Full lesson notes
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\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
Differential calculus is the mathematics of change. It answers one question: how fast is this changing at this exact instant?
In this lesson you will learn to:
- find an average gradient between two points;
- differentiate from first principles;
- use the rules of differentiation confidently;
- find the equation of a tangent to a curve;
- sketch a cubic graph completely;
- find the equation of a cubic from its graph;
- solve optimisation (maximum and minimum) problems; and
- work with rates of change, including motion.
Differential calculus carries roughly 35 marks in Paper 1 — more than any other single topic.
1. Limits & Average Gradient
\[ m_{\text{av}} = \frac{y_2-y_1}{x_2-x_1} = \frac{f(b)-f(a)}{b-a} \]
The average gradient is the gradient of the straight line (a secant) joining two points on the curve.
For \(f(x)=x^{2}\) between \(x=1\) and \(x=3\):
\(m_{\text{av}} = \dfrac{f(3)-f(1)}{3-1} = \dfrac{9-1}{2} = 4\)
This is the
average rate of change over that interval — not the gradient at any one point.
\[ \lim_{h \to 0} \]
Now bring the second point closer and closer to the first, a distance \(h\) away:
| \(h\) | Interval | Average gradient of \(f(x)=x^{2}\) at \(x=1\) |
|---|
| 1 | \(1 \to 2\) | 3 |
| 0,5 | \(1 \to 1{,}5\) | 2,5 |
| 0,1 | \(1 \to 1{,}1\) | 2,1 |
| 0,01 | \(1 \to 1{,}01\) | 2,01 |
The values close in on \(2\). We write \(\lim\limits_{h \to 0} = 2\). That limit is the gradient of the tangent at \(x=1\).
\[ \text{secant} \;\longrightarrow\; \text{tangent} \]
| Average gradient | Instantaneous gradient |
|---|
| Line | Secant through two points | Tangent touching at one point |
| Over | An interval | A single instant |
| Formula | \(\dfrac{f(b)-f(a)}{b-a}\) | \(f'(x)\) |
\[ f'(x) \text{ is a limit of average gradients} \]
Key idea: the derivative is not a new invention — it is the gradient formula you already know, pushed to the instant where the two points meet.
2. First Principles
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
This is the formal definition. If a question says “from first principles” you MUST use it — using the rules earns zero.
Write the limit notation on every line until you actually take the limit. Marks are awarded for it.
\[ f(x)=x^{2} \;\Rightarrow\; f'(x)=2x \]
\(f(x+h) = (x+h)^{2} = x^{2}+2xh+h^{2}\)
\(f(x+h)-f(x) = 2xh+h^{2}\)
\(\dfrac{2xh+h^{2}}{h} = \dfrac{h(2x+h)}{h} = 2x+h\)
\(f'(x) = \lim\limits_{h \to 0}(2x+h) = 2x\)
Note the key step: factorise out \(h\) and cancel. You may only substitute \(h=0\) after the \(h\) in the denominator has gone.
\[ f(x)=2x^{2}-3x \;\Rightarrow\; f'(x)=4x-3 \]
\(f(x+h) = 2(x+h)^{2}-3(x+h) = 2x^{2}+4xh+2h^{2}-3x-3h\)
\(f(x+h)-f(x) = 4xh+2h^{2}-3h\)
\(\dfrac{h(4x+2h-3)}{h} = 4x+2h-3\)
\(f'(x) = \lim\limits_{h \to 0}(4x+2h-3) = 4x-3\)
\[ \text{expand} \rightarrow \text{subtract} \rightarrow \text{factorise } h \rightarrow \text{cancel} \rightarrow \text{limit} \]
Key idea: five steps, every time. The only difficulty is the algebra — the structure never changes.
3. Rules of Differentiation
\[ \frac{d}{dx}\left[ax^{n}\right] = nax^{\,n-1} \]
| Function | Derivative | Note |
|---|
| \(x^{5}\) | \(5x^{4}\) | Multiply by the power, subtract 1 |
| \(3x^{4}\) | \(12x^{3}\) | The coefficient comes along |
| \(7x\) | \(7\) | \(x^{1} \to 1 \cdot x^{0}=1\) |
| \(12\) | \(0\) | A constant has no gradient |
| \(\dfrac{1}{x}=x^{-1}\) | \(-x^{-2}=-\dfrac{1}{x^{2}}\) | Rewrite with a negative exponent first |
| \(\sqrt{x}=x^{\frac{1}{2}}\) | \(\dfrac{1}{2}x^{-\frac{1}{2}}\) | Rewrite surds as powers first |
\[ f'(x) \qquad \frac{dy}{dx} \qquad D_x \]
All three mean exactly the same thing:
\(f'(x)\) — used when the function is named \(f\)
\(\dfrac{dy}{dx}\) — used when the equation is written \(y = \ldots\)
\(D_x\left[\ldots\right]\) — the instruction “differentiate with respect to \(x\)”
Answer in the notation the question uses.
\[ \text{no fractions, no surds, no brackets} \]
You may only apply the power rule to terms of the form \(ax^{n}\). Rewrite first:
\(\dfrac{x^{3}-2x}{x} = x^{2}-2 \quad\Rightarrow\quad \dfrac{dy}{dx}=2x\)
\((x+2)(x-3) = x^{2}-x-6 \quad\Rightarrow\quad \dfrac{dy}{dx}=2x-1\)
\(\dfrac{3}{x^{2}} = 3x^{-2} \quad\Rightarrow\quad \dfrac{dy}{dx}=-6x^{-3}\)
There is no product or quotient rule in the CAPS curriculum — simplify to separate terms first.
\[ \text{simplify first, then differentiate term by term} \]
Key idea: most lost marks here are algebra errors made before the differentiation even starts.
4. Tangents
\[ m_{\text{tangent}} = f'(a) \]
To find the equation of a tangent you need a point and a gradient.
- Differentiate to get \(f'(x)\).
- Substitute the \(x\)-value to get the gradient \(m=f'(a)\).
- Find the \(y\)-value on the CURVE (not the derivative).
- Use \(y-y_1=m(x-x_1)\).
\[ y = 4x - 8 \]
Find the tangent to \(y=x^{3}-2x^{2}\) at \(x=2\).
\(y = (2)^{3}-2(2)^{2} = 8-8 = 0 \quad\Rightarrow\quad \text{point } (2;\ 0)\)
\(\dfrac{dy}{dx} = 3x^{2}-4x\)
\(m = 3(2)^{2}-4(2) = 12-8 = 4\)
\(y-0 = 4(x-2) \quad\Rightarrow\quad y = 4x-8\)
\[ f'(a) \ne f(a) \]
- Substituting into \(f'(x)\) to find the \(y\)-value. The point lies on the CURVE, so use \(f(x)\).
- A horizontal tangent means \(f'(x)=0\) — that is how turning points are found.
- For a tangent PARALLEL to a given line, set \(f'(x)\) equal to that line's gradient.
- For a tangent PERPENDICULAR to a line of gradient \(m\), set \(f'(x) = -\dfrac{1}{m}\).
\[ y - y_1 = f'(a)(x - x_1) \]
Key idea: the derivative supplies the gradient; the original function supplies the point.
5. Cubic Graphs
1 2 3 4 2 4 6 x y
To sketch \(f(x)=x^{3}-6x^{2}+9x\) you need four things:
- the \(y\)-intercept (set \(x=0\));
- the \(x\)-intercepts (factorise and set \(f(x)=0\));
- the turning points (solve \(f'(x)=0\));
- the point of inflection (solve \(f''(x)=0\)).
\[ f(x) = x^{3}-6x^{2}+9x = x(x-3)^{2} \]
\(y\)-intercept: \(f(0)=0 \quad\Rightarrow\quad (0;\ 0)\)
Factorise: \(x(x^{2}-6x+9) = x(x-3)^{2}\)
\(x\)-intercepts: \(x=0\) and \(x=3\)
\((x-3)^{2}\) is a repeated factor, so the curve TOUCHES the \(x\)-axis at \(x=3\) and turns — it does not cross. That point is therefore also a turning point.
(1; 4) max (3; 0) min 1 2 3 4 2 4 6 x y
\(f'(x) = 3x^{2}-12x+9 = 3(x-1)(x-3)\)
\(f'(x)=0 \quad\Rightarrow\quad x=1 \text{ or } x=3\)
\(f(1) = 1-6+9 = 4 \quad\Rightarrow\quad (1;\ 4)\)
\(f(3) = 27-54+27 = 0 \quad\Rightarrow\quad (3;\ 0)\)
\[ f''(x) 0 \Rightarrow \text{minimum} \]
\(f''(x) = 6x-12\)
At \(x=1\): \(f''(1) = -6 < 0\) → local maximum at \((1;\ 4)\)
At \(x=3\): \(f''(3) = 6 > 0\) → local minimum at \((3;\ 0)\)
A sign table for \(f'(x)\) works just as well. Use whichever the question asks for.
(1; 4) max (3; 0) min (2; 2) 1 2 3 4 2 4 6 x y
\(f''(x) = 6x-12 = 0 \quad\Rightarrow\quad x=2\)
\(f(2) = 8-24+18 = 2 \quad\Rightarrow\quad (2;\ 2)\)
At the point of inflection the curve changes
concavity — from concave down to concave up here. It sits exactly halfway between the two turning points.
\[ f''(x) 0 \;\text{concave up} \]
| Interval | \(f''(x)\) | Shape |
|---|
| \(x<2\) | Negative | Concave down (like a frown) |
| \(x=2\) | Zero | Point of inflection |
| \(x>2\) | Positive | Concave up (like a smile) |
\[ \text{intercepts} + \text{turning points} + \text{inflection} \]
Key idea: label every one of those points on your sketch with its coordinates. The marks are for the labelled points, not the artistry.
6. Finding the Equation
\[ y = a(x-x_1)(x-x_2)(x-x_3) \]
If the graph shows three \(x\)-intercepts, write the factorised form and use one more point to find \(a\).
Intercepts at \(-1;\ 2;\ 3\) and the curve passes through \((0;\ 12)\):
\(y = a(x+1)(x-2)(x-3)\)
\(12 = a(1)(-2)(-3) = 6a \quad\Rightarrow\quad a = 2\)
\(y = 2(x+1)(x-2)(x-3)\)
\[ y = a(x-p)^{2}(x-q) \]
If the curve TOUCHES the \(x\)-axis at \(x=p\), that factor is squared.
Touches at \(x=3\), cuts at \(x=0\), passes through \((1;\ 4)\):
\(y = ax(x-3)^{2}\)
\(4 = a(1)(-2)^{2} = 4a \quad\Rightarrow\quad a = 1\)
\(y = x(x-3)^{2} = x^{3}-6x^{2}+9x\)
That is exactly the cubic we sketched — a useful check that the two methods agree.
\[ f'(x) = 0 \text{ at each turning point} \]
Given \(f(x)=ax^{3}+bx^{2}+cx+d\) and information about turning points, build simultaneous equations:
- Each turning point \((p;q)\) gives TWO equations: \(f(p)=q\) and \(f'(p)=0\).
- Each point on the curve gives one equation.
- Solve for \(a,\ b,\ c,\ d\).
\[ \text{count the unknowns, then count the equations} \]
Key idea: you need as many independent facts as unknowns. A repeated root is worth remembering — it is the most commonly missed clue.
7. Optimisation
\[ \text{maximum or minimum} \;\Rightarrow\; \frac{dA}{dx}=0 \]
- Write an expression for the quantity to be optimised.
- Use the given constraint to reduce it to ONE variable.
- Differentiate and set the derivative equal to zero.
- Solve, then confirm it is a maximum or minimum.
- Answer the question that was actually asked.
\[ A(x) = 60x - 2x^{2} \]
A farmer has 60 m of fencing for a rectangular enclosure against an existing wall (so only three sides need fencing). Find the maximum area.
Let the two equal sides be \(x\). The remaining side \(= 60-2x\).
\(A(x) = x(60-2x) = 60x-2x^{2}\)
\(A'(x) = 60-4x = 0 \quad\Rightarrow\quad x = 15\)
\(A''(x) = -4 < 0\) → maximum confirmed
\[ A_{\max} = 450 \text{ m}^2 \]
Width \(= 15\) m, length \(= 60-2(15) = 30\) m
\(A_{\max} = 15 \times 30 = 450\text{ m}^{2}\)
Read the question carefully: if it asks for the DIMENSIONS, give 15 m by 30 m. If it asks for the AREA, give 450 m². Answering the wrong one costs the final mark.
\[ \text{one variable} \rightarrow \text{differentiate} \rightarrow \text{set to } 0 \]
Key idea: the calculus is easy. The difficulty is setting up the expression and using the constraint to eliminate the second variable.
8. Rates of Change
\[ \frac{dy}{dx} = \text{the rate at which } y \text{ changes per unit } x \]
Any derivative is a rate of change.
\(\dfrac{dV}{dt}\) — how fast volume changes with time
\(\dfrac{dC}{dx}\) — how fast cost changes per extra item
“Rate of change” in a question is a direct instruction to differentiate.
\[ s(t) \;\xrightarrow{\;\frac{d}{dt}\;}\; v(t) \;\xrightarrow{\;\frac{d}{dt}\;}\; a(t) \]
| Quantity | Symbol | How to get it |
|---|
| Displacement | \(s(t)\) | Given |
| Velocity | \(v(t)=s'(t)\) | Differentiate displacement |
| Acceleration | \(a(t)=v'(t)=s''(t)\) | Differentiate velocity |
Velocity has direction: a negative velocity means moving backwards. Speed is its magnitude.
\[ s(t)=t^{3}-6t^{2}+9t \]
A particle's displacement (metres) after \(t\) seconds is \(s(t)=t^{3}-6t^{2}+9t\).
\(v(t) = 3t^{2}-12t+9 = 3(t-1)(t-3)\)
\(a(t) = 6t-12\)
The particle is at rest when \(v=0\): \(t=1\) s and \(t=3\) s
Acceleration is zero when \(6t-12=0\): \(t=2\) s
\(t\) \(s(t)\) \(v(t)\) \(a(t)\) What is happening 0 0 9 \(-12\) Moving forward, slowing 1 4 0 \(-6\) At rest — furthest forward so far 2 2 \(-3\) 0 Moving backward at constant velocity 3 0 0 6 At rest again, back at the start 4 4 9 12 Moving forward, speeding up
Notice this is the same cubic we sketched earlier — the turning points of \(s(t)\) are exactly the moments when the particle is at rest, and the point of inflection is where acceleration is zero.
\[ v=s' \qquad a=v'=s'' \]
Key idea: “at rest” means \(v=0\), not \(s=0\). “Returns to the start” means \(s=0\).
\[ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \qquad \frac{d}{dx}\left[ax^{n}\right]=nax^{\,n-1} \]
You can now:
- find average gradients and understand the limit that defines the derivative;
- differentiate from first principles using the five-step structure;
- apply the power rule after simplifying fractions, surds and brackets;
- find the equation of a tangent at a point;
- sketch a cubic with intercepts, turning points and the point of inflection;
- find the equation of a cubic from its graph, including repeated roots;
- solve optimisation problems by reducing to one variable; and
- handle rates of change and motion with \(v=s'\) and \(a=s''\).
Practise first principles until it is automatic — it appears almost every year and is the easiest full-mark question in the topic.