Further Mathematics · Lesson 14

Calculus toolkit

WAECNECO

What the syllabus asks

WAEC: Calculus: limits; differentiation from first principles; derivatives of algebraic and trigonometric functions; sum, product, quotient and chain rules; second derivatives; tangents and normals; stationary points, maxima and minima; rates of change; kinematics; integration as the reverse of differentiation; definite integrals; area under a curve and between curves

NECO: Calculus: limits; differentiation from first principles; derivatives of algebraic and trigonometric functions; sum, product, quotient and chain rules; second derivatives; tangents and normals; stationary points, maxima and minima; rates of change; kinematics; integration as the reverse of differentiation; definite integrals; area under a curve and between curves

What you will learn

Explain the derivative as a gradient and find it from first principles for simple functions.
Differentiate polynomials, powers, and sin and cos using sum, chain, product and quotient rules.
Find equations of tangents and normals.
Locate and classify stationary points; solve maximum and minimum problems.
Solve rates-of-change and kinematics problems.
Integrate polynomials and simple composite powers; evaluate definite integrals; find areas.

Key points

dy/dx = lim (f(x + h) − f(x))/h is the gradient of the tangent.
d/dx(xⁿ) = nxⁿ⁻¹; rewrite roots and fractions as powers first.
Chain: outside derivative × inside derivative. Product: u v′ + v u′. Quotient: (v u′ − u v′)/v².
d/dx sin x = cos x; d/dx cos x = −sin x (x in radians).
Tangent: y − y₁ = m(x − x₁); normal gradient −1/m.
Stationary: dy/dx = 0; d²y/dx² < 0 max, > 0 min.
Optimise: write the quantity in one variable, differentiate, set to 0, check.
v = ds/dt, a = dv/dt; connected rates by the chain rule.
∫xⁿ dx = xⁿ⁺¹/(n + 1) + c; definite integrals F(b) − F(a).
Area = ∫ y dx (split at roots); between curves ∫(upper − lower) dx.

Start here: The idea of a derivative

On a straight line the gradient is the same everywhere. On a curve such as y = x², the steepness keeps changing: the curve is flat at the origin and gets steeper as x grows. Calculus answers the question: exactly how steep is the curve at one particular point?

We define the gradient at a point as the gradient of the tangent there. To find it, take a point P(x, f(x)) and a nearby point Q(x + h, f(x + h)). The gradient of the chord PQ is (f(x + h) − f(x))/h. As Q slides towards P (h → 0), the chord becomes the tangent.

dy/dx = f′(x) = lim (h→0) [f(x + h) − f(x)] / h
This is differentiation from first principles.

Worked example 14.1

Differentiate y = x² from first principles.

f(x + h) = (x + h)² = x² + 2xh + h².
f(x + h) − f(x) = 2xh + h². Divide by h: 2x + h. We cancel h before letting it shrink, because we cannot divide by zero.
As h → 0, this approaches 2x.

dy/dx = 2x

So the gradient of y = x² at x = 3 is 6, as shown in the first infographic. At x = 0 it is 0 (flat), and at x = −2 it is −4 (falling).

Worked example 14.2

Differentiate y = 3x + 5 from first principles.

f(x + h) − f(x) = 3(x + h) + 5 − 3x − 5 = 3h.
Divide by h: 3. The limit is 3, a constant, as expected for a straight line of gradient 3.

dy/dx = 3

⚠️ Common mistake: substituting h = 0 before cancelling, which gives 0/0.

Try it yourself. Use first principles to show that the derivative of y = x² + 4x is 2x + 4.

Show solution

f(x + h) − f(x) = 2xh + h² + 4h; divide by h: 2x + h + 4 → 2x + 4.

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Practice questions

1. Differentiate y = 2x⁵ − 3x² + 4.

Show worked answer

10x⁴ − 6x.

2. Differentiate y = (4x − 1)⁵.

Show worked answer

20(4x − 1)⁴.

3. Differentiate y = (x + 2)/(x − 2).

Show worked answer

−4/(x − 2)².

4. Find the tangent to y = x² − 3x at x = 4.

Show worked answer

Point (4, 4), m = 5: y = 5x − 16.

5. Find and classify the stationary points of y = x³ − 12x.

Show worked answer

Max (−2, 16); min (2, −16).

Key words

Limit: The value a function approaches as x approaches a number.
Derivative: The gradient function dy/dx or f′(x).
Differentiation: The process of finding the derivative.
First principles: Finding the derivative from the limit of (f(x + h) − f(x))/h.
Tangent: A line touching a curve at a point with the same gradient.
Normal: The line perpendicular to the tangent at the point of contact.
Stationary point: A point where dy/dx = 0.
Maximum / minimum: A turning point where the curve changes from rising to falling, or falling to rising.
Second derivative: d²y/dx², the derivative of dy/dx.
Rate of change: How fast one quantity changes with respect to another.
Velocity: Rate of change of displacement, v = ds/dt.
Acceleration: Rate of change of velocity, a = dv/dt.

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