Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Differentiation Techniques!
Differentiation is the cornerstone of calculus — it measures how functions change. In Cambridge A-Level Pure Mathematics 1, you move beyond the basics to master a full toolkit: first principles, the power rule, chain rule, product and quotient rules, differentiation of trigonometric and exponential functions, and the powerful techniques of implicit and parametric differentiation.
Differentiate from first principles using the limit definition
Apply the power rule to polynomials and fractional/negative powers
Use the chain rule for composite functions
Apply the product rule and quotient rule correctly
Differentiate sin x, cos x, tan x, eˣ, e^f(x), ln x, ln(f(x))
Perform implicit differentiation with respect to x
Use parametric differentiation: dy/dx = (dy/dt)/(dx/dt)
Find equations of tangents and normals using derivatives
Identify stationary points and determine their nature
Apply second derivatives to confirm max/min
First Principles
The limit definition — where all differentiation starts
Power Rule
d/dx(xⁿ) = nxⁿ⁻¹ for any rational n
Chain Rule
Composite functions — outer × inner derivative
Product Rule
d/dx(uv) = u'v + uv'
Quotient Rule
d/dx(u/v) = (u'v − uv') / v²
Trig & Exp
sin, cos, tan, eˣ, ln — with chain rule
Implicit
Differentiate both sides w.r.t. x, chain on y
Parametric
dy/dx = (dy/dt) ÷ (dx/dt)
Learn 1 — First Principles & Power Rule
Definition of the Derivative
The derivative of a function f(x) measures the instantaneous rate of change. It is defined as the limit of the gradient of a chord as the chord's width shrinks to zero:
f'(x) = lim[h→0] (f(x + h) − f(x)) / h
This is called differentiation from first principles. The process always follows the same structure: expand f(x+h), subtract f(x), divide by h, then let h → 0.
Always rewrite expressions as powers of x before differentiating. Fractions like 3/x² become 3x⁻², and roots like √x become x^(1/2). This is the single most important preparation step in A-Level differentiation.
Learn 2 — Chain Rule
What Is a Composite Function?
A composite function is a function applied inside another function — for example (3x+1)⁵, sin(x²), or e^(2x+1). The chain rule handles all such cases.
dy/dx = (dy/du) × (du/dx)
Equivalently: differentiate the outer function (leaving the inner unchanged), then multiply by the derivative of the inner function.
Step-by-Step Method
1. Identify the outer function and the inner function u.
2. Differentiate the outer with respect to u (leaving u in place).
3. Multiply by du/dx (the derivative of the inner function).
4. Substitute back if needed.
Example: y = (3x + 1)⁵
Let u = 3x + 1, so y = u⁵
dy/du = 5u⁴ du/dx = 3
dy/dx = 5u⁴ × 3 = 15(3x + 1)⁴
Example: y = sin(x²)
Let u = x², so y = sin u
dy/du = cos u du/dx = 2x
dy/dx = cos(x²) × 2x = 2x cos(x²)
Example: y = e^(2x + 1)
Let u = 2x + 1, so y = eᵘ
dy/du = eᵘ du/dx = 2
dy/dx = e^(2x+1) × 2 = 2e^(2x+1)
Example: y = (x² − 4)^(3/2)
Let u = x² − 4, so y = u^(3/2)
dy/du = (3/2)u^(1/2) du/dx = 2x
dy/dx = (3/2)(x²−4)^(1/2) × 2x = 3x(x²−4)^(1/2)
Example: y = ln(3x² + 1)
Let u = 3x² + 1, so y = ln u
dy/du = 1/u du/dx = 6x
dy/dx = 6x / (3x² + 1)
The most common chain rule error is forgetting to multiply by the inner derivative. In (3x+1)⁵, the 3 from differentiating 3x+1 must appear. Always ask: "What is inside, and what is its derivative?"
Learn 3 — Product & Quotient Rules
The Product Rule
When two differentiable functions are multiplied together, use the product rule:
d/dx (uv) = u'v + uv'
In words: derivative of first × second + first × derivative of second.
When to Use the Product Rule
Use it when the function is a product of two parts that cannot be easily expanded, e.g.:
x² sin x, (2x+1)³ ln x, eˣ cos x, x^(1/2)(x−3)²
Product Rule Examples
Differentiate y = x³ sin x
u = x³, v = sin x u' = 3x², v' = cos x
dy/dx = 3x² sin x + x³ cos x
Differentiate y = x² ln x
u = x², v = ln x u' = 2x, v' = 1/x
dy/dx = 2x ln x + x² · (1/x) = 2x ln x + x = x(2 ln x + 1)
The Quotient Rule
When one function is divided by another, use the quotient rule:
d/dx (u/v) = (u'v − uv') / v²
Memory aid: "low d-high minus high d-low, over the square of what's below".
Quotient Rule Examples
Differentiate y = sin x / x²
u = sin x, v = x² u' = cos x, v' = 2x
dy/dx = (cos x · x² − sin x · 2x) / x⁴ = x(x cos x − 2 sin x) / x⁴ = (x cos x − 2 sin x) / x³
Note: u/v = u · v⁻¹, so you can always use the product rule with v⁻¹ instead of the quotient rule. Both give the same answer. Use whichever feels cleaner.
In the quotient rule, the sign is u'v MINUS uv' — never the other way around. A plus sign here is the single most common error. Also, always fully simplify the numerator by expanding and collecting like terms.
Example: y = ln(5x)
dy/dx = 5/(5x) = 1/x (same as ln x — the constant inside vanishes!)
Example: y = ln(x² + 3)
dy/dx = 2x / (x² + 3)
Example: y = ln(sin x)
dy/dx = cos x / sin x = cot x
Combined Examples
y = e^(sin x)
Chain rule: outer = eᵘ, inner u = sin x
dy/dx = cos x · e^(sin x)
y = sin(eˣ)
Outer = sin u, inner u = eˣ
dy/dx = eˣ cos(eˣ)
y = ln(cos x)
Outer = ln u, inner u = cos x
dy/dx = (−sin x) / cos x = −tan x
For ln(ax) where a is a constant: d/dx ln(ax) = a/(ax) = 1/x. The constant a always cancels. For ln(ax + b): d/dx = a/(ax+b). These are the two most tested ln forms in Cambridge 9709.
Learn 5 — Implicit & Parametric Differentiation
Implicit Differentiation
When a curve is defined by an equation in both x and y (not rearranged to y = f(x)), differentiate both sides with respect to x. For any term involving y, apply the chain rule: differentiate with respect to y, then multiply by dy/dx.
When a curve is defined by x = f(t) and y = g(t), the gradient dy/dx is found by:
dy/dx = (dy/dt) ÷ (dx/dt)
Never take the reciprocal — divide dy/dt by dx/dt.
Parametric Example: x = t², y = 2t
dx/dt = 2t dy/dt = 2
dy/dx = 2 / (2t) = 1/t
At t = 3: dy/dx = 1/3. Point: (9, 6). Tangent: y − 6 = (1/3)(x − 9) → y = x/3 + 3.
Parametric Example: x = cos t, y = sin 2t
dx/dt = −sin t dy/dt = 2 cos 2t
dy/dx = 2 cos 2t / (−sin t)
Using cos 2t = 1 − 2sin²t:
dy/dx = 2(1 − 2sin²t) / (−sin t) = −2(1 − 2sin²t) / sin t
Second Parametric Derivative
d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] / (dx/dt)
For x = t², y = 2t: dy/dx = 1/t
d/dt(1/t) = −1/t²
d²y/dx² = (−1/t²) / (2t) = −1/(2t³)
The most common parametric error is writing dy/dx = (dx/dt)/(dy/dt) — inverting the fraction. Always write (dy/dt) on top. Also, for implicit differentiation, every y term needs its dy/dx factor. Never forget it for terms like sin y, y³, or ln y.
Worked Examples
Eight fully worked examples covering all key differentiation techniques for Cambridge A-Level 9709.
Example 1 — First Principles for f(x) = x²
Q: Differentiate f(x) = x² from first principles.
M1: Write the definition: f'(x) = lim[h→0] [(x+h)² − x²] / h
M1: Expand: [(x² + 2xh + h²) − x²] / h = [2xh + h²] / h = 2x + h
A1: Let h → 0: f'(x) = 2x3 marks
Example 2 — Chain Rule on (2x − 1)⁴
Q: Find dy/dx when y = (2x − 1)⁴.
M1: Identify outer function uⁿ and inner u = 2x − 1.
M1: dy/dx = 4(2x−1)³ × d/dx(2x−1)
A1: dy/dx = 4(2x−1)³ × 2 = 8(2x−1)³3 marks
Example 3 — Product Rule with Trig
Q: Differentiate y = x² cos x.
M1: Let u = x², v = cos x → u' = 2x, v' = −sin x
M1: dy/dx = u'v + uv' = 2x cos x + x²(−sin x)
A1: dy/dx = 2x cos x − x² sin x = x(2 cos x − x sin x) 3 marks
Write tan x = sin x / cos x. Let u = sin x, v = cos x.
u' = cos x, v' = −sin x
d/dx(tan x) = (cos x · cos x − sin x · (−sin x)) / cos²x
= (cos²x + sin²x) / cos²x
Using the identity cos²x + sin²x = 1:
= 1 / cos²x = sec²x Q.E.D.
Proof 4 — Parametric dy/dx Formula
If x = f(t) and y = g(t) are differentiable, and dx/dt ≠ 0, then by the chain rule:
dy/dt = (dy/dx) · (dx/dt)
(since we can treat y as a function of x, which is a function of t)
Dividing both sides by dx/dt (≠ 0): dy/dx = (dy/dt) / (dx/dt) Q.E.D.
Tangent Line Visualiser
Select a function and click a point on the curve to see the tangent line and gradient value at that point.
Click anywhere on the curve to see the tangent and gradient.
Exercise 1 — Power Rule & Polynomial Differentiation (10 Questions)
The curve C has equation x² + 2y² + xy = 11. Find dy/dx in terms of x and y. Hence find the coordinates of the points on C where the tangent is parallel to the x-axis.
Differentiate y = ln(sin x) + sin(ln x) with respect to x.
d/dx[ln(sin x)] = cos x / sin x = cot x [M1 A1 chain rule on ln]
d/dx[sin(ln x)] = cos(ln x) × (1/x) [M1 chain rule on sin] dy/dx = cot x + cos(ln x)/x [A1]
Q6 [5 marks]
A curve is defined parametrically by x = 3t², y = 6t. Find the equation of the normal to the curve at the point where t = 2.
dx/dt = 6t, dy/dt = 6 [B1]
dy/dx = 6/(6t) = 1/t [M1]
At t = 2: dy/dx = 1/2. Normal gradient = −2 [M1]
Point: x = 12, y = 12 [B1]
Normal: y − 12 = −2(x − 12) → y = −2x + 36 [A1]
Q7 [4 marks]
Find the x-coordinates of the stationary points of y = x³e^(−x).
dy/dx = 3x²e^(−x) + x³(−e^(−x)) [M1 product rule]
= e^(−x)(3x² − x³) = x²e^(−x)(3 − x) [M1 factorise]
Set dy/dx = 0: x² = 0 or 3 − x = 0 (e^(−x) > 0 always) [M1] x = 0 or x = 3 [A1]
Q8 [6 marks]
The equation of a curve is x² + 4xy + y² = −4. (i) Find dy/dx in terms of x and y. (ii) Find the coordinates of the two points where the tangent is vertical (i.e. dx/dy = 0).