Further Differentiation — A-Level Pure 3

Welcome to Further Differentiation (Pure 3)

Pure 3 differentiation extends the toolkit to cover every standard function found in A-Level Mathematics. You already know how to differentiate polynomials, exponentials e^x, logarithms ln x, and basic trig. This module adds four new families of derivatives — exponentials to any base a^x, logarithms to any base log_a(x), reciprocal trig functions (sec, cosec, cot), and inverse trig functions (arcsin, arccos, arctan) — together with harder implicit and parametric problems that combine all these techniques.

d/dx(a^x) = a^x · ln a | d/dx(log_a x) = 1/(x · ln a) | d/dx(arctan x) = 1/(1+x²) | d/dx(arcsin x) = 1/√(1−x²)

Learning Objectives

Differentiate a^x for any base a > 0 using the formula d/dx(a^x) = a^x · ln a
Differentiate log_a(x) using change of base: d/dx(log_a x) = 1/(x · ln a)
Differentiate sec x, cosec x, and cot x from first principles using the quotient rule
Differentiate arcsin(f(x)) and arctan(f(x)) using the chain rule
Combine all new derivatives with the chain, product, and quotient rules fluently
Use all techniques for implicitly defined curves, including those with trig or exponentials
Find stationary points of curves defined implicitly by setting dy/dx = 0
Apply parametric differentiation to harder parametric equations involving trig or inverse trig
Find d²y/dx² for parametric curves using the chain rule on dy/dx
Solve connected rates of change problems using the chain rule

Topics in This Module

Differentiating a^x

Write a^x = e^(x·ln a) and differentiate to get a^x · ln a

Reciprocal Trig Derivatives

sec x, cosec x, cot x — derived from sin/cos using quotient rule

Inverse Trig Derivatives

arcsin, arccos, arctan — each with domain restrictions and chain rule applications

Implicit Curves

Harder curves with products, trig terms, finding stationary points implicitly

Parametric Harder

Parametric equations with trig, exponentials, or inverse trig; second derivatives

Applications

Related rates, connected rates of change, tangents and normals to harder curves

Learn 1 — Differentiating a^x and log_a(x)

When the base of an exponential is not e, or the base of a logarithm is not e, we need a technique to convert. The key is the change of base formula and the fact that e^(ln a) = a.

Deriving d/dx(a^x)

Write y = a^x in terms of e:
Since a = e^(ln a), we have a^x = e^(x · ln a)
Differentiate using the chain rule:
dy/dx = ln a · e^(x · ln a) = ln a · a^x
Result: d/dx(a^x) = a^x · ln a

d/dx(a^x) = a^x · ln a

Special Cases

d/dx(2^x) = 2^x · ln 2 ≈ 2^x · 0.693
d/dx(10^x) = 10^x · ln 10 ≈ 10^x · 2.303
d/dx(e^x) = e^x · ln e = e^x · 1 = e^x ✓ (consistent with known result)

With Chain Rule: d/dx(a^(f(x))) = f'(x) · ln a · a^(f(x))

Example: Differentiate y = 3^(2x)
Let u = 2x, so y = 3^u
dy/dx = dy/du · du/dx = (3^u · ln 3) · 2 = 2 · ln 3 · 3^(2x)

Example: Differentiate y = 5^(x²)
dy/dx = 2x · ln 5 · 5^(x²)

Deriving d/dx(log_a x)

Use change of base: log_a x = ln x / ln a
Since ln a is a constant:
d/dx(log_a x) = (1/ln a) · d/dx(ln x) = (1/ln a) · (1/x)
Result: d/dx(log_a x) = 1/(x · ln a)

d/dx(log_a x) = 1/(x · ln a)

Examples

d/dx(log_2 x) = 1/(x · ln 2)
d/dx(log_10 x) = 1/(x · ln 10) (this is the common log derivative)
d/dx(log_2(3x)) = 1/(3x · ln 2) · 3 = 1/(x · ln 2) (chain rule; constants cancel)

Memory tip: Both formulas involve ln a — the natural log of the base. If a = e, ln e = 1 and both reduce to the familiar e^x and ln x derivatives.

Common trap: d/dx(a^x) ≠ x · a^(x−1). That rule only applies when x is the BASE (like x^n), not when x is the EXPONENT. For a^x, the exponent is x, so the power rule does NOT apply.

Learn 2 — Differentiating Reciprocal Trig Functions

The reciprocal trig functions — sec x, cosec x, and cot x — are defined in terms of sin and cos, so their derivatives are found using the quotient rule.

Differentiating sec x = 1/cos x

sec x = 1/cos x = (cos x)^(−1)
Using quotient rule with u=1, v=cos x:
d/dx(sec x) = (0·cos x − 1·(−sin x)) / cos²x = sin x / cos²x
= (1/cos x)·(sin x/cos x) = sec x · tan x
d/dx(sec x) = sec x · tan x

Differentiating cosec x = 1/sin x

cosec x = 1/sin x
Using quotient rule with u=1, v=sin x:
d/dx(cosec x) = (0·sin x − 1·cos x) / sin²x = −cos x / sin²x
= −(1/sin x)·(cos x/sin x) = −cosec x · cot x
d/dx(cosec x) = −cosec x · cot x

Differentiating cot x = cos x/sin x

Using quotient rule with u=cos x, v=sin x:
d/dx(cot x) = (−sin x·sin x − cos x·cos x) / sin²x
= −(sin²x + cos²x) / sin²x = −1/sin²x
d/dx(cot x) = −cosec²x

d/dx(sec x) = sec x · tan x | d/dx(cosec x) = −cosec x · cot x | d/dx(cot x) = −cosec²x

With Chain Rule

d/dx(sec(2x)) = sec(2x) · tan(2x) · 2 = 2·sec(2x)·tan(2x)
d/dx(cosec(3x)) = −cosec(3x)·cot(3x) · 3 = −3·cosec(3x)·cot(3x)
d/dx(cot(x²)) = −cosec²(x²) · 2x = −2x·cosec²(x²)

Pattern: sec and cosec derivatives both involve a product of two trig functions. sec x gives sec·tan, cosec x gives −cosec·cot. The minus sign for cosec and cot is always present — think of them as the "negative pair".

Do not confuse: d/dx(tan x) = sec²x, but d/dx(sec x) = sec x · tan x — they are DIFFERENT. Similarly d/dx(cot x) = −cosec²x, not −cot²x.

Learn 3 — Differentiating Inverse Trig Functions

Inverse trig functions arise naturally as antiderivatives of algebraic expressions. Their derivatives are derived using implicit differentiation.

d/dx(arcsin x)

Let y = arcsin x, so sin y = x. Differentiate implicitly:
cos y · (dy/dx) = 1 → dy/dx = 1/cos y
Since sin y = x, cos y = √(1 − x²) (positive because y ∈ [−π/2, π/2])
d/dx(arcsin x) = 1/√(1−x²), valid for |x| < 1

d/dx(arccos x)

Let y = arccos x, so cos y = x. Differentiate implicitly:
−sin y · (dy/dx) = 1 → dy/dx = −1/sin y = −1/√(1−x²)
d/dx(arccos x) = −1/√(1−x²), valid for |x| < 1

d/dx(arctan x)

Let y = arctan x, so tan y = x. Differentiate implicitly:
sec²y · (dy/dx) = 1 → dy/dx = 1/sec²y = cos²y
Since tan y = x, sec²y = 1 + tan²y = 1 + x² → cos²y = 1/(1+x²)
d/dx(arctan x) = 1/(1+x²), valid for all x

d/dx(arcsin x) = 1/√(1−x²) | d/dx(arccos x) = −1/√(1−x²) | d/dx(arctan x) = 1/(1+x²)

With Chain Rule: d/dx(arctan(f(x))) = f'(x)/(1+(f(x))²)

d/dx(arctan(3x)) = 3 / (1 + (3x)²) = 3/(1+9x²)
d/dx(arctan(x/3)) = (1/3) / (1 + x²/9) = (1/3) · 9/(9+x²) = 3/(9+x²)
d/dx(arcsin(x/2)) = (1/2) / √(1 − x²/4) = 1 / √(4−x²)
d/dx(arcsin(2x+1)) = 2 / √(1 − (2x+1)²)

Standard Integral Results (from these derivatives)

∫ 1/√(1−x²) dx = arcsin x + C
∫ 1/(1+x²) dx = arctan x + C
∫ 1/√(a²−x²) dx = arcsin(x/a) + C
∫ 1/(a²+x²) dx = (1/a)·arctan(x/a) + C

Critical distinction: arctan uses (1+x²) in the denominator, while arcsin uses √(1−x²) in the denominator. Mixing these up is the single most common error in P3 exams.

Learn 4 — Harder Implicit Differentiation

In Pure 2 you met basic implicit differentiation. In Pure 3 the curves are harder — they include products of x and y terms, trig and exponential functions, and require you to find stationary points and second derivatives.

Key Rule: Differentiating f(y) with Respect to x

d/dx(f(y)) = f'(y) · dy/dx (chain rule)
Examples:
d/dx(y²) = 2y · dy/dx
d/dx(sin y) = cos y · dy/dx
d/dx(e^y) = e^y · dy/dx

Differentiating Products Involving y

d/dx(xy²): use product rule — think of it as x · (y²)
= (1)·y² + x · d/dx(y²) = y² + x · 2y · dy/dx = y² + 2xy · dy/dx

d/dx(x²y): = 2x·y + x²·dy/dx
d/dx(x·sin y): = sin y + x·cos y·dy/dx
d/dx(y·e^x): = (dy/dx)·e^x + y·e^x

Example: Find dy/dx for x²y³ + y·sin(x) = 0

Differentiate each term with respect to x:
d/dx(x²y³) = 2x·y³ + x²·3y²·dy/dx
d/dx(y·sin x) = (dy/dx)·sin x + y·cos x
So: 2xy³ + 3x²y²·dy/dx + sin(x)·dy/dx + y·cos(x) = 0
dy/dx(3x²y² + sin x) = −2xy³ − y·cos x
dy/dx = −(2xy³ + y·cos x) / (3x²y² + sin x)

Finding Stationary Points Implicitly

Stationary points occur where dy/dx = 0 (and the curve exists).
Set the numerator of dy/dx = 0, then solve simultaneously with the curve equation.

Example: Curve 4x² + y² − 2xy = 5
Differentiating: 8x + 2y·dy/dx − 2y − 2x·dy/dx = 0
dy/dx(2y − 2x) = 2y − 8x
dy/dx = (2y − 8x)/(2y − 2x) = (y − 4x)/(y − x)
Set dy/dx = 0: y − 4x = 0 → y = 4x
Sub into curve: 4x² + 16x² − 8x² = 5 → 12x² = 5 → x = ±√(5/12)
Stationary points: x = ±√(5/12), y = ±4√(5/12)

Parametric Second Derivative d²y/dx²

For parametric equations x=f(t), y=g(t):
dy/dx = (dy/dt)/(dx/dt)

For d²y/dx²: differentiate dy/dx with respect to t, then divide by dx/dt
d²y/dx² = d/dt(dy/dx) ÷ (dx/dt)

Example: x = t², y = t³
dy/dx = 3t²/(2t) = 3t/2
d/dt(3t/2) = 3/2
d²y/dx² = (3/2) ÷ (2t) = 3/(4t)

When finding the second parametric derivative, you differentiate dy/dx (which is a function of t) with respect to t, and divide by dx/dt again. Never differentiate dy/dx with respect to x directly — it's expressed in terms of t.

Learn 5 — Applications and Mixed Problems

Combining Multiple Rules in One Problem

Example: Differentiate y = x²·arctan(x)
Product rule: u = x², v = arctan(x)
dy/dx = 2x·arctan(x) + x²·(1/(1+x²))
= 2x·arctan(x) + x²/(1+x²)

Example: Differentiate y = sec²(arctan x)
Chain rule: outer = u², inner = sec(arctan x), then chain again
Let u = arctan x: d/dx(arctan x) = 1/(1+x²)
Let v = sec u: d/du(sec u) = sec u · tan u
y = (sec u)²: dy/du = 2·sec u · sec u · tan u = 2·sec²u·tan u
dy/dx = 2·sec²(arctan x)·tan(arctan x) · 1/(1+x²)
Since tan(arctan x) = x:
dy/dx = 2x·sec²(arctan x)/(1+x²)

Tangents and Normals to Harder Curves

For a curve defined parametrically, the tangent gradient at parameter t is dy/dx at that t.
For an implicit curve, substitute the point into dy/dx to find the gradient.

Example: Find the tangent to x = t², y = arctan(t) at t = 1.
dy/dt = 1/(1+t²), dx/dt = 2t
At t=1: dy/dx = [1/(1+1)] / (2·1) = (1/2)/2 = 1/4
Point: (1, arctan(1)) = (1, π/4)
Tangent: y − π/4 = (1/4)(x − 1)

Related Rates of Change

The chain rule connects rates: dV/dt = dV/dr · dr/dt

Example: Sphere, V = (4/3)πr³, A = 4πr². If dA/dt = 6 cm²/s when r = 3 cm, find dr/dt and dV/dt.
dA/dr = 8πr. At r=3: dA/dr = 24π
dr/dt = (dA/dt) / (dA/dr) = 6/(24π) = 1/(4π) cm/s
dV/dr = 4πr². At r=3: dV/dr = 36π
dV/dt = 36π · 1/(4π) = 9 cm³/s

Connected Rates — General Strategy

Step 1: Write down the formula connecting the two quantities (volume, area, length, etc.).

Step 2: Differentiate implicitly with respect to time t to get the rate equation.

Step 3: Substitute the given values (rates and dimensions) to find the unknown rate.

Step 4: State units clearly in your answer.

In exam problems, "at the instant when…" is your cue to substitute specific values after differentiating the general relationship. Never substitute before differentiating.

Worked Examples

Example 1 — Differentiating a^x and log_a(x)

Find dy/dx for y = 3^x + log_2(x).

M1 Differentiate 3^x using d/dx(a^x) = a^x · ln a: d/dx(3^x) = 3^x · ln 3

M1 Differentiate log_2(x) using d/dx(log_a x) = 1/(x·ln a): d/dx(log_2 x) = 1/(x · ln 2)

A1 dy/dx = 3^x · ln 3 + 1/(x · ln 2)

Example 2 — Differentiating sec with Chain Rule

Find dy/dx for y = sec(3x² + 1).

M1 Identify outer function: sec(u) where u = 3x² + 1

M1 d/du(sec u) = sec u · tan u; du/dx = 6x

A1 dy/dx = 6x · sec(3x² + 1) · tan(3x² + 1)

Example 3 — Differentiating arctan

Find dy/dx for y = arctan(x/3).

M1 Let u = x/3, so y = arctan(u). Then d/du(arctan u) = 1/(1+u²) and du/dx = 1/3

M1 Chain rule: dy/dx = (1/3) · 1/(1+(x/3)²) = (1/3) · 1/(1 + x²/9)

A1 Simplify: dy/dx = (1/3) · 9/(9+x²) = 3/(9+x²)

Example 4 — Product Rule with arcsin

Differentiate y = x · arcsin(x).

M1 Product rule: u = x, v = arcsin(x). So u' = 1, v' = 1/√(1−x²)

A1 dy/dx = 1 · arcsin(x) + x · 1/√(1−x²) = arcsin(x) + x/√(1−x²)

Example 5 — Stationary Points via Implicit Differentiation

Find the stationary points of 4x² + y² − 2xy = 5.

M1 Differentiate implicitly: 8x + 2y·dy/dx − 2y − 2x·dy/dx = 0

M1 Collect dy/dx terms: dy/dx(2y − 2x) = 2y − 8x → dy/dx = (y − 4x)/(y − x)

M1 For stationary points, set numerator = 0: y − 4x = 0 → y = 4x

M1 Substitute y = 4x into curve: 4x² + 16x² − 8x² = 5 → 12x² = 5 → x² = 5/12

A1 x = ±√(5/12) = ±√15/6; y = ±4√15/6 = ±2√15/3
Stationary points: (√15/6, 2√15/3) and (−√15/6, −2√15/3)

Example 6 — Parametric with arctan

Given x = t², y = arctan(t), find dy/dx as a function of t.

M1 Differentiate: dx/dt = 2t; dy/dt = 1/(1+t²)

A1 dy/dx = (dy/dt)/(dx/dt) = [1/(1+t²)] / (2t) = 1/(2t(1+t²))

Example 7 — Related Rates of Change

A circular disc has area A = πr². If dA/dt = 6 cm²/s when r = 3 cm, find dr/dt.

M1 Chain rule: dA/dt = dA/dr · dr/dt

M1 dA/dr = 2πr. At r = 3: dA/dr = 6π

A1 dr/dt = (dA/dt)/(dA/dr) = 6/(6π) = 1/π ≈ 0.318 cm/s

Example 8 — Chain Rule on (sec x)³

Differentiate y = sec³(x).

M1 Write as y = (sec x)³. Outer function: u³, inner: sec x

M1 d/du(u³) = 3u² and d/dx(sec x) = sec x · tan x

A1 dy/dx = 3·sec²(x) · sec(x)·tan(x) = 3·sec³(x)·tan(x)

Common Mistakes

Mistake 1 — Confusing a^x with x^a

✗ d/dx(3^x) = x · 3^(x−1) (WRONG — this is the POWER RULE for x^n)

✓ d/dx(3^x) = 3^x · ln 3 (use the exponential rule: a^x · ln a)

Mistake 2 — Forgetting ln a in d/dx(a^x)

✗ d/dx(2^x) = 2^x (WRONG — ln 2 factor is missing)

✓ d/dx(2^x) = 2^x · ln 2 ≈ 2^x · 0.693

Mistake 3 — Confusing d/dx(sec x) with d/dx(tan x)

✗ d/dx(sec x) = sec²x (WRONG — that's d/dx(tan x))

✓ d/dx(sec x) = sec x · tan x

Mistake 4 — Mixing arctan and arcsin denominators

✗ d/dx(arctan x) = 1/√(1−x²) (WRONG — that's d/dx(arcsin x))

✓ d/dx(arctan x) = 1/(1+x²) ✓ d/dx(arcsin x) = 1/√(1−x²)

Mistake 5 — Forgetting the inner derivative with arcsin/arctan

✗ d/dx(arcsin(3x)) = 1/√(1−9x²) (WRONG — chain rule inner derivative missing)

✓ d/dx(arcsin(3x)) = 3/√(1−9x²) (multiply by 3 from differentiating 3x)

Mistake 6 — Implicit: omitting dy/dx when differentiating y terms

✗ d/dx(y²) = 2y (WRONG — chain rule on y requires dy/dx factor)

✓ d/dx(y²) = 2y · dy/dx (y is a function of x, so chain rule applies)

Mistake 7 — Forgetting the minus sign in d/dx(arccos x)

✗ d/dx(arccos x) = 1/√(1−x²) (WRONG — the negative sign is essential)

✓ d/dx(arccos x) = −1/√(1−x²) (the minus sign distinguishes it from arcsin)

Key Formulas — Complete Reference

Exponential and Logarithmic

Function	Derivative	With Chain Rule (inner u)
a^x	a^x · ln a	a^u · ln a · u'
log_a(x)	1/(x · ln a)	u'/(u · ln a)
e^x	e^x	e^u · u'
ln x	1/x	u'/u

Reciprocal Trig

Function	Derivative	With Chain Rule
sec x	sec x · tan x	sec(u) · tan(u) · u'
cosec x	−cosec x · cot x	−cosec(u) · cot(u) · u'
cot x	−cosec²x	−cosec²(u) · u'
tan x (reminder)	sec²x	sec²(u) · u'

Inverse Trig

Function	Derivative	With Chain Rule
arcsin x	1/√(1−x²)	u'/√(1−u²)
arccos x	−1/√(1−x²)	−u'/√(1−u²)
arctan x	1/(1+x²)	u'/(1+u²)

General Differentiation Rules

Rule	Formula
Product Rule	d/dx(uv) = u'v + uv'
Quotient Rule	d/dx(u/v) = (u'v − uv')/v²
Chain Rule	d/dx(f(g(x))) = f'(g(x)) · g'(x)
Implicit (y terms)	d/dx(f(y)) = f'(y) · dy/dx
Parametric dy/dx	dy/dx = (dy/dt)/(dx/dt)
Parametric d²y/dx²	d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

Proof Bank

Proof 1 — d/dx(a^x) = a^x · ln a

Let y = a^x. Since a = e^(ln a) for any a > 0, we can write:
y = a^x = (e^(ln a))^x = e^(x · ln a)
Now differentiate using the chain rule (ln a is a constant):
dy/dx = e^(x · ln a) · ln a
But e^(x · ln a) = a^x, so:
dy/dx = a^x · ln a □
Note: When a = e, ln e = 1 and this gives d/dx(e^x) = e^x, consistent with the known result.

Proof 2 — d/dx(sec x) = sec x · tan x

Write sec x = 1/cos x = (cos x)^(−1).
Apply the quotient rule with u = 1, v = cos x:
d/dx(1/cos x) = (u'v − uv') / v² = (0 · cos x − 1 · (−sin x)) / cos²x
= sin x / cos²x
Now rewrite using trig identities:
sin x / cos²x = (sin x / cos x) · (1/cos x) = tan x · sec x
d/dx(sec x) = sec x · tan x □

Proof 3 — d/dx(arctan x) = 1/(1+x²)

Let y = arctan x. By definition of arctan, this means tan y = x.
Differentiate both sides with respect to x (implicit differentiation):
sec²y · (dy/dx) = 1
dy/dx = 1/sec²y = cos²y
Now use the identity sec²y = 1 + tan²y, and since tan y = x:
sec²y = 1 + x²
Therefore cos²y = 1/sec²y = 1/(1+x²)
d/dx(arctan x) = 1/(1+x²) □
Note: This proof works for all x since arctan is defined for all real x (y ∈ (−π/2, π/2) keeps cos y positive).

Visualiser — Comparing a^x Curves

This canvas plots y = 2^x (blue), y = e^x (red), and y = 3^x (green) simultaneously. Select a curve to highlight its tangent line at x = 1, and see the derivative value annotated on the graph.

Highlight curve:

y = 2^x, dy/dx = 2^x · ln 2 y = eˣ, dy/dx = eˣ y = 3^x, dy/dx = 3^x · ln 3

Select a curve and press Redraw to see its tangent at x = 1.

Function: —

Derivative at x=1: —

Tangent slope: —

Exercise 1 — Differentiating a^x and log_a(x)

Exercise 2 — Reciprocal Trig Derivatives

Exercise 3 — Inverse Trig Derivatives

Exercise 4 — Harder Implicit Differentiation

Exercise 5 — Mixed Applications

Practice — 30 Mixed Questions

Challenge — 15 Harder Questions

Exam Style Questions

These questions are structured in Cambridge A-Level format with mark allocations. Show all working. Use the mark scheme button to check after attempting.

Q1 — [4 marks]

Differentiate y = 5^(2x) · arctan(x) with respect to x. Simplify your answer.

M1: Product rule: u = 5^(2x), v = arctan(x)
M1: u' = 2·ln5·5^(2x), v' = 1/(1+x²)
A1: dy/dx = 2·ln5·5^(2x)·arctan(x) + 5^(2x)/(1+x²)
A1: Factor if desired: 5^(2x)[2·ln5·arctan(x) + 1/(1+x²)]

Q2 — [5 marks]

A curve is defined implicitly by x³ + y³ = 6xy. Find dy/dx in terms of x and y, and find the coordinates of the point on the curve where dy/dx = −1 and x > 0.

M1: Differentiate implicitly: 3x² + 3y²·(dy/dx) = 6y + 6x·(dy/dx)
M1: Rearrange: dy/dx(3y² − 6x) = 6y − 3x²
A1: dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)
M1: Set dy/dx = −1: 2y − x² = −(y² − 2x) → 2y − x² = −y² + 2x → y² + 2y − x² − 2x = 0
M1: Also x³ + y³ = 6xy. Try x = y (from symmetry of dy/dx = −1): 2x − x² = −(x² − 2x) → 0 = 0 ✓
A1: On x = y and x³ + x³ = 6x², so 2x³ = 6x² → x = 3. Point: (3, 3)

Q3 — [4 marks]

The parametric equations of a curve are x = sec t, y = tan t for −π/2 < t < π/2. Find dy/dx in terms of t, and show that d²y/dx² = −cosec³(t)·cot(t)... wait — find d²y/dx² in terms of t.

M1: dx/dt = sec t · tan t; dy/dt = sec²t
A1: dy/dx = sec²t / (sec t · tan t) = sec t / tan t = 1/(sin t) = cosec t
M1: d²y/dx² = [d/dt(cosec t)] / (dx/dt) = [−cosec t · cot t] / (sec t · tan t)
A1: = −cosec t · cot t / (sec t · tan t) = −cosec t · (cos t/sin t) / (tan t / cos t)
Simplify: = −cosec t · cos²t / (sin t · tan t) = −cos²t / (sin²t · tan t) = −cos³t / sin³t = −cot³t

Q4 — [3 marks]

Find the exact value of d/dx(arcsin(x/√2)) at x = 1.

M1: Let u = x/√2. d/dx(arcsin u) = (du/dx)/√(1−u²)
M1: du/dx = 1/√2; at x=1: u = 1/√2, 1−u² = 1 − 1/2 = 1/2
A1: d/dx = (1/√2)/√(1/2) = (1/√2)/(1/√2) = 1

Q5 — [4 marks]

Given that y = log_3(sin x), find dy/dx and the exact value of dy/dx when x = π/6.

M1: Use change of base: y = ln(sin x)/ln 3
M1: Chain rule: dy/dx = (1/ln 3) · (cos x/sin x) = cot x / ln 3
A1: dy/dx = cot x / ln 3
A1: At x = π/6: cot(π/6) = cos(π/6)/sin(π/6) = (√3/2)/(1/2) = √3
dy/dx = √3/ln 3

Q6 — [5 marks]

The volume of a cone is V = (1/3)πr²h. Given that the height h is always equal to twice the radius, find the rate of change of the volume when r = 4 cm and the radius is increasing at 0.5 cm/s.

M1: Substitute h = 2r: V = (1/3)πr²(2r) = (2/3)πr³
M1: Differentiate: dV/dr = 2πr²
M1: Chain rule: dV/dt = dV/dr · dr/dt
M1: At r=4: dV/dr = 2π(16) = 32π
A1: dV/dt = 32π × 0.5 = 16π ≈ 50.3 cm³/s

Q7 — [4 marks]

Differentiate y = cosec²(2x) + cot(2x) with respect to x.

M1: d/dx(cosec²(2x)) = 2·cosec(2x) · d/dx(cosec(2x)) = 2·cosec(2x)·(−cosec(2x)·cot(2x))·2
A1: = −4·cosec²(2x)·cot(2x)
M1: d/dx(cot(2x)) = −cosec²(2x) · 2 = −2·cosec²(2x)
A1: dy/dx = −4·cosec²(2x)·cot(2x) − 2·cosec²(2x) = −2·cosec²(2x)(2cot(2x)+1)

Q8 — [6 marks]

A curve has parametric equations x = e^t + e^(−t), y = e^t − e^(−t). (i) Find dy/dx in terms of t. (ii) Find the equation of the tangent at t = 0.

M1: dx/dt = e^t − e^(−t); dy/dt = e^t + e^(−t)
A1: dy/dx = (e^t + e^(−t))/(e^t − e^(−t))
M1: At t=0: e^0=1, so dy/dx = (1+1)/(1−1) = 2/0 — undefined (vertical tangent)!
Note: at t=0, x = 1+1=2, y=1−1=0. dx/dt = 0 at t=0 → vertical tangent at (2,0).
A1: Vertical tangent: equation is x = 2
B1: Note that as t→0, the curve passes through (2,0) with a vertical tangent.
B1: dy/dx → ∞ as t → 0, confirming vertical tangent x = 2.

Past Paper Questions — Cambridge 9709 Style

These questions are modelled on Cambridge A-Level 9709 Paper 3 past papers. Attempt each question fully before revealing the mark scheme.

PP1 — 9709/31/M/J/19 Style [4 marks]

Find the gradient of the curve x² + 2y² + xy = 7 at the point (1, √(3/2)). Give your answer in exact form.

M1: Differentiate implicitly: 2x + 4y·(dy/dx) + y + x·(dy/dx) = 0
M1: Collect: dy/dx(4y + x) = −2x − y
A1: dy/dx = −(2x+y)/(4y+x)
A1: At (1, √(3/2)): gradient = −(2 + √(3/2))/(4√(3/2) + 1) = −(2 + √6/2)/(2√6 + 1)

PP2 — 9709/32/O/N/20 Style [3 marks]

Differentiate y = arctan(e^x) with respect to x.

M1: Let u = e^x. d/du(arctan u) = 1/(1+u²)
M1: du/dx = e^x
A1: dy/dx = e^x/(1+(e^x)²) = e^x/(1+e^(2x))

PP3 — 9709/33/M/J/21 Style [5 marks]

Given that x = 2arctan(t) and y = ln(1+t²), find dy/dx in terms of t. Hence show that dy/dx = t.

M1: dx/dt = 2/(1+t²)
M1: dy/dt = 2t/(1+t²)
M1: dy/dx = [2t/(1+t²)] / [2/(1+t²)]
A1: = 2t/(1+t²) × (1+t²)/2 = t
A1: dy/dx = t □

PP4 — 9709/31/O/N/22 Style [5 marks]

A curve has equation y = (2^x)/(1 + 2^x). Show that dy/dx = y(1−y)·ln 2.

M1: Quotient rule: u = 2^x, v = 1+2^x. u' = 2^x·ln2, v' = 2^x·ln2
M1: dy/dx = [2^x·ln2·(1+2^x) − 2^x·2^x·ln2]/(1+2^x)²
A1: = ln2·2^x[(1+2^x) − 2^x]/(1+2^x)² = ln2·2^x/(1+2^x)²
M1: Note y = 2^x/(1+2^x) → y(1−y) = [2^x/(1+2^x)] · [1/(1+2^x)] = 2^x/(1+2^x)²
A1: Therefore dy/dx = y(1−y)·ln2 □

PP5 — 9709/32/M/J/23 Style [6 marks]

The parametric equations of a curve are x = t − sin t, y = 1 − cos t for 0 < t < 2π. Find dy/dx in terms of t and find the values of t where the tangent is horizontal. State the coordinates of those points.

M1: dx/dt = 1 − cos t; dy/dt = sin t
A1: dy/dx = sin t/(1 − cos t)
M1: Horizontal tangent: sin t = 0 in (0, 2π) → t = π
M1: Check denominator at t=π: 1−cos π = 1+1 = 2 ≠ 0 ✓
A1: At t=π: x = π − sin π = π; y = 1 − cos π = 2
A1: Only horizontal tangent in given range: (π, 2)

Further Differentiation A-Level Pure 3