Pure 1 — Differentiation | Full Lesson

Pure Mathematics 1

Differentiation Lesson

Hello. I am going to teach you everything you need to know about differentiation. I will assume you know nothing — and by the end, you will be differentiating confidently. Let us begin.

What is Differentiation?

Differentiation is a mathematical tool that finds the rate of change of a quantity. In geometry, it tells you the gradient (slope) of a curve at any point.

You already know how to find the gradient of a straight line: pick any two points and use Δy/Δx. But a curve does not have a constant gradient — it changes at every point.

Imagine driving along a winding road. At each moment, your speed (rate of change of position) is different. Differentiation is like looking at your speedometer at a single instant.

Gradient of a Curve

The gradient of a curve at a point is the gradient of the tangent — the straight line that just touches the curve at that point.

The Idea of a Limit f'(x) = lim_h→0 (f(x+h) − f(x)) / h

As h gets smaller and smaller, the chord between two points becomes the tangent. We call this the derivative from first principles. In practice, we use rules that skip the limit and give us the answer directly.

Notation

There are two common ways to write the derivative:

🔹 dy/dx — Leibniz notation (says "the derivative of y with respect to x")

🔹 f'(x) — Lagrange notation (says "f-prime of x")

They mean the same thing. I will use both throughout this lesson.

Teacher's Tip: Differentiation is the first big idea of calculus. It is all about measuring change. When you hear "differentiate", think "find the gradient function".

Differentiating Powers of x

The most important rule in differentiation is the power rule. It tells you how to differentiate any power of x.

The Power Rule If y = xⁿ, then dy/dx = n xⁿ⁻¹

In words: "bring the power down, then reduce the power by one".

Positive Powers

Example 1: y = x⁵, find dy/dx

Bring the power (5) down: 5x

Reduce the power by one: 5x⁴

dy/dx = 5x⁴

Example 2: y = 4x³, find dy/dx

Multiply by the power: 4 × 3 = 12

Reduce the power by one: 12x²

dy/dx = 12x²

Negative Powers

The power rule works for any power, including negative ones. The method is exactly the same.

Example 3: y = x⁻², find dy/dx

Bring the power (−2) down: −2x

Reduce the power by one: −2x⁻³

dy/dx = −2x⁻³ = −2/x³

Example 4: y = 6x⁻¹, find dy/dx

6 × (−1) = −6

Reduce −1 to −2: −6x⁻²

dy/dx = −6x⁻² = −6/x²

Fractional Powers

Example 5: y = x^{1/2} (which is √x), find dy/dx

Bring 1/2 down: (1/2)x

Reduce power by one: (1/2)x^{-1/2}

Rewrite as a fraction: 1/(2√x)

dy/dx = 1/(2√x)

Example 6: y = 3x² + 2x − 5, find dy/dx

Differentiate each term separately:

d/dx(3x²) = 6x

d/dx(2x) = 2 (remember: 2x = 2x¹, so 2 × 1 × x⁰ = 2)

d/dx(−5) = 0 (constants differentiate to zero)

dy/dx = 6x + 2

Teacher's Tip: The power rule is the most used rule in calculus. Practise it until it is instant: n xⁿ⁻¹. Positive, negative, fractional — it does not matter. Same rule, always.

Tangents and Normals

Once you can differentiate, you can find the gradient at any point on a curve. This lets you find the equation of the tangent (the line that just touches the curve) and the normal (the line perpendicular to the tangent).

Equation of a Tangent

If you know the gradient m at a point (x₀, y₀), the equation of the tangent is:

Tangent y − y₀ = m(x − x₀)

Where m = f'(x₀), the derivative evaluated at x₀.

Example 1: Find the equation of the tangent to y = x² + 3x at x = 2

Step 1: Find the derivative: dy/dx = 2x + 3

Step 2: Gradient at x = 2: m = 2(2) + 3 = 7

Step 3: y-coordinate at x = 2: y = 4 + 6 = 10

Step 4: Use y − y₀ = m(x − x₀):

y − 10 = 7(x − 2)

y − 10 = 7x − 14

y = 7x − 4

Equation of a Normal

The normal is perpendicular to the tangent. If the tangent has gradient m, the normal has gradient −1/m.

Normal y − y₀ = (−1/m)(x − x₀)

Where m = f'(x₀) and m ≠ 0.

Example 2: Find the equation of the normal to y = x² + 1 at x = 2

Step 1: Derivative: dy/dx = 2x

Step 2: Tangent gradient at x = 2: m = 4

Step 3: y-coordinate: y = 4 + 1 = 5

Step 4: Normal gradient = −1/4

Step 5: y − 5 = −(1/4)(x − 2)

y − 5 = −x/4 + 1/2

y = −x/4 + 11/2

Teacher's Tip: The tangent and normal are perpendicular, so m_tangent × m_normal = −1. If m = 0 (horizontal tangent), the normal is vertical (x = x₀).

Stationary Points

A stationary point is a point on a curve where the gradient is zero. At a stationary point, the tangent is horizontal.

Stationary Point Condition f'(x) = 0

There are three types of stationary point:

🔵 Local Maximum — the highest point in its neighbourhood (peak)

🟢 Local Minimum — the lowest point in its neighbourhood (trough)

🟠 Point of Inflection — the curve changes from concave to convex (or vice versa), but it is not a peak or trough

How to Find Stationary Points

Step 1: Find f'(x) (the derivative)

Step 2: Set f'(x) = 0 and solve for x

Step 3: Substitute each x back into f(x) to find the y-coordinate

Example 1: Find the stationary point(s) of y = x² − 6x + 10

Step 1: dy/dx = 2x − 6

Step 2: Set = 0: 2x − 6 = 0

2x = 6 → x = 3

Step 3: y = 9 − 18 + 10 = 1

Stationary point at (3, 1)

Example 2: Find the stationary points of y = 2x³ − 9x² + 12x − 3

Step 1: dy/dx = 6x² − 18x + 12

= 6(x² − 3x + 2)

= 6(x − 1)(x − 2)

Step 2: Set = 0: x = 1 or x = 2

Step 3: When x = 1: y = 2 − 9 + 12 − 3 = 2

When x = 2: y = 16 − 36 + 24 − 3 = 1

Stationary points at (1, 2) and (2, 1)

Teacher's Tip: Setting dy/dx = 0 is the key step. Factorise fully to find all solutions — do not lose any roots. Every stationary point satisfies dy/dx = 0.

Second Derivative

The second derivative is simply the derivative of the derivative. It tells you about the shape of the curve — whether it is curving upwards or downwards.

Second Derivative f''(x) = d²y/dx² = d/dx (dy/dx)

To find it: differentiate the function, then differentiate the result.

Using the Second Derivative to Classify Stationary Points

Once you have found a stationary point, the second derivative tells you its nature:

🟢 If f''(x) > 0 → the curve is concave up → minimum point

🔵 If f''(x) < 0 → the curve is concave down → maximum point

🟠 If f''(x) = 0 → the test is inconclusive (could be a point of inflection — check further)

Example: For y = 2x³ − 9x² + 12x − 3, classify the stationary points

We found stationary points at x = 1 and x = 2.

First derivative: f'(x) = 6x² − 18x + 12

Second derivative: f''(x) = 12x − 18

At x = 1: f''(1) = 12(1) − 18 = −6

Since f''(1) < 0 → maximum at (1, 2)

At x = 2: f''(2) = 12(2) − 18 = 6

Since f''(2) > 0 → minimum at (2, 1)

(1, 2) is a maximum, (2, 1) is a minimum

Teacher's Tip: Memorise: positive second derivative → minimum (smiley face ∪). Negative second derivative → maximum (frowny face ∩). The second derivative is like the "curvature detector".

Increasing and Decreasing Functions

The derivative tells you whether a function is increasing (going up) or decreasing (going down) at any point.

Increasing / Decreasing Test f'(x) > 0 → function is increasing
f'(x) < 0 → function is decreasing
f'(x) = 0 → stationary point

A function is increasing on an interval if f'(x) > 0 for all x in that interval — as x increases, y increases.

A function is decreasing on an interval if f'(x) < 0 for all x in that interval — as x increases, y decreases.

Example: For f(x) = x² − 4x + 5, find where it is increasing and where decreasing

Step 1: f'(x) = 2x − 4

Step 2: Find where f'(x) = 0: 2x − 4 = 0 → x = 2

Step 3: Test on each side:

When x < 2 (e.g. x = 0): f'(0) = −4 < 0 → decreasing

When x > 2 (e.g. x = 4): f'(4) = 4 > 0 → increasing

Decreasing for x < 2, increasing for x > 2

Example: For f(x) = x³ − 3x, find the intervals where it is increasing

Step 1: f'(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1)

Step 2: f'(x) = 0 when x = −1 or x = 1

Step 3: Test intervals:

x < −1 (e.g. x = −2): f'(−2) = 3(4 − 1) = 9 > 0 → increasing

−1 < x < 1 (e.g. x = 0): f'(0) = −3 < 0 → decreasing

x > 1 (e.g. x = 2): f'(2) = 3(4 − 1) = 9 > 0 → increasing

Increasing for x < −1 and x > 1

Teacher's Tip: Think of the derivative as a "direction detector". If f'(x) is positive, the function is climbing. If negative, it is descending. The sign of f'(x) tells you everything.

Optimisation Problems

Optimisation is one of the most powerful real-world uses of differentiation. It finds the maximum or minimum value of a quantity — like the largest area you can enclose, the cheapest cost, or the fastest time.

The Method

1. Write down what you want to maximise/minimise as a formula

2. Use any constraints to express it in terms of one variable

3. Differentiate and set the derivative equal to zero

4. Solve to find the critical value(s)

5. Use the second derivative to confirm it is a max or min

6. Answer the original question

Example: A farmer has 100 m of fencing to make a rectangular enclosure against a barn (only 3 sides needed). Find the maximum area.

Step 1: Let width = x (perpendicular to barn), length = (100 − 2x).

Area: A = x(100 − 2x) = 100x − 2x²

Step 2: Differentiate: dA/dx = 100 − 4x

Step 3: Set = 0: 100 − 4x = 0 → x = 25

Step 4: Second derivative: d²A/dx² = −4 < 0 → maximum

Step 5: Length = 100 − 2(25) = 50. Area = 25 × 50 = 1250 m²

Maximum area = 1250 m² (25 m × 50 m)

Example: A closed rectangular box with a square base has volume 32 cm³. Find the minimum surface area.

Step 1: Let base side = x, height = h.

Volume: x²h = 32 → h = 32/x²

Surface area: S = 2x² + 4xh = 2x² + 4x(32/x²) = 2x² + 128/x

Step 2: Differentiate: dS/dx = 4x − 128/x²

Step 3: Set = 0: 4x = 128/x² → 4x³ = 128 → x³ = 32 → x = ∛32 = 2∛4

Hmm, this gives a messy cube root. Let me use a cleaner example.

(Revised) Let base side = x, volume = 32.

h = 32/x²

S = 2x² + 128/x

dS/dx = 4x − 128/x² = 0

4x³ = 128 → x³ = 32 → x = ∛32 ≈ 3.17

Alternatively, choose nicer numbers: let volume = 8.

h = 8/x²

S = 2x² + 32/x

dS/dx = 4x − 32/x² = 0 → 4x³ = 32 → x³ = 8 → x = 2

h = 8/4 = 2

d²S/dx² = 4 + 64/x³ → at x = 2: 4 + 64/8 = 12 > 0 → minimum

Minimum surface area = 2(4) + 4(2)(2) = 8 + 16 = 24 cm²

Teacher's Tip: Optimisation problems all follow the same recipe. The hardest part is often writing the formula in one variable. Use the constraint to eliminate the extra variable, then differentiate as usual.

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Interactive Derivative Visualiser

Drag the point along the curve to see how the tangent changes. The gradient at each point is displayed below.

Function: f(x) = x²/4 Point: x = 0.0

Gradient f'(x) = 0.00

Practice — Step by Step

Solve each problem one step at a time. Type your answer for each step and click Check Step. You must get each step correct before moving to the next.

Mastery Kit

The art of doing mathematics consists in finding that special case which contains all the germs of generality.

— David Hilbert

⚡ Quick Hacks

Power rule in one breath: "n x to the n-minus-one". Say it every time you differentiate. n times x to the n-minus-one.
Tangent / normal memory: Tangent = same gradient as derivative. Normal = negative reciprocal (−1/m). Perpendicular means multiply to −1.
Stationary shortcut: For a quadratic f(x) = ax² + bx + c, the stationary point is always at x = −b/(2a). No need to differentiate every time.
Second derivative trick: f''(x) > 0 → ∪ shape (minimum). f''(x) < 0 → ∩ shape (maximum). Think of the sign as which way the curve bends.
Check your answer: After finding a stationary point, quickly test one x on each side to verify it is a max or min by checking the sign of f'(x).

📌 Key Notes

Differentiation finds the gradient function: f'(x) gives the gradient of the tangent at any point x. It is a function, not a single number.
Power rule works for all powers: Positive, negative, fractional — the rule n xⁿ⁻¹ never changes. This is the most important single rule.
Constants vanish: The derivative of any constant term is zero. A constant does not change, so its rate of change is zero.
Tangent vs normal: Tangent uses m = f'(x₀). Normal uses m = −1/f'(x₀). If m = 0, the normal is vertical.
Second derivative confirms, not finds: Use f'(x) = 0 to find stationary points. Use f''(x) to classify them as max, min, or inflection.

⚠️ Common Mistakes

Forgetting to reduce the power by exactly one: x⁵ → 5x⁴, not 5x⁵. The power drops by 1 every time.
Forgetting the constant factor: If y = 3x⁴, then dy/dx = 12x³ (multiply 3 × 4 = 12), not 3x³.
Mixing up tangent and normal: The normal is perpendicular to the tangent. Normal gradient = −1/f'(x₀), not f'(x₀).
Not checking the second derivative sign: f''(x) > 0 means minimum, < 0 means maximum. Getting this backwards loses marks.
Losing a solution when solving f'(x) = 0: If f'(x) factorises, write down all roots. Do not cancel a factor that contains x.

🧠 Memory Aid — Differentiation Cheat Sheet

Power rule: d/dx(xⁿ) = n xⁿ⁻¹

Tangent: y − y₀ = f'(x₀)(x − x₀)

Normal: y − y₀ = −1/f'(x₀) (x − x₀)

Stationary: f'(x) = 0

2nd derivative test: f''(x) > 0 → min | f''(x) < 0 → max

Increasing: f'(x) > 0 Decreasing: f'(x) < 0

💡

The Golden Rule of Differentiation

Differentiation is just the power rule, applied over and over. Every function you meet can be broken into powers of x — constant times xⁿ. Learn the power rule cold, and everything else follows. Practise it until it is automatic.

You Have Completed the Differentiation Lesson

Here is what you have learned:

✅ What differentiation is — gradient of a curve, rate of change, the idea of a limit
✅ The power rule: d/dx(xⁿ) = nxⁿ⁻¹ for any power (positive, negative, fractional)
✅ Tangents (use f'(x₀)) and Normals (use −1/f'(x₀))
✅ Stationary points: set f'(x) = 0 to find them
✅ Second derivative: f''(x) > 0 is minimum, f''(x) < 0 is maximum
✅ Increasing (f'(x) > 0) and decreasing (f'(x) < 0) functions
✅ Optimisation problems — using differentiation to find max/min in real-world contexts

Well done. You now know differentiation. This is the gateway to all of calculus — and you have mastered the fundamentals.