CBC / KICD Curriculum Alignment: Differentiation falls under the Pure Mathematics strand of the Senior Secondary school curriculum as defined by the Kenya Institute of Curriculum Development (KICD). It is examined in the KCSE Mathematics Paper 2 and features across all three Senior Secondary pathways: STEM, Social Sciences, and Arts & Sports Science. This topic also supports core competencies of the Basic Education Curriculum Framework (BECF) — specifically critical thinking, problem solving, and communication of mathematical reasoning.
Introduction

What is Differentiation?

All Levels

Differentiation is the branch of calculus that studies how quantities change. It gives us the exact rate at which one variable changes with respect to another — the mathematical language of motion, growth, and optimisation.

Consider a car travelling along a straight road. Its position changes over time. The speed at any instant is the rate of change of position with respect to time. Differentiation gives us a precise way to calculate that instantaneous rate — not an average over an interval, but the exact value at a single moment. This idea underpins engineering, economics, physics, biology, and every quantitative science taught at Kenyan universities.

In the CBC Senior Secondary curriculum, differentiation is introduced as part of the Pure Mathematics strand. Students are expected to understand the derivative conceptually, apply the standard rules fluently, and use differentiation to solve practical problems involving rates of change, optimisation, and curve sketching.

The Gradient of a Curve

For a straight line, the gradient (slope) is constant: rise ÷ run. For a curve, the gradient is different at every point. The derivative of a function gives the gradient of the tangent line to the curve at any chosen point.

Geometric Interpretation
If y = f(x), then f′(x) = gradient of the tangent to the curve at x

f′(x) > 0 → curve is rising (increasing) at x
f′(x) < 0 → curve is falling (decreasing) at x
f′(x) = 0 → tangent is horizontal → possible stationary point

Notation for the Derivative

Several equivalent notations are used across different contexts and textbooks:

Derivative Notation — All Forms
Leibniz notationdy/dx   (read: "dee y by dee x")
Lagrange / primef′(x)   (read: "f prime of x")
Operator notationDf(x) or D[f(x)]
Newton / dot (physics)ẋ   (used when differentiating with respect to time)
Second derivatived²y/dx²   or   f′′(x)
nth derivativedⁿy/dxⁿ   or   f⁽ⁿ⁾(x)
KCSE Exam Note: In Kenya Certificate of Secondary Education (KCSE) and CBC Senior Secondary assessments, Leibniz notation (dy/dx) is the most commonly used and expected in solutions. Always state your working using this notation unless the question specifies otherwise.
Foundations

Differentiation from First Principles

Secondary / A-Level

First principles shows where the derivative comes from. Before applying any rules, every student should understand this foundational definition — it is the formal mathematical foundation upon which all differentiation rests.

The idea begins with a secant line — a straight line connecting two points on a curve. If those two points are very close together, the secant line approximates the tangent. As the horizontal distance between the points approaches zero, the secant line becomes the tangent, and its gradient becomes the derivative.

Definition — The Derivative from First Principles
f′(x) = limh→0 [f(x + h) − f(x)] / h

Here, h represents the small horizontal change (sometimes written as Δx or δx). As h approaches 0, we are making the interval smaller and smaller until we arrive at the instantaneous rate of change at exactly the point x.

Worked Examples — First Principles
Ex 1 Differentiate f(x) = x² from first principles
Step 1
Write f(x + h):   f(x + h) = (x + h)² = x² + 2xh + h²
Step 2
Form f(x + h) − f(x):   (x² + 2xh + h²) − x² = 2xh + h²
Step 3
Divide by h:   (2xh + h²) / h = 2x + h
Step 4
Take the limit as h → 0:   lim(2x + h) = 2x
∴ f′(x) = 2x
Ex 2 Differentiate f(x) = 3x² − 5x + 2 from first principles
Step 1
f(x + h) = 3(x+h)² − 5(x+h) + 2 = 3x² + 6xh + 3h² − 5x − 5h + 2
Step 2
f(x+h) − f(x) = 6xh + 3h² − 5h
Step 3
Divide by h: 6x + 3h − 5
Step 4
As h → 0: 6x − 5
∴ f′(x) = 6x − 5
Ex 3 Differentiate f(x) = 1/x from first principles
Step 1
f(x+h) = 1/(x+h)
Step 2
f(x+h) − f(x) = 1/(x+h) − 1/x = [x − (x+h)] / [x(x+h)] = −h / [x(x+h)]
Step 3
Divide by h: −1 / [x(x+h)]
Step 4
As h → 0: −1/x²
∴ f′(x) = −1/x²
Exam Requirement: In KCSE and CBC Senior Secondary assessments, when a question says "differentiate from first principles", you must show the limit definition. Simply applying a rule (e.g. power rule) will earn zero marks even if the answer is correct.
Core Rules

Basic Differentiation Rules

Secondary

Once first principles are understood, these four rules allow us to differentiate almost any polynomial function instantly — without repeating the limit process each time.

The Four Basic Rules
Power Ruled/dx(xⁿ) = nxⁿ⁻¹   (for any real n)
Constant Ruled/dx(c) = 0   (the derivative of any constant is zero)
Constant Multipled/dx[c·f(x)] = c·f′(x)
Sum / Differenced/dx[f(x) ± g(x)] = f′(x) ± g′(x)

Applying the Power Rule — Special Cases

The power rule applies not only to positive integer powers but also to negative powers, fractional powers, and roots. The key step is to rewrite the expression in the form xⁿ first.

Rewriting Before Differentiating
Square root√x = x^(1/2)  →  d/dx = (1/2)x^(−1/2) = 1/(2√x)
Cube root∛x = x^(1/3)  →  d/dx = (1/3)x^(−2/3)
Reciprocal1/x = x^(−1)  →  d/dx = −x^(−2) = −1/x²
Reciprocal square1/x² = x^(−2)  →  d/dx = −2x^(−3) = −2/x³
Worked Examples — Basic Rules
Ex 1 Differentiate y = 4x⁵ − 3x³ + 7x − 2
Apply
Differentiate term by term using power rule + constant rule:
Working
d/dx(4x⁵) = 20x⁴  ·  d/dx(−3x³) = −9x²  ·  d/dx(7x) = 7  ·  d/dx(−2) = 0
dy/dx = 20x⁴ − 9x² + 7
Ex 2 Differentiate y = √x + 1/x²
Rewrite
y = x^(1/2) + x^(−2)
Apply
d/dx(x^(1/2)) = (1/2)x^(−1/2)  ·  d/dx(x^(−2)) = −2x^(−3)
Simplify
= 1/(2√x) − 2/x³
dy/dx = 1/(2√x) − 2/x³
Ex 3 Differentiate y = (3x² − 2x + 5) / x   (simplify first)
Simplify
Divide each term by x:   y = 3x − 2 + 5x^(−1)
Apply
d/dx(3x) = 3  ·  d/dx(−2) = 0  ·  d/dx(5x^(−1)) = −5x^(−2)
dy/dx = 3 − 5/x²
Standard Results

Derivatives of Trig, Exponential & Log Functions

Secondary / A-Level

These are standard results that must be memorised. They are proved using first principles or limits of trigonometric functions, but at secondary level you are expected to know and apply them directly.

Function f(x)
Derivative f′(x)
sin x
cos x
cos x
−sin x
tan x
sec²x
sec x
sec x · tan x
cosec x
−cosec x · cot x
cot x
−cosec²x
eˣ
eˣ   (the exponential function is its own derivative)
aˣ (a > 0, a ≠ 1)
aˣ · ln a
ln x (x > 0)
1/x
log_a x
1 / (x · ln a)
Worked Examples — Standard Derivatives
Ex 1 Find d/dx [e^(3x)]
Rule
This is a composite function: outer = eˣ, inner = 3x. Use chain rule.
Apply
d/dx[e^(3x)] = e^(3x) × d/dx(3x) = e^(3x) × 3
= 3e^(3x)
Ex 2 Find d/dx [ln(x² + 1)]
Rule
d/dx[ln(u)] = (1/u) × u′ by chain rule. Here u = x² + 1.
Apply
= (1/(x²+1)) × 2x
= 2x/(x² + 1)
Ex 3 Find d/dx [sin(3x) + cos(x²)]
Term 1
d/dx[sin(3x)] = cos(3x) × 3 = 3cos(3x)
Term 2
d/dx[cos(x²)] = −sin(x²) × 2x = −2x·sin(x²)
= 3cos(3x) − 2x·sin(x²)
Ex 4 Differentiate y = 5 · 2ˣ
Rule
d/dx(aˣ) = aˣ · ln a. Here a = 2.
Apply
dy/dx = 5 × 2ˣ × ln 2
dy/dx = 5 · 2ˣ · ln 2
Composite Rules

Product, Quotient & Chain Rules

Secondary / A-Level

These three rules extend differentiation to combinations of functions — products, quotients, and compositions. They are among the most frequently examined rules in both KCSE and CBC Senior Secondary mathematics.

The Three Composite Rules
Product Ruled/dx(uv) = u′v + uv′    [where u and v are functions of x]
Quotient Ruled/dx(u/v) = (u′v − uv′) / v²    [v ≠ 0]
Chain Ruled/dx[f(g(x))] = f′(g(x)) · g′(x)    (function of a function)
Chain Rule (Leibniz)dy/dx = (dy/du) × (du/dx)

How to Choose Which Rule to Use

🔵 Product Rule
Use when differentiating two functions multiplied together, e.g. y = x² · sin x or y = e^x · ln x. Neither can be simplified away.
🟠 Quotient Rule
Use when differentiating one function divided by another, e.g. y = sin x / x². Can also be done via product rule with a negative power.
🟢 Chain Rule
Use when differentiating a function inside another function, e.g. y = (3x+1)⁵ or y = sin(x²). The inner function is the key.
Worked Examples — Product Rule
Ex 1 Differentiate y = x² · sin x
Identify
u = x² → u′ = 2x    v = sin x → v′ = cos x
Apply
dy/dx = u′v + uv′ = (2x)(sin x) + (x²)(cos x)
dy/dx = 2x sin x + x² cos x
Ex 2 Differentiate y = e^x · ln x
Identify
u = eˣ → u′ = eˣ    v = ln x → v′ = 1/x
Apply
dy/dx = eˣ · ln x + eˣ · (1/x)
Factor
= eˣ(ln x + 1/x)
dy/dx = eˣ(ln x + 1/x)
Worked Examples — Quotient Rule
Ex 3 Differentiate y = (x² + 1)/(x − 2)
Identify
u = x²+1 → u′ = 2x    v = x−2 → v′ = 1
Apply
dy/dx = [2x(x−2) − (x²+1)(1)] / (x−2)²
Expand
= [2x² − 4x − x² − 1] / (x−2)² = (x² − 4x − 1)/(x−2)²
dy/dx = (x² − 4x − 1) / (x − 2)²
Ex 4 Differentiate y = sin x / x²
Identify
u = sin x → u′ = cos x    v = x² → v′ = 2x
Apply
dy/dx = [x² cos x − sin x · 2x] / x⁴
Factor
= x[x cos x − 2 sin x] / x⁴ = (x cos x − 2 sin x) / x³
dy/dx = (x cos x − 2 sin x) / x³
Worked Examples — Chain Rule
Ex 5 Differentiate y = (3x² + 5)⁷
Identify
Outer: u⁷, Inner: u = 3x²+5 → u′ = 6x
Apply
dy/dx = 7(3x²+5)⁶ × 6x
dy/dx = 42x(3x² + 5)⁶
Ex 6 Differentiate y = √(4x³ − x)
Rewrite
y = (4x³ − x)^(1/2). Inner: u = 4x³−x → u′ = 12x²−1
Apply
dy/dx = (1/2)(4x³−x)^(−1/2) × (12x²−1)
dy/dx = (12x² − 1) / [2√(4x³ − x)]
Applications

Stationary Points, Maxima & Minima

Secondary / A-Level

A stationary point is any point where the gradient equals zero — that is, where the tangent line is horizontal. Identifying and classifying stationary points is one of the most important applications of differentiation in KCSE and CBC Senior Secondary assessments.

Procedure: Finding & Classifying Stationary Points
Step 1Find f′(x) and set f′(x) = 0. Solve for x.
Step 2Find y-coordinates: substitute each x back into f(x).
Step 3Find f′′(x) — the second derivative.
Classify: f′′ < 0Local maximum (curve is concave down at that point)
Classify: f′′ > 0Local minimum (curve is concave up at that point)
Classify: f′′ = 0Inconclusive — check sign of f′(x) on either side (first derivative test)
Worked Examples — Stationary Points
Ex 1 Find and classify all stationary points of f(x) = x³ − 3x + 2
Step 1
f′(x) = 3x² − 3. Set f′(x) = 0:   3x² − 3 = 0  →  x² = 1  →  x = 1 or x = −1
Step 2
f(1) = 1 − 3 + 2 = 0  ·  f(−1) = −1 + 3 + 2 = 4
Step 3
f′′(x) = 6x
Classify
At x = 1: f′′(1) = 6 > 0 → local minimum (1, 0)
At x = −1: f′′(−1) = −6 < 0 → local maximum (−1, 4)
✓ Check
The function is a cubic with positive leading coefficient, so it should have a maximum to the left and a minimum to the right. ✔
Local maximum at (−1, 4); Local minimum at (1, 0)
Ex 2 Find the turning points of y = 2x³ + 3x² − 12x + 1 and determine their nature
Step 1
dy/dx = 6x² + 6x − 12 = 0  →  6(x² + x − 2) = 0  →  6(x+2)(x−1) = 0
x values
x = −2 or x = 1
y values
y(−2) = 2(−8)+3(4)−12(−2)+1 = −16+12+24+1 = 21
y(1) = 2+3−12+1 = −6
d²y/dx²
= 12x + 6
Classify
At x = −2: 12(−2)+6 = −18 < 0 → maximum
At x = 1: 12(1)+6 = 18 > 0 → minimum
Maximum at (−2, 21); Minimum at (1, −6)
Point of Inflection: A point of inflection occurs where the curve changes concavity (concave up to concave down or vice versa). It is found where f′′(x) = 0 AND the sign of f′′(x) changes. If f′′(x) = 0 but does not change sign, it is NOT a point of inflection.
Applications

Rates of Change & Optimisation

Secondary / A-Level

Rates of change and optimisation are the most practically valuable applications of differentiation. They appear in every field from engineering and physics to economics and biology — and in every KCSE Paper 2 examination.

Connected Rates of Change

When two quantities are related through an equation, their rates of change are connected through the chain rule. If y depends on x which depends on t (time), then:

Connected Rates of Change
dy/dt = (dy/dx) × (dx/dt)
Worked Examples — Rates of Change & Optimisation
Ex 1 The radius of a circle is increasing at 3 cm/s. Find the rate at which the area increases when the radius is 5 cm.
Given
dr/dt = 3 cm/s. Find dA/dt when r = 5.
Formula
A = πr²  →  dA/dr = 2πr
Chain rule
dA/dt = (dA/dr) × (dr/dt) = 2πr × 3 = 6πr
Substitute
When r = 5:   dA/dt = 6π × 5 = 30π cm²/s
Rate of increase of area = 30π ≈ 94.25 cm²/s
Ex 2 — Optimisation A farmer has 120 m of fencing. He wants to enclose a rectangular field against a straight wall (no fence needed on the wall side). Find the maximum area he can enclose.
Let
Length parallel to wall = L. Width = w. Only three sides need fencing: L + 2w = 120 → L = 120 − 2w
Area
A = L × w = (120 − 2w)w = 120w − 2w²
Maximise
dA/dw = 120 − 4w = 0  →  w = 30 m
Check
d²A/dw² = −4 < 0  →  maximum confirmed. L = 120 − 60 = 60 m.
Maximum area = 60 × 30 = 1800 m²
Ex 3 — Optimisation A box with a square base and open top has volume 256 cm³. Find the dimensions that minimise the surface area.
Let
Base side = x, height = h. Volume: V = x²h = 256 → h = 256/x²
SA
SA = x² + 4xh = x² + 4x(256/x²) = x² + 1024/x
Minimise
d(SA)/dx = 2x − 1024/x² = 0  →  2x³ = 1024  →  x³ = 512  →  x = 8 cm
Height
h = 256/64 = 4 cm
✓
d²(SA)/dx² = 2 + 2048/x³. At x=8: = 2 + 4 = 6 > 0 → minimum ✔
Minimum surface area when x = 8 cm, h = 4 cm. SA = 64 + 128 = 192 cm²
Applications

Equations of Tangents & Normals

Secondary / A-Level

The derivative gives the gradient of the tangent line to a curve at any point. The normal is perpendicular to the tangent at that same point. Both are regularly examined in KCSE Paper 2.

Key Relationships
Gradient of tangentm_T = f′(x₁)   evaluated at the given point x₁
Gradient of normalm_N = −1/m_T   (perpendicular → product of gradients = −1)
Equation of tangenty − y₁ = m_T (x − x₁)
Equation of normaly − y₁ = m_N (x − x₁)
Worked Examples — Tangents & Normals
Ex 1 Find the equation of the tangent to y = x³ − 2x at the point (2, 4)
Step 1
dy/dx = 3x² − 2. At x = 2: m = 3(4) − 2 = 10
Step 2
Tangent: y − 4 = 10(x − 2)  →  y = 10x − 20 + 4
Tangent: y = 10x − 16
Ex 2 Find the equation of the normal to y = x² + 3x at x = 1
y-coord
y(1) = 1 + 3 = 4. Point: (1, 4)
Tangent m
dy/dx = 2x+3. At x=1: m_T = 5
Normal m
m_N = −1/5
Equation
y − 4 = −(1/5)(x − 1)  →  5y − 20 = −x + 1  →  x + 5y = 21
Normal: x + 5y = 21
Advanced

Implicit & Parametric Differentiation

A-Level / University

Not all functions are written as y = f(x). Implicit and parametric forms require specialised differentiation techniques that extend beyond the standard rules.

Implicit Differentiation

An implicit function is one where y cannot easily be isolated — e.g. x² + y² = 25. To differentiate, apply d/dx to both sides. When you differentiate a term involving y, multiply by dy/dx using the chain rule (since y is a function of x).

Key Principle
d/dx[f(y)] = f′(y) · dy/dx    (chain rule applied to y)
Worked Examples — Implicit Differentiation
Ex 1 Find dy/dx for x² + y² = 25
Differentiate
d/dx(x²) + d/dx(y²) = d/dx(25)  →  2x + 2y(dy/dx) = 0
Solve
2y(dy/dx) = −2x  →  dy/dx = −x/y
dy/dx = −x/y
Ex 2 Find dy/dx for x³ + y³ = 6xy
Differentiate
3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)
Collect
3y²(dy/dx) − 6x(dy/dx) = 6y − 3x²
Factor
(dy/dx)(3y² − 6x) = 6y − 3x²
Solve
dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)
dy/dx = (2y − x²)/(y² − 2x)

Parametric Differentiation

In parametric equations, both x and y are expressed as functions of a third variable (the parameter), usually t. To find dy/dx, use the chain rule in the form:

Parametric Derivatives
First derivativedy/dx = (dy/dt) ÷ (dx/dt)
Second derivatived²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)
Worked Examples — Parametric Differentiation
Ex 3 Given x = t², y = t³, find dy/dx and d²y/dx² in terms of t
Derivatives
dx/dt = 2t    dy/dt = 3t²
dy/dx
dy/dx = 3t²/(2t) = 3t/2
d²y/dx²
d/dt(dy/dx) = d/dt(3t/2) = 3/2.   d²y/dx² = (3/2) ÷ (2t) = 3/(4t)
dy/dx = 3t/2    d²y/dx² = 3/(4t)
Ex 4 The parametric equations of a curve are x = 2cos t, y = 3sin t. Find the gradient at t = π/4.
Derivatives
dx/dt = −2sin t    dy/dt = 3cos t
dy/dx
dy/dx = 3cos t / (−2sin t) = −(3/2)cot t
At t = π/4
cot(π/4) = 1  →  dy/dx = −3/2
Gradient at t = π/4 is −3/2
University Level

Partial Derivatives & Multivariable Calculus

University

When a function depends on more than one variable — e.g. temperature depends on both position and time — we need partial derivatives to study how the function changes with respect to each variable independently while holding the others fixed.

Partial Derivative Rules
∂f/∂xDifferentiate with respect to x, treating all other variables as constants
∂f/∂yDifferentiate with respect to y, treating all other variables as constants
Gradient vector∇f = (∂f/∂x, ∂f/∂y)   — points in the direction of steepest increase
Critical pointsSet ∂f/∂x = 0 AND ∂f/∂y = 0 simultaneously
Worked Examples — Partial Derivatives
Ex 1 Find all partial derivatives of f(x, y) = x²y + 3xy² − 5y
∂f/∂x
Treat y as constant:   ∂f/∂x = 2xy + 3y²
∂f/∂y
Treat x as constant:   ∂f/∂y = x² + 6xy − 5
∂²f/∂x²
Differentiate ∂f/∂x again w.r.t. x:   = 2y
∂²f/∂y²
Differentiate ∂f/∂y again w.r.t. y:   = 6x
Mixed ∂²f/∂x∂y
Differentiate ∂f/∂x w.r.t. y:   = 2x + 6y
∂f/∂x = 2xy + 3y²  ·  ∂f/∂y = x² + 6xy − 5
Ex 2 Find the critical points of f(x, y) = x³ − 3x + y² − 4y
∂f/∂x = 0
3x² − 3 = 0  →  x = ±1
∂f/∂y = 0
2y − 4 = 0  →  y = 2
Critical points at (1, 2) and (−1, 2)

The Second Derivative Test for Two Variables

Discriminant D = f_xx · f_yy − (f_xy)²
D > 0, f_xx > 0Local minimum
D > 0, f_xx < 0Local maximum
D < 0Saddle point (neither max nor min)
D = 0Inconclusive — further analysis needed
University Level

Taylor & Maclaurin Series

University

Taylor and Maclaurin series express functions as infinite polynomials using their derivatives. This powerful technique allows computers to compute transcendental functions like sin, cos, and eˣ numerically, and underpins much of numerical analysis and physics.

Taylor & Maclaurin Series
Taylor Series (about x = a)f(x) = f(a) + f′(a)(x−a) + f′′(a)(x−a)²/2! + f′′′(a)(x−a)³/3! + ···
Maclaurin Series (a = 0)f(x) = f(0) + f′(0)x + f′′(0)x²/2! + f′′′(0)x³/3! + ···

Common Maclaurin Series — Must Know

eˣ (all x)
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ···
= Σ xⁿ/n! (n=0 to ∞)
sin x (all x)
sin x = x − x³/3! + x⁵/5! − x⁷/7! + ···
= Σ (−1)ⁿ x^(2n+1)/(2n+1)!
cos x (all x)
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + ···
= Σ (−1)ⁿ x^(2n)/(2n)!
ln(1+x) (|x| ≤ 1)
ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + ···
Valid for −1 < x ≤ 1
Worked Example — Maclaurin Series
Ex 1 Find the Maclaurin series for f(x) = eˣ up to the x³ term
Evaluate
f(0) = e⁰ = 1. f′(x) = eˣ → f′(0) = 1. f′′(x) = eˣ → f′′(0) = 1. f′′′(x) = eˣ → f′′′(0) = 1.
Series
eˣ ≈ 1 + x + x²/2! + x³/3! = 1 + x + x²/2 + x³/6
eˣ ≈ 1 + x + x²/2 + x³/6 + ···
Reference

Complete Differentiation Formula Sheet

All Levels

Use this as your quick-reference summary for all rules and standard results covered in this lesson.

First Principles
f′(x) = lim[h→0] [f(x+h)−f(x)]/h
Power & Basic Rules
d/dx(xⁿ) = nxⁿ⁻¹
d/dx(c) = 0
d/dx[cf(x)] = c·f′(x)
d/dx[f ± g] = f′ ± g′
Composite Rules
Product: (uv)′ = u′v + uv′
Quotient: (u/v)′ = (u′v−uv′)/v²
Chain: d/dx[f(g)] = f′(g)·g′
Trig Derivatives
d/dx(sin x) = cos x
d/dx(cos x) = −sin x
d/dx(tan x) = sec²x
d/dx(sec x) = sec x tan x
Exp & Log
d/dx(eˣ) = eˣ
d/dx(aˣ) = aˣ ln a
d/dx(ln x) = 1/x
d/dx(log_a x) = 1/(x ln a)
Parametric
dy/dx = (dy/dt)/(dx/dt)
d²y/dx² = [d(dy/dx)/dt]/(dx/dt)
Tangents & Normals
Tangent: y−y₁ = f′(x₁)(x−x₁)
Normal: m_N = −1/f′(x₁)
Stationary Points
Set f′(x) = 0 to find
f′′ > 0 → minimum
f′′ < 0 → maximum
f′′ = 0 → use 1st deriv. test
Exam Preparation

Common Errors & How to Avoid Them

All Levels

These are the errors most frequently seen by examiners in KCSE and CBC Senior Secondary marking scripts. Recognising and avoiding these errors can mean the difference between a B and an A.

✗ Wrong: d/dx(x⁵) = x⁴
✓ Right: d/dx(x⁵) = 5x⁴
Forgot to bring down the power. Power rule: multiply by n, then reduce power by 1.
✗ Wrong: d/dx(sin x²) = cos x²
✓ Right: d/dx(sin(x²)) = 2x·cos(x²)
Chain rule was not applied. The inner function x² gives a factor of 2x.
✗ Wrong: d/dx(x² · sin x) = 2x · cos x
✓ Right: = 2x·sin x + x²·cos x
Product rule forgotten. Both terms must appear: u′v AND uv′.
✗ Wrong: d/dx(e^(x²)) = e^(x²)
✓ Right: = 2x·e^(x²)
Chain rule: the inner function x² gives a factor of 2x. eˣ is its own derivative only when the exponent is exactly x.
✗ Wrong: d/dx(7) = 1
✓ Right: d/dx(7) = 0
The derivative of any constant is zero, not 1. Confusing with integration.
✗ Wrong: f′′ = 0 means inflection point
✓ Right: Also check sign change of f′′
f′′(x₀) = 0 alone does not confirm inflection. The sign of f′′ must change at that point.