A secant line cuts through a curve at two points. A tangent line just kisses it at one. By letting those two points slide closer and closer together — using a limit — we can turn the gradient of a secant into the gradient of a tangent. That limit has a special name: the derivative.
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Imagine driving along a curved road. Your average speed between two points is easy to calculate. But what if you want to know your exact speed at one precise moment — say, the instant you pass a speed camera? How might you use the idea of average speed over smaller and smaller time intervals to find your instantaneous speed?
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Wrong: Integration and differentiation produce identical results.
Right: Integration is the reverse of differentiation, but includes an arbitrary constant (+C) for indefinite integrals.
📚 Core Content
Consider a function $y = f(x)$. The gradient of the straight line joining two points on the curve — a secant — is:
$$\frac{\text{rise}}{\text{run}} = \frac{f(x+h) - f(x)}{h}$$
If we let $h$ get smaller and smaller, the second point slides along the curve toward the first. The secant line becomes the tangent line. The gradient of this tangent is given by the limit:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
This expression is called the derivative of $f$ with respect to $x$. It tells us the instantaneous rate of change of $f$ at any point $x$.
Finding the derivative using the limit definition is called differentiation from first principles. The process always follows the same steps:
For polynomials, this usually involves expanding brackets and collecting like terms before cancelling $h$.
Once you master first principles, you will discover a shortcut: for $f(x) = x^n$, the derivative is $f'(x) = nx^{n-1}$. But understanding why this works — through limits — is what separates rote learning from real comprehension.
🧮 Worked Examples
🧪 Activities
1 $f(x) = x^2 + 3x$
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2 $f(x) = 2x - 5$
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3 $f(x) = x^3$
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4 Find the gradient of the tangent to $y = x^2 + 3x$ at $x = 1$.
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1 If $s(t)$ represents the displacement of a particle at time $t$, what does $s'(4) = 6$ mean physically?
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2 A student says: "Because a secant uses two points and a tangent uses one point, the tangent gradient must always be half the secant gradient." Explain why this is incorrect.
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Earlier you thought about finding your exact speed at the moment you pass a speed camera.
Just as average speed over smaller and smaller time intervals approaches your instantaneous speed, the gradient of a secant over smaller and smaller $h$ approaches the gradient of the tangent. That limit is the derivative — and it is exactly how physicists define instantaneous velocity from a displacement function.
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✍️ Short Answer
6. (a) Find $f'(x)$ for $f(x) = x^2 + 3x$ from first principles. 2 MARKS
(b) Hence find the gradient of the tangent to $y = x^2 + 3x$ at $x = 2$. 1 MARK
(c) A line is drawn perpendicular to the tangent at $x = 2$. What is the gradient of this perpendicular line? 1 MARK
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7. Explain why we need the limit $h \to 0$ in the definition of the derivative, and why we cannot simply set $h = 0$ in the difference quotient $\frac{f(x+h) - f(x)}{h}$. 3 MARKS
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8. The height of a rocket above the ground is given by $h(t) = 5t^2$, where $t$ is in seconds and $h$ is in metres. 3 MARKS
(a) Find $h'(t)$ from first principles.
(b) Find the instantaneous velocity of the rocket at $t = 3$ seconds, and explain what this value means in context.
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1. $f'(x) = 2x + 3$
2. $f'(x) = 2$
3. $f'(x) = 3x^2$ (using $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$)
4. $f'(1) = 2(1) + 3 = 5$
1. At $t = 4$, the particle's instantaneous velocity is $6$ units per second. It tells us the speed and direction of the particle at exactly that moment.
2. The tangent gradient is not half the secant gradient. As the two points get closer, the secant approaches the tangent, and their gradients become equal in the limit — not half.
1. A — By definition, $f'(x) = \lim_{h o 0} \frac{f(x+h) - f(x)}{h}$.
2. C — First principles on $x^3$ gives $\lim_{h o 0} \frac{(x+h)^3 - x^3}{h} = 3x^2$.
3. C — $y' = 4x$, so at $x = 1$ the gradient is $4$.
4. B — $f'(a)$ represents the instantaneous rate of change (gradient of the tangent) at $x = a$.
5. B — A constant function has zero gradient everywhere, so its derivative is $0$.
6. A — The gradient of a straight line $y = mx + c$ is $m$ everywhere.
7. B — In the limit, the secant through two coincident points becomes the tangent at that point.
Q6 (4 marks): (a) $\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 + 3(x+h) - (x^2 + 3x)}{h} = \frac{2xh + h^2 + 3h}{h} = 2x + 3 + h$ [1]. Taking the limit as $h \to 0$ gives $f'(x) = 2x + 3$ [1]. (b) $f'(2) = 2(2) + 3 = 7$ [1]. (c) $m_{\perp} = -\frac{1}{7}$ [1].
Q7 (3 marks): The derivative is the gradient of the tangent at a single point, found by letting the two points of a secant slide infinitely close together [1]. Setting $h = 0$ directly gives $\frac{f(x) - f(x)}{0} = \frac{0}{0}$, which is undefined [1]. The limit $h \to 0$ lets us simplify the difference quotient algebraically first, then evaluate the result as $h$ approaches zero to obtain the exact tangent gradient [1].
Q8 (3 marks): (a) $h'(t) = \lim_{h \to 0} \frac{5(t+h)^2 - 5t^2}{h} = \lim_{h \to 0} \frac{10th + 5h^2}{h} = \lim_{h \to 0} (10t + 5h) = 10t$ [2]. (b) $h'(3) = 10(3) = 30$ m/s [1]. This means that exactly 3 seconds after launch, the rocket is rising at a rate of 30 metres per second.
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