/dt = 2/9
dSA/dt = dSA/dr × dr/dt = 8πr × 2/9 = 16πr/9
At r=3: dSA/dt = 16π cm²/s

Exam Tip: For connected rates involving more than two variables (V, SA, r all related), apply the chain rule step by step. First find dr/dt using the volume rate, then use dr/dt to find dSA/dt.

Q6 [4 marks] — A particle moves in a straight line such that its velocity at time t is v = 3t² − 12t + 9 m/s (t ≥ 0). Find: (a) the times when the particle is at rest (b) the acceleration when t = 3 (c) the distance travelled in the first 3 seconds.

(a) v=0: 3t²−12t+9=0 → t²−4t+3=0 → (t−1)(t−3)=0
t = 1 s and t = 3 s

(b) a = dv/dt = 6t−12. At t=3: a=18−12=6 m/s²

(c) s = ∫v dt = t³−6t²+9t (+c, but find distance not displacement)
s(0)=0, s(1)=1−6+9=4, s(3)=27−54+27=0
From t=0→1: moves 4 m forward. From t=1→3: moves from s=4 to s=0, so 4 m back.
Total distance = 4+4 = 8 m

Exam Tip: Distance ≠ displacement. When a particle changes direction (v=0), calculate the distance in each section separately and ADD. Displacement can be negative but distance is always positive.

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