Ch 9 · E2.12 Differentiation · IGCSE Mathematics 0580
Topic 9 of 9 in Algebra and Graphs — IGCSE Mathematics 0580 — 48 lessons.
Introduction
The gradient of a curve changes from point to point. Drawing a tangent gives an estimate; differentiation gives the exact gradient anywhere, and finds the turning points (the tops and bottoms of the curve) without drawing at all. (Syllabus section E2.12, IGCSE 0580 Extended only.)
Key ideas — Gradient by a tangent
Draw a ruled line that just touches the curve at the point, then find its gradient from two points far apart on the line (rise ÷ run). This is an estimate.
🧑🏫 Teacher's explanation
Imagine you are driving a car along a winding mountain road. If you look at your speedometer at one exact moment, you are finding your 'instantaneous velocity'. On a graph, a curve is like that winding road. Because the curve is always changing direction, we can't use a single straight line to describe it. Instead, we use a 'tangent'—a straight line that just grazes the curve at a single point, mimicking the direction of the curve at that exact instant.
📖 Definitions
🔍 Worked example — Easy
Estimate the gradient of the curve y = x^2 at the point where x = 2 by drawing a tangent.
🔍 Worked example — Standard
Find dy/dx for the function y = 3x^2 + 5x and find the gradient at x = 1.
🔍 Worked example — Exam-style
A curve has the equation y = x^3 - 3x. Find the coordinates of the stationary points and determine their nature (maximum or minimum).
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — The rule for differentiating
y = ax^n ⇒ dy/dx = nax^n-1
🧑🏫 Teacher's explanation
Imagine you are driving a car along a winding mountain road. If you look at your speedometer at one specific moment, that speed is your 'instantaneous' rate of change. In mathematics, the derivative, written as dy/dx, is exactly that: it tells you the exact speed (or gradient) of a function at any single point.
📖 Definitions
🔍 Worked example — Easy
Find dy/dx for y = 5x^4.
🔍 Worked example — Standard
Find the gradient of the curve y = 2x^3 - 4x + 5 at the point where x = 2.
🔍 Worked example — Exam-style
A curve has the equation y = x^3 - 3x^2 + 2. Find the coordinates of the stationary points.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find dy/dx. (b) Find the coordinates of the stationary points. [4 marks]
Key ideas — Stationary (turning) points
At a turning point the gradient is zero: solve dy/dx = 0, then substitute the x-values into y to get the coordinates.
🧑🏫 Teacher's explanation
Imagine you are driving a car along a hilly road. As you go up a hill, your elevation increases; as you go down, it decreases. At the very peak of the hill (the maximum) or at the very bottom of a valley (the minimum), there is a split second where you are neither going up nor going down. For that tiny moment, your vertical speed is zero. In calculus, we represent this 'vertical speed' or slope as the derivative dy/dx.
📖 Definitions
🔍 Worked example — Easy
Find the coordinates of the stationary point for the curve y = x^2 - 4x + 5.
🔍 Worked example — Standard
A curve has the equation y = 2x^3 - 3x^2 - 12x + 5. Find the coordinates of its stationary points.
🔍 Worked example — Exam-style
The function y = ax^2 + bx + c has a minimum point at (3, -4). Find the values of a and b if c = 5.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find dy/dx. (b) Find the coordinates of the stationary points. (c) State, with reason, which stationary point is a maximum. [5 marks]
Key ideas — Maximum or minimum?
Any valid method:
🧑🏫 Teacher's explanation
Imagine you are driving a car along a hilly road. A 'turning point' is that exact moment when you stop going uphill and start going downhill (a peak), or stop going downhill and start going uphill (a valley). At that precise split second, your car is neither ascending nor descending; your vertical speed is zero. In calculus, we represent this 'vertical speed' as the gradient, dy/dx.
📖 Definitions
🔍 Worked example — Easy
Find the coordinates of the stationary points for the curve y = x^2 - 4x + 5.
🔍 Worked example — Standard
A curve is defined by y = 2x^3 - 3x^2 - 12x + 5. Determine the nature of its stationary points.
🔍 Worked example — Exam-style
The function y = x^3 - 3x has stationary points. Use the second derivative method to identify which is the maximum.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find the coordinates of the two stationary points. [3] (b) Determine, using the second derivative, which point is a maximum. [2] [5 marks]
Worked examples
y = x^3 - 6x^2 + 9x. dy/dx = 3x^2 - 12x + 9; at x = 2 the gradient is 12 - 24 + 9 = -3.
🔍 Worked example — Easy
Find the gradient of the curve y = x^3 - 6x^2 + 9x at the point where x = 2.
🔍 Worked example — Standard
A curve has the equation y = 5 + 8x - 2x^2. Find the coordinates of the stationary point and determine if it is a maximum or a minimum.
🔍 Worked example — Exam-style
Find the coordinates of the stationary points for the curve y = x^3 - 6x^2 + 9x and classify them as maxima or minima.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 8x^3 - 6x + 1. (2) 2x - 5 = 3. (3) 2x + 6 = 0 ⇒ (-3, -11); minimum (d^2y/dx^2 = 2 > 0). (4) 6x^2 - 6x - 12 = 0 ⇒ x = 2 or -1: (2, -20) minimum and (-1, 7) maximum.