Mathematics · Lesson 23

Differentiation: the slope machine

JAMB

What the syllabus asks

JAMB: Calculus: differentiation of algebraic and simple trigonometric functions (sine, cosine and tangent), rate of change, maxima and minima

What you will learn

Find the gradient of a straight line and explain why the gradient of a curve changes.
Use function notation and substitute values.
Explain the gradient of a curve at a point as the gradient of the tangent, using chords and the idea of a limit.
Differentiate simple functions from first principles.
Use the power rule, including negative and fractional powers, and differentiate sums term by term.
Use the chain, product and quotient rules.
Differentiate sin x, cos x and tan x (x in radians) and simple compositions.
Find the gradient at a point and the equations of tangents and normals.
Find and classify stationary points using the second derivative.
Solve maximum and minimum problems.
Apply differentiation to rates of change and to motion.

Key points

Gradient = rise ÷ run. Positive slopes up, negative down, zero is flat. A curve’s gradient changes from point to point.
The gradient of a curve at a point is the gradient of the tangent there; chord gradients get closer and closer to it (the limit).
First principles: [f(x + h) − f(x)] ÷ h, simplify, divide by h, then let h → 0. For y = x², dy/dx = 2x.
Power rule: axⁿ → anxⁿ⁻¹; constant → 0; term by term. Rewrite 1/xⁿ = x⁻ⁿ and √x = x^½; expand brackets and split fractions first.
Gradient at a point: substitute x into dy/dx, not into y.
Tangent: y − y₁ = m(x − x₁) with m = dy/dx at x₁. Normal: gradient −1/m.
Chain rule: dy/dx = dy/du × du/dx (outside, then × derivative of the inside).
Product rule: u dv/dx + v du/dx. Quotient rule: (v du/dx − u dv/dx) ÷ v².
With x in radians: sin x → cos x, cos x → −sin x, tan x → sec²x; sin(ax + b) → a cos(ax + b).
Stationary points: dy/dx = 0. d²y/dx² > 0 minimum (smile), < 0 maximum (frown).
Max/min problems: one formula in one letter, differentiate, set to 0, test, then answer the question.
dy/dx is a rate of change. Motion: v = ds/dt, a = dv/dt; at rest when v = 0.

Start here: What is a gradient?

Ride an okada up one of the hilly roads in Jos or Enugu. Some stretches are gentle; some are so steep the engine strains. Differentiation is the maths that measures steepness at every single point, and with it how fast anything changes: how fast a car speeds up, how fast a population grows, and where a profit is highest.

In exams, differentiation is in the JAMB UTME Mathematics syllabus (under Calculus) and in WAEC and NECO Further Mathematics. It is not part of the General Mathematics core papers. It is very learnable, so we start from zero.

Quick refresher: coordinates. A point is written (x, y): x tells you how far across, y how far up. The point (4, 8) is 4 across and 8 up.

The gradient of a straight line measures its steepness:

Gradient = rise ÷ run = (change in y) ÷ (change in x) = (y₂ − y₁) ÷ (x₂ − x₁).

Fig. 23.1: The gradient of a straight line is rise ÷ run.

Worked example 23.1

Find the gradient of the line through (1, 2) and (4, 8).

Rise (change in y): 8 − 2 = 6.
Run (change in x): 4 − 1 = 3.
Gradient = 6 ÷ 3 = 2. For every 1 step across, the line goes 2 up.

The sign tells you the direction, reading from left to right: a positive gradient slopes up, a negative gradient slopes down, and a gradient of 0 is flat.

Worked example 23.2

Find the gradient of the line through (−1, 5) and (3, −3).

Rise: −3 − 5 = −8.
Run: 3 − (−1). Two minus signs together make a plus: 3 + 1 = 4.
Gradient = −8 ÷ 4 = −2. Negative, so the line slopes down.

−2

Try it yourself. Find the gradient of the line through (2, 1) and (6, 3).

Show solution

Rise = 3 − 1 = 2. Run = 6 − 2 = 4. Gradient = 2 ÷ 4 = ½.

Remember: gradient = rise ÷ run = change in y ÷ change in x. Positive slopes up, negative slopes down, zero is flat.

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Practice questions

1. Find the gradient of the line through (−2, 3) and (4, 15).

Show worked answer

Rise = 15 − 3 = 12. Run = 4 − (−2) = 4 + 2 = 6. Gradient = 12 ÷ 6 = 2.

2. Differentiate y = x² + 4x from first principles.

Show worked answer

f(x + h) = (x + h)² + 4(x + h) = x² + 2xh + h² + 4x + 4h. Subtract f(x) = x² + 4x: 2xh + h² + 4h. Divide by h: 2x + h + 4. Let h → 0: dy/dx = 2x + 4.

3. Differentiate y = 7x³ − 4x + 1/x.

Show worked answer

Rewrite 1/x = x⁻¹. 7x³ → 21x²; −4x → −4; x⁻¹ → −1 × x⁻² = −x⁻². So dy/dx = 21x² − 4 − 1/x².

4. Differentiate y = 6√x − 2/x³.

Show worked answer

Rewrite: y = 6x^½ − 2x⁻³. 6x^½ → 6 × ½ x^(−½) = 3x^(−½). −2x⁻³ → −2 × (−3) x⁻⁴ = 6x⁻⁴. So dy/dx = 3/√x + 6/x⁴.

5. Differentiate y = (3x − 2)(x + 4).

Show worked answer

Expand: 3x² + 12x − 2x − 8 = 3x² + 10x − 8. Differentiate: 6x + 10.

Key words

δx (delta x): Another way some books write h: a very small change in x.
Coordinates: A pair (x, y) giving the position of a point: x across, y up.
Gradient: Steepness of a line: rise ÷ run, the change in y divided by the change in x.
Function: A rule that turns each input number x into one output number, e.g. y = x².
f(x): A name for a function of x. f(3) means: put 3 in place of x.
Substitute: Put a number in place of a letter and work out the value.
Tangent: A straight line that touches a curve at a point and has the same steepness as the curve there (it may cross the curve somewhere else).
Chord: A straight line joining two points on a curve.
Limit: The value an answer gets closer and closer to as something (such as h) gets closer and closer to 0.
Derivative: The formula for the gradient of a curve at any point, written dy/dx or f′(x).
dy/dx: Read “dee y by dee x”: the derivative of y with respect to x.
f′(x): Read “f dash of x”: the derivative of f(x).

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