This note introduces the idea of a limit and uses it to define the derivative as the gradient of a curve at a point and as an instantaneous rate of change. You will learn the notation and how to differentiate simple functions from first principles. Everything else in calculus is built on this idea.
Key ideas
A straight line has one gradient everywhere. A curve does not: its steepness changes from point to point. To measure the steepness at a single point, we draw a straight line that just touches the curve there. This line is the tangent, and its gradient is the gradient of the curve at that point.
We cannot find a gradient from one point alone, so we start with a chord: a line joining two points on the curve. For the points \((x, f(x))\) and \((x+h, f(x+h))\), the chord has gradient
\[ \frac{f(x+h)-f(x)}{(x+h)-x} = \frac{f(x+h)-f(x)}{h}. \]Now let the second point slide towards the first, so that \(h\) gets smaller and smaller. The chord turns into the tangent. We cannot put \(h=0\) directly (that would give \(\tfrac{0}{0}\)), so we simplify first and then see what value the expression approaches.
Limits
The limit of an expression as \(h\) tends to 0 is the value it gets closer and closer to as \(h\) gets closer to 0. We write \(\lim_{h\to 0}\). For example, \(\lim_{h\to 0}(6+3h)=6\), because \(3h\) becomes negligible. Here you only need limits that become clear after expanding and cancelling an \(h\).
The derivative
The derivative of \(f\) is the function that gives the gradient of the curve at every \(x\):
\[ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. \]Finding it this way is called differentiating from first principles. The same function can be written \(\dfrac{dy}{dx}\) if \(y=f(x)\).
Rate of change
The derivative also measures how fast one quantity changes with respect to another. If \(s\) is distance in metres and \(t\) is time in seconds, then \(\dfrac{ds}{dt}\) is the rate of change of distance with time, in m s\(^{-1}\). The units of a derivative are the units of \(y\) divided by the units of \(x\).
Sign of the gradient
- \(f'(a)\gt 0\): the curve is rising at \(x=a\).
- \(f'(a)\lt 0\): the curve is falling at \(x=a\).
- \(f'(a)=0\): the tangent is horizontal at \(x=a\).
For a positive integer power, the expansion needed is \((x+h)^n = x^n + n x^{n-1}h + (\text{terms containing } h^2)\). All the terms with \(h^2\) or higher vanish in the limit. This is why the derivative of \(x^n\) turns out to be \(nx^{n-1}\), which you will use as a rule in the next notes.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{f(x+h)-f(x)}{h}\) | Gradient of the chord between \(x\) and \(x+h\). It is an average rate of change. |
| \(f'(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\) | Definition of the derivative: the gradient of the tangent, or instantaneous rate of change. |
| \(f'(x)\), \(\dfrac{dy}{dx}\) | Two notations for the derivative. |
| \(f'(a)\) | The gradient of the curve at the point where \(x=a\). |
| Tangent | A straight line touching the curve at one point with the same gradient as the curve there. |
| \(f(x)=ax^n\Rightarrow f'(x)=anx^{n-1}\) | Result that first principles gives for \(n\) a positive integer (used as a rule in note 5.3). |
Worked examples
Example 1 Use first principles to find the derivative of \(f(x)=3x^2-5x\), and hence the gradient at \(x=4\).
Write out \(f(x+h)\): \(f(x+h)=3(x+h)^2-5(x+h)=3x^2+6xh+3h^2-5x-5h\).
Subtract \(f(x)\): \(f(x+h)-f(x)=6xh+3h^2-5h\).
Divide by \(h\): \(\dfrac{f(x+h)-f(x)}{h}=6x+3h-5\).
Let \(h\to 0\): \(f'(x)=\lim_{h\to 0}(6x+3h-5)=6x-5\).
At \(x=4\): \(f'(4)=6(4)-5=19\).
Answer: \(f'(x)=6x-5\); the gradient at \(x=4\) is 19.
Example 2 For \(f(x)=x^3\), find the gradient of the chord from \(x=2\) to \(x=2+h\). Evaluate it for \(h=0.1\), then state the gradient of the curve at \(x=2\).
\(f(2)=8\) and \(f(2+h)=(2+h)^3=8+12h+6h^2+h^3\).
Chord gradient \(=\dfrac{12h+6h^2+h^3}{h}=12+6h+h^2\).
For \(h=0.1\): \(12+0.6+0.01=12.61\).
As \(h\to 0\), the terms \(6h\) and \(h^2\) tend to 0, so the limit is 12.
Answer: chord gradient \(12+6h+h^2\), which is 12.61 when \(h=0.1\); the gradient of the curve at \(x=2\) is 12.
Example 3 A drone climbs so that its height after \(t\) seconds is \(h(t)=4t^2+2t\) metres. Find its instantaneous rate of climb after 3 seconds, using first principles. Use \(\delta\) for the small time step.
\(h(t+\delta)=4(t+\delta)^2+2(t+\delta)=4t^2+8t\delta+4\delta^2+2t+2\delta\).
\(h(t+\delta)-h(t)=8t\delta+4\delta^2+2\delta\).
Divide by \(\delta\): \(8t+4\delta+2\).
Let \(\delta\to 0\): \(h'(t)=8t+2\).
At \(t=3\): \(h'(3)=8(3)+2=26\).
Answer: 26 m s\(^{-1}\).
Common errors
- Common error: substituting \(h=0\) straight into \(\dfrac{f(x+h)-f(x)}{h}\). This gives \(\tfrac00\). Expand, cancel the \(h\), and only then let \(h\to 0\).
- Common error: forgetting to subtract \(f(x)\), or forgetting a bracket so that \(-5(x+h)\) becomes \(-5x+h\).
- Common error: dropping the \(\lim\) notation part-way through, or writing \(f'(x)=\) before the limit has been taken.
- Common error: mixing up the chord gradient (an average rate over an interval) with the derivative (the rate at one instant).
- Common error: leaving out units when the derivative is a rate in a real context.
Exam tip: in a first-principles question, write the three lines in order: \(f(x+h)\), then \(f(x+h)-f(x)\), then the quotient after dividing by \(h\). Only then write the limit.
Check yourself
- Use first principles to find \(f'(x)\) for \(f(x)=5x^2\).
Answer
\(f(x+h)-f(x)=5(x^2+2xh+h^2)-5x^2=10xh+5h^2\). Dividing by \(h\) gives \(10x+5h\). As \(h\to 0\), \(f'(x)=10x\). - For \(f(x)=x^2\), find the gradient of the chord from \(x=3\) to \(x=3.1\), and use it to estimate the gradient of the curve at \(x=3\).
Answer
\(f(3)=9\), \(f(3.1)=9.61\). Chord gradient \(=\dfrac{9.61-9}{0.1}=6.1\). As the second point moves closer, the gradient approaches 6, so the gradient at \(x=3\) is about 6. - Given that \(f(x)=x^3\) has first-principles chord gradient \(3x^2+3xh+h^2\), write down \(f'(x)\) and find the gradient at \(x=-2\).
Answer
Let \(h\to 0\): \(f'(x)=3x^2\). At \(x=-2\): \(3(-2)^2=12\). - The temperature of a cooling drink is \(T\) °C after \(t\) minutes, and \(T'(4)=-2\). Explain what this means.
Answer
After 4 minutes the temperature is falling at the rate of 2 °C per minute. The negative sign shows that it is decreasing.