This lesson covers all the advanced differentiation techniques required for Cambridge 9709 Pure 2 and Pure 3. Master these skills and you'll be able to differentiate almost any function you encounter.
What you'll learn:
✦ Learn 1 — Derivatives of eˣ, ln x, sin x, cos x, tan x and the chain rule
✦ Learn 2 — Implicit differentiation (equations in x and y)
✦ Learn 3 — Parametric differentiation (dy/dx and d²y/dx²)
✦ Learn 4 — Product rule and quotient rule with trig/log/exp
✦ Learn 5 — Related rates of change (dr/dt, dA/dt, dV/dt)
How to use this lesson: Work through Learn 1–5 in order, then tackle the Examples tab. Use the Visualiser to see tangent lines live. Complete Ex 1–5 for topic practice, then test yourself on Practice, Challenge, Exam Style, and Past Paper Qs.
Prerequisites: You should already know basic differentiation (polynomials, power rule), and be comfortable with algebra including logarithm laws and trigonometric identities.
Quick Reference
d/dx(eˣ) = eˣ | d/dx(ln x) = 1/x | d/dx(sin x) = cos x | d/dx(cos x) = −sin x | d/dx(tan x) = sec²x
Parametric: dy/dx = (dy/dt)/(dx/dt) | Implicit: differentiate both sides w.r.t. x, using d/dx(f(y)) = f′(y)·dy/dx
Learn 1: Standard Derivatives & Chain Rule
The Five Key Derivatives
These results must be memorised. They form the foundation of all further differentiation:
Function f(x)
Derivative f′(x)
Note
eˣ
eˣ
Unique: its own derivative
eᵃˣ
a·eᵃˣ
Chain rule built in
ln x
1/x
x > 0 required
ln(ax)
1/x
Same as ln x (log laws)
sin x
cos x
x in radians
cos x
−sin x
Note the minus sign
tan x
sec²x
= 1/cos²x
The Chain Rule
If y = f(g(x)), then dy/dx = f′(g(x)) · g′(x)
Equivalently: if y = f(u) and u = g(x), then dy/dx = (dy/du) · (du/dx)
Think of it as: "Derivative of the outside (leaving inside alone) × derivative of the inside"
Chain Rule with Standard Functions
When the argument is a function of x (not just x itself), always apply the chain rule:
d/dx[e^(f(x))] = f′(x)·e^(f(x))
d/dx[ln(f(x))] = f′(x)/f(x)
d/dx[sin(f(x))] = f′(x)·cos(f(x))
d/dx[cos(f(x))] = −f′(x)·sin(f(x))
d/dx[tan(f(x))] = f′(x)·sec²(f(x))
Worked Examples
Example 1: Differentiate y = e^(3x²)
Let u = 3x², so y = eᵘ
du/dx = 6x, dy/du = eᵘ
dy/dx = 6x · e^(3x²)
Example 2: Differentiate y = ln(sin x)
d/dx = cos x / sin x = cot x
Example 3: Differentiate y = sin(x² + 1)
d/dx = 2x · cos(x² + 1)
Example 4: Differentiate y = (2x + 3)⁵
dy/dx = 5(2x + 3)⁴ · 2 = 10(2x + 3)⁴
Common error: Forgetting to multiply by the derivative of the inner function. For example, d/dx[sin(3x)] = 3cos(3x), NOT just cos(3x).
Learn 2: Implicit Differentiation
What is Implicit Differentiation?
An implicit equation defines a relationship between x and y without explicitly writing y as a function of x. For example: x² + y² = 25 or x³ + y³ = 6xy.
We differentiate both sides with respect to x, using the chain rule for any term involving y.
The Key Rule
d/dx[f(y)] = f′(y) · dy/dx
This comes directly from the chain rule: y is a function of x, so when differentiating a function of y with respect to x, we must multiply by dy/dx.
Important Cases
Expression
d/dx
y²
2y · dy/dx
y³
3y² · dy/dx
sin y
cos y · dy/dx
eʸ
eʸ · dy/dx
ln y
(1/y) · dy/dx
xy
y + x · dy/dx (product rule)
x²y
2xy + x² · dy/dx (product rule)
Method: Finding dy/dx Implicitly
Step 1: Differentiate every term on both sides with respect to x Step 2: Remember: d/dx(y) = dy/dx; d/dx(y²) = 2y·dy/dx; etc. Step 3: Collect all dy/dx terms on one side Step 4: Factorise out dy/dx Step 5: Divide to get dy/dx = ...
Example 2: x = 2cos t, y = 3sin t (ellipse)
dx/dt = −2sin t, dy/dt = 3cos t
dy/dx = 3cos t / (−2sin t) = −(3/2)cot t
At t = π/4: dy/dx = −(3/2)·1 = −3/2
Example 3: x = t + 1/t, y = t − 1/t
dx/dt = 1 − 1/t², dy/dt = 1 + 1/t²
dy/dx = (1 + 1/t²)/(1 − 1/t²) = (t² + 1)/(t² − 1)
Finding Tangents and Normals
At a given value of t:
1. Find dy/dx at that t-value
2. Find the (x, y) coordinates by substituting t into x = f(t) and y = g(t)
3. Use y − y₁ = m(x − x₁) for the tangent, or y − y₁ = −(1/m)(x − x₁) for the normal
Warning: If dx/dt = 0 at a point, then dy/dx is undefined — this corresponds to a vertical tangent on the curve.
Learn 4: Product Rule & Quotient Rule
Product Rule
If y = u·v, then dy/dx = u′v + uv′
When to use: When the function is a product of two different functions that can't be simplified first. Memory aid: "First times derivative of second, plus second times derivative of first"
Product Rule — Examples
Example 1: y = xeˣ
u = x → u′ = 1 v = eˣ → v′ = eˣ
dy/dx = 1·eˣ + x·eˣ = eˣ(1 + x)
Example 2: y = x²·sin x
u = x² → u′ = 2x v = sin x → v′ = cos x
dy/dx = 2x·sin x + x²·cos x = x(2sin x + x·cos x)
Example 3: y = eˣ·ln x
u = eˣ → u′ = eˣ v = ln x → v′ = 1/x
dy/dx = eˣ·ln x + eˣ·(1/x) = eˣ(ln x + 1/x)
Quotient Rule
If y = u/v, then dy/dx = (u′v − uv′) / v²
Important: The order in the numerator is u′v MINUS uv′ (NOT u′v + uv′). The minus sign is crucial!
Quotient Rule — Examples
Example 4: y = sin x / x
u = sin x → u′ = cos x v = x → v′ = 1
dy/dx = (cos x · x − sin x · 1) / x² = (x·cos x − sin x) / x²
Example 6: y = ln x / x² (verify using quotient rule)
u = ln x → u′ = 1/x v = x² → v′ = 2x
dy/dx = [(1/x)·x² − ln x·2x] / x⁴ = [x − 2x·ln x] / x⁴ = (1 − 2ln x) / x³
Choosing Between Product and Quotient
Often you can use either! For example, y = sin x / eˣ can be written as y = sin x · e^(−x) and use the product rule instead. Whichever you find easier is fine — but be consistent.
Combined Rules (Chain + Product)
Example 7: y = x²·e^(3x)
u = x² → u′ = 2x v = e^(3x) → v′ = 3e^(3x) [chain rule]
dy/dx = 2x·e^(3x) + x²·3e^(3x) = xe^(3x)(2 + 3x)
Learn 5: Related Rates of Change
What are Related Rates?
When two quantities are related by an equation, their rates of change are also related. If we know how fast one quantity changes, we can find how fast another changes using the chain rule.
If y = f(x) and x changes with time t, then: dy/dt = (dy/dx) · (dx/dt)
Common Geometry Formulas
Shape
Formula
Useful derivative
Circle: Area
A = πr²
dA/dr = 2πr
Circle: Circumference
C = 2πr
dC/dr = 2π
Sphere: Volume
V = (4/3)πr³
dV/dr = 4πr²
Sphere: Surface Area
S = 4πr²
dS/dr = 8πr
Cylinder: Volume
V = πr²h
dV/dr = 2πrh (h fixed)
Cone: Volume
V = (1/3)πr²h
Varies by context
Method for Related Rates Problems
Step 1: Identify the two quantities that are changing (e.g., radius r and volume V) Step 2: Write down the equation connecting them (e.g., V = (4/3)πr³) Step 3: Differentiate with respect to time t, using the chain rule Step 4: Substitute the given values (rate and/or specific value of the variable) Step 5: Solve for the required rate
Worked Examples
Example 1 — Expanding circle:
A circular oil slick has area A = πr². The radius increases at 0.5 m/s. Find dA/dt when r = 4 m.
A = πr² → dA/dr = 2πr
dA/dt = dA/dr · dr/dt = 2πr · 0.5 = πr
When r = 4: dA/dt = 4π ≈ 12.57 m²/s
Example 2 — Inflating sphere:
Air is pumped into a sphere at 10 cm³/s (dV/dt = 10). Find dr/dt when r = 5 cm.
Sign convention: A positive rate means increasing, a negative rate means decreasing. Always check whether the quantity increases or decreases and assign the sign accordingly before starting.
Exam tip: Always state the units of your answer for rates of change. If r is in cm and t is in s, then dr/dt is in cm/s.
Fully Worked Examples
Example 1 — Chain Rule with Exponential
Differentiate y = e^(x² − 3x)
Let u = x² − 3x, so y = eᵘ M1
du/dx = 2x − 3 A1
dy/dx = (2x − 3)e^(x² − 3x) A1
Example 2 — Chain Rule with Logarithm
Differentiate y = ln(3x² + 1)
Using d/dx[ln(f(x))] = f′(x)/f(x) M1
f(x) = 3x² + 1, f′(x) = 6x A1
dy/dx = 6x / (3x² + 1) A1
Example 3 — Product Rule (trig × exponential)
Differentiate y = e^(2x)·cos x
u = e^(2x), u′ = 2e^(2x) [chain rule] v = cos x, v′ = −sin x M1
✓ RIGHT: d²y/dx² = [d/dt(dy/dx)] / (dx/dt) — differentiate dy/dx w.r.t. t, then divide by dx/dt
Mistake 7 — Not collecting dy/dx terms in implicit differentiation
✗ WRONG: Leaving dy/dx on both sides and stopping
✓ RIGHT: Move ALL dy/dx terms to one side, factorise dy/dx, then divide to isolate it
Key Formulas Reference Table
Rule / Function
Formula
eˣ
d/dx(eˣ) = eˣ
eᵃˣ
d/dx(eᵃˣ) = a·eᵃˣ
e^(f(x))
d/dx[e^(f(x))] = f′(x)·e^(f(x))
ln x
d/dx(ln x) = 1/x
ln(f(x))
d/dx[ln(f(x))] = f′(x)/f(x)
sin x
d/dx(sin x) = cos x
cos x
d/dx(cos x) = −sin x
tan x
d/dx(tan x) = sec²x
sin(f(x))
d/dx[sin(f(x))] = f′(x)·cos(f(x))
cos(f(x))
d/dx[cos(f(x))] = −f′(x)·sin(f(x))
Chain Rule
dy/dx = (dy/du)·(du/dx)
Product Rule
d/dx(uv) = u′v + uv′
Quotient Rule
d/dx(u/v) = (u′v − uv′)/v²
Implicit: d/dx(y)
dy/dx
Implicit: d/dx(yⁿ)
n·yⁿ⁻¹·(dy/dx)
Parametric dy/dx
(dy/dt) / (dx/dt)
Parametric d²y/dx²
[d/dt(dy/dx)] / (dx/dt)
Related Rates
dy/dt = (dy/dx)·(dx/dt)
Proof Bank
Proof 1: d/dx(eˣ) = eˣ
Using the definition of the derivative:
d/dx(eˣ) = lim[h→0] (e^(x+h) − eˣ) / h
= lim[h→0] eˣ(eʰ − 1) / h
= eˣ · lim[h→0] (eʰ − 1) / h
Now we need: lim[h→0] (eʰ − 1)/h
Using the series expansion: eʰ = 1 + h + h²/2! + ... so eʰ − 1 = h + h²/2! + ...
Therefore (eʰ − 1)/h = 1 + h/2! + ... → 1 as h → 0
Therefore: d/dx(eˣ) = eˣ · 1 = eˣ ✓
Proof 2: d/dx(ln x) = 1/x via Inverse Function
Using the inverse function rule:
Let y = ln x, which means x = eʸ
Differentiate both sides with respect to y:
dx/dy = eʸ = x
Taking the reciprocal (inverse function rule):
dy/dx = 1/(dx/dy) = 1/x
Therefore: d/dx(ln x) = 1/x ✓
Proof 3: Derivation of Implicit Differentiation
Why does d/dx[f(y)] = f′(y)·dy/dx?
Suppose y is a differentiable function of x. Then f(y) is a composite function:
f(y(x)) — f applied to y, and y is a function of x.
By the chain rule:
d/dx[f(y(x))] = f′(y(x)) · y′(x) = f′(y) · dy/dx
This technique works for any differentiable equation in x and y, because differentiating both sides of a valid equation preserves the equality.
Interactive Derivative Visualiser
Select a curve below. The tangent line is drawn at a point you can control using the slider. The gradient (dy/dx) is shown live.
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Exercise 1 — Standard Derivatives & Chain Rule
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Exercise 2 — Implicit Differentiation
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Exercise 3 — Parametric Differentiation
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Exercise 4 — Product & Quotient Rule
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Exercise 5 — Related Rates of Change
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Practice — 30 Mixed Questions
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Challenge — 15 Harder Questions
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Exam-Style Questions
Question 1 [4 marks]
A curve has equation y = x·e^(2x). Find dy/dx and hence show that the curve has a stationary point at x = −1/2. Find the y-coordinate of this stationary point.
M1: Product rule — u = x, v = e^(2x)
A1: dy/dx = e^(2x) + 2x·e^(2x) = e^(2x)(1 + 2x)
M1: Set dy/dx = 0 → e^(2x)(1 + 2x) = 0 → 1 + 2x = 0 → x = −1/2
A1: y = (−1/2)·e^(−1) = −1/(2e) ≈ −0.184
Question 2 [5 marks]
The curve C has equation x³ + y³ − 3x²y = 3. Find dy/dx in terms of x and y. Hence find the coordinates of the point(s) where the tangent is horizontal.
M1: Differentiate implicitly: 3x² + 3y²·(dy/dx) − [6xy + 3x²·(dy/dx)] = 0
A1: dy/dx(3y² − 3x²) = 6xy − 3x²
A1: dy/dx = (6xy − 3x²)/(3y² − 3x²) = x(2y − x)/(y² − x²)
M1: For horizontal tangent, dy/dx = 0 → x(2y − x) = 0 → x = 0 or y = x/2
A1: x = 0: 0 + y³ − 0 = 3 → y = ∛3. Point: (0, ∛3). Substitute y = x/2 to find further points.
Question 3 [6 marks]
A curve is defined parametrically by x = t² + 1, y = t³ − 3t. Find dy/dx and d²y/dx² in terms of t. Determine the nature of the stationary points of the curve.
B1: dx/dt = 2t, dy/dt = 3t² − 3
A1: dy/dx = (3t² − 3)/(2t) = 3(t − 1)(t + 1)/(2t)
Stationary points: dy/dx = 0 → t = ±1
M1: d/dt(dy/dx) = [6t·(2t) − (3t²−3)·2]/(4t²) = (6t²+6)/(4t²)
A1: d²y/dx² = [(6t²+6)/(4t²)] / (2t) = 3(t²+1)/(4t³)
A1: t = 1: d²y/dx² = 3(2)/4 = 3/2 > 0 → minimum; t = −1: d²y/dx² = 3(2)/(−4) = −3/2 < 0 → maximum
A1: Min at (2, −2), Max at (2, 2)
Question 4 [4 marks]
Differentiate y = ln(x² + 4) / (x + 1) with respect to x, simplifying your answer.
M1: Quotient rule: u = ln(x² + 4), u′ = 2x/(x² + 4); v = x + 1, v′ = 1
M1: dy/dx = [2x(x+1)/(x²+4) − ln(x²+4)] / (x+1)²
A1: = [2x(x+1) − (x²+4)ln(x²+4)] / [(x²+4)(x+1)²]
A1: Correct simplified form
Question 5 [5 marks]
Water leaks from a conical tank (vertex down) at 2 cm³/s. The cone has height 30 cm and base radius 10 cm. Find the rate at which the water level falls when the depth is 12 cm.
A curve is given parametrically by x = cos²t, y = sin t (0 ≤ t ≤ π/2). Find the equation of the tangent at the point where t = π/6. Express your answer in the form ax + by = c.
B1: dx/dt = −2cos t sin t = −sin(2t); dy/dt = cos t
M1: dy/dx = cos t / (−sin 2t) = cos t / (−2sin t cos t) = −1/(2sin t)
A1: At t = π/6: dy/dx = −1/(2·1/2) = −1; point: (3/4, 1/2)
M1: Tangent: y − 1/2 = −1(x − 3/4) → y = −x + 5/4
A1: 4x + 4y = 5
Question 8 [6 marks]
A spherical balloon is inflated so that its volume increases at a constant rate of 50 cm³/s. (a) Find the rate of increase of radius when r = 5 cm. (b) Find the rate of increase of surface area at the same instant.
B1: V = (4/3)πr³ → dV/dr = 4πr²
M1: dr/dt = dV/dt ÷ dV/dr = 50/(4πr²)
A1: When r = 5: dr/dt = 50/(100π) = 1/(2π) ≈ 0.159 cm/s
B1: S = 4πr² → dS/dr = 8πr
M1: dS/dt = dS/dr · dr/dt = 8πr · 50/(4πr²) = 100/r
A1: When r = 5: dS/dt = 100/5 = 20 cm²/s
Past Paper Questions (Cambridge 9709 Style)
Past Paper 1 — 9709/32/O/N/18 style [5 marks]
The curve with equation e^(x+y) = x + y² passes through the point (0, 0). Find dy/dx at this point and hence find the equation of the tangent to the curve at (0, 0).
A curve is defined by x = 3t − t³, y = 3t². (a) Find dy/dx in terms of t. (b) Find the coordinates of the stationary point(s) of the curve and determine their nature.
B1: dx/dt = 3 − 3t², dy/dt = 6t
A1: dy/dx = 6t/(3 − 3t²) = 2t/(1 − t²)
M1: Stationary points where dy/dx = 0 → t = 0
A1: t = 0: x = 0, y = 0. Stationary point at (0, 0)
M1: Check sign of dy/dx either side of t = 0: t = −0.1 gives dy/dx negative, t = 0.1 gives positive → minimum
A1: Minimum at (0, 0)
Past Paper 3 — 9709/32/M/J/20 style [5 marks]
The variables x and y satisfy the equation y·sin x = x·cos y. Find dy/dx in terms of x and y.
M1: Differentiate left side: d/dx(y·sin x) = (dy/dx)·sin x + y·cos x [product rule]
M1: Differentiate right side: d/dx(x·cos y) = cos y + x·(−sin y)·(dy/dx) [product rule + chain]
A1: (dy/dx)·sin x + y·cos x = cos y − x·sin y·(dy/dx)
M1: Collect dy/dx: (dy/dx)(sin x + x·sin y) = cos y − y·cos x
A1: dy/dx = (cos y − y·cos x) / (sin x + x·sin y)
Past Paper 4 — 9709/31/O/N/21 style [4 marks]
A colony of bacteria grows such that its mass M grams satisfies M = 5e^(0.3t) where t is time in hours. Find dM/dt when t = 2. Interpret your answer in context.
M1: Differentiate: dM/dt = 5 × 0.3 × e^(0.3t) = 1.5e^(0.3t)
A1: When t = 2: dM/dt = 1.5e^(0.6) = 1.5 × 1.8221... = 2.733 g/hr (3 s.f.)
B1: Correct calculation shown
B1: Interpretation — the mass of bacteria is increasing at approximately 2.73 grams per hour at t = 2 hours.
Past Paper 5 — 9709/32/O/N/22 style [7 marks]
The curve C has equation x²e^y + y = 3. (a) Show that dy/dx = −2xe^y/(x²e^y + 1). (b) Find the equation of the tangent at the point (1, 0).