The Derivative as the Gradient of a Tangent
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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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? Without using any formulas, how might you use the idea of average speed over smaller and smaller time intervals to find your instantaneous speed?
There are only two ideas in this entire lesson — and they both emerge from one limit. The derivative is defined as the limit of the difference quotient, and geometrically it represents the gradient of the tangent to a curve at a point.
Every derivative problem in this module travels along one of two roads: algebraic — using the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ to find the derivative from first principles, or geometric — using the derivative to find the gradient of a tangent line and hence its equation.
$m_{\text{tangent}} = f'(a)$
Key facts
- The limit definition of the derivative (first principles)
- The notations $f'(x)$ and $\frac{dy}{dx}$
- The difference between a secant and a tangent
- How to expand and simplify $f(x+h)$ for polynomial functions
Concepts
- The derivative as the gradient of the tangent to a curve
- The connection between average rate of change (secant) and instantaneous rate of change (tangent)
- Why the limit $h \to 0$ is needed instead of setting $h = 0$
Skills
- Find the derivative of simple polynomials from first principles
- Find the gradient of a tangent at a specific point
- Find the equation of a tangent line using point-gradient form
Consider a function $y = f(x)$. The gradient of the straight line joining two points on the curve — a secant — is:
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:
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 $x$.
Geometrically, as the two points of the secant slide together, the secant approaches the tangent. The limit captures the exact moment when the two points become one — and the secant becomes the tangent.
Notation. We write the derivative as $f'(x)$ (Lagrange notation) or $\frac{dy}{dx}$ (Leibniz notation). Both mean the same thing: the rate at which $y$ changes with respect to $x$ at a specific point.
From first principles, we can find the derivative of $f(x) = x^2$:
- Expand $(x+h)^2 = x^2 + 2xh + h^2$
- Form the difference quotient: $\frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h$
- Take the limit as $h \to 0$: $f'(x) = 2x$
Similarly, for $f(x) = x^3$, expanding $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ gives $f'(x) = 3x^2$.
Once we have $f'(x)$, the equation of the tangent line at $x = a$ uses the point-gradient form:
The secant gradient between two points on $y = f(x)$ is $\frac{f(x+h)-f(x)}{h}$; As $h \to 0$, the secant becomes the tangent; its gradient is the derivative
Pause — copy the secant gradient formula $\dfrac{f(x+h)-f(x)}{h}$ and the geometric argument that as $h \to 0$ the secant becomes the tangent into your book.
Quick check: What does the expression $\frac{f(x+h)-f(x)}{h}$ represent geometrically?
True or false: For $f(x) = x^2$, the derivative from first principles gives $f'(x) = 2x$.
Worked examples · 3 in a row, reveal as you go
Find $f'(x)$ for $f(x) = x^2$ using first principles.
Find $f'(x)$ for $f(x) = 3x + 1$ from first principles.
Find the equation of the tangent to $y = x^2$ at the point where $x = 2$.
Think-talk-listen: In the first-principles process for $f(x) = x^2$, why can we factor out and cancel $h$ — but only when $h \neq 0$?
Common errors · the 3 traps that cost marks
Odd one out: Which statement about the tangent and normal is incorrect?
Quick-fire practice · 5 reps
Find $f'(x)$ for $f(x) = x^2$ from first principles.
Find $f'(x)$ for $f(x) = 2x + 3$ from first principles.
Find the gradient of the tangent to $y = x^3$ at $x = 1$.
Find the equation of the tangent to $y = x^2$ at $(1, 1)$.
Find the equation of the normal to $y = x^2$ at $(2, 4)$.
Fill the gaps: drag each token into the correct blank.
The difference quotient gives the gradient of the ___ line. As $h \to$ ___, this approaches the gradient of the ___ line. This process of taking the ___ defines the derivative.
Earlier you were asked: Imagine driving along a curved road. How might you use average speed over smaller and smaller time intervals to find your instantaneous speed?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the derivative of $f(x) = x^2 + 3x$ from first principles. 4 MARKS
Q2. Find the equation of the tangent to $y = x^3$ at the point where $x = 1$. 3 MARKS
Q3. Explain geometrically why the derivative of a linear function $f(x) = mx + c$ is constant and equal to $m$. 3 MARKS
📖 Comprehensive answers (click to reveal)
Drill 1: $f'(x) = 2x$
Drill 2: $f'(x) = 2$
Drill 3: $y' = 3x^2$, so at $x = 1$ the gradient is $3$.
Drill 4: $f'(x) = 2x$, so $f'(1) = 2$. Tangent: $y - 1 = 2(x - 1) \Rightarrow y = 2x - 1$.
Drill 5: $f'(2) = 4$, so tangent gradient is $4$. Normal gradient is $-1/4$. Normal: $y - 4 = -\frac{1}{4}(x - 2) \Rightarrow y = -\frac{1}{4}x + \frac{9}{2}$.
Q1 (4 marks): $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$ [1]. Difference quotient: $\frac{2xh + h^2 + 3h}{h} = 2x + 3 + h$ [2]. Limit: $f'(x) = 2x + 3$ [1].
Q2 (3 marks): $f'(x) = 3x^2$, so $f'(1) = 3$ [1]. Point is $(1, 1)$ [0.5]. Tangent: $y - 1 = 3(x - 1) \Rightarrow y = 3x - 2$ [1.5].
Q3 (3 marks): A linear function $f(x) = mx + c$ is a straight line [1]. The tangent to a straight line at any point is the line itself [1]. Therefore the gradient of the tangent is always $m$, so $f'(x) = m$ for all $x$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚒ Enter the arenaClimb platforms by answering derivative and tangent questions. Lighter alternative to the boss.
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