Move the second point closer and closer and watch the secant settle onto the tangent. The derivative is what that settling converges to.
First principles can feel like algebra for its own sake. Watch a secant slide towards a tangent on a screen and the limit stops being a symbol: you can see the number it is heading for.
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Recall, your gut answer first
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For $f(x) = x^2$, find the gradient of the chord from $x = 2$ to $x = 3$, then to $x = 2.5$, then to $x = 2.1$. What number do your answers seem to be approaching?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Watch the secant become the tangent
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Watch the secant become the tangent
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A secant through $(a, f(a))$ and $(a+h, f(a+h))$ has gradient $\dfrac{f(a+h) - f(a)}{h}$. As $h$ shrinks towards $0$ the secant converges on the tangent, and that limiting value is $f'(a)$.
secant gradient $= \dfrac{f(a+h) - f(a)}{h}$ as $h \to 0$ this tends to $f'(a)$
Halve $h$ each row
A spreadsheet column of $h = 1, 0.5, 0.25, 0.125, \ldots$ makes the convergence obvious, because each row roughly halves the distance to the limit.
Approach from both sides
Use negative $h$ as well as positive. If the two sides converge on different numbers the derivative does not exist there, which is what happens at a sharp corner.
Very small $h$ goes wrong
Below about $h = 10^{-10}$ a calculator or spreadsheet loses accuracy, because it is subtracting two nearly equal numbers. The estimate gets worse, not better.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A secant joins two points on a curve; a tangent touches at one.
The gradient of the secant from $x = a$ to $x = a + h$ is $\dfrac{f(a+h) - f(a)}{h}$.
As $h \to 0$ the secant gradient converges on the derivative $f'(a)$.
A numerical estimate degrades once $h$ becomes extremely small, through rounding error.
Understand
Concepts
Why shrinking $h$ makes the secant approximate the tangent better.
Why approaching from both sides matters when a function has a corner.
Why an extremely small $h$ makes a spreadsheet estimate worse rather than better.
Can do
Skills
Build a table of secant gradients for decreasing $h$.
Estimate a derivative numerically and say how confident the estimate is.
Use graphing software to observe a secant converging on a tangent.
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Key terms
SecantA straight line cutting a curve at two points. Like this: the line through $(2, 4)$ and $(3, 9)$ on $y = x^2$ is a secant of gradient $5$.
ConvergenceValues settling towards one number as a variable changes. Like this: secant gradients $5$, $4.5$, $4.1$, $4.01$ converge on $4$.
Step size $h$How far the second point sits from the first. Like this: from $x = 2$ with $h = 0.1$, the second point is at $x = 2.1$.
Numerical derivativeAn estimate of $f'(a)$ from a small $h$ rather than from algebra. Like this: $\dfrac{f(2.001) - f(2)}{0.001} \approx 4.001$ for $f(x) = x^2$.
Rounding errorLoss of accuracy when a calculator subtracts two nearly equal numbers. Like this: at $h = 10^{-14}$ a spreadsheet may return $0$ for a gradient that is really $4$.
Two-sided checkEstimating with $+h$ and $-h$ to see whether both agree. Like this: at a sharp corner the two sides give different answers, so no derivative exists.
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Building the table
Work through the core explanation before applying it.
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Building the table
core concept
Set up a column of $h$ values that shrink: $1$, $0.5$, $0.25$, $0.125$, and so on. Beside each, compute the secant gradient $\dfrac{f(a+h) - f(a)}{h}$.
For $f(x) = x^2$ at $a = 2$: $h = 1$ gives $5$, $h = 0.5$ gives $4.5$, $h = 0.25$ gives $4.25$, $h = 0.1$ gives $4.1$. The pattern is unmistakable and the limit is $4$.
The algebra confirms it: $\dfrac{(2+h)^2 - 4}{h} = \dfrac{4h + h^2}{h} = 4 + h$, which tends to $4$ as $h \to 0$. The table and the algebra are the same statement.
The table is evidence, not proof. It shows the limit convincingly and it builds the intuition, but an exam answer asking you to differentiate from first principles still wants the algebra.
Quick check: for $f(x) = x^2$ at $a = 2$, the secant gradient simplifies to $4 + h$. What is the derivative at $x = 2$?
The secant gradient from $a$ to $a + h$ is $\dfrac{f(a+h) - f(a)}{h}$. Shrinking $h$ in a table shows it converging on $f'(a)$, and for $f(x) = x^2$ at $a = 2$ it simplifies to $4 + h$, which tends to $4$.
Pause, copy the secant formula, the table of $5, 4.5, 4.25, 4.1$, and the algebra giving $4 + h$, into your book.
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Watching it on a screen
core concept
We just saw the numbers converge. That raises a question: what does that look like? This card answers it → drag the second point towards the first and the secant visibly rotates onto the tangent.
In graphing software, plot the curve, mark a fixed point, and mark a second point you can drag. Draw the line through both and display its gradient.
Drag the second point towards the fixed one. The line rotates, and the displayed gradient settles on a single value. That value is the derivative at the fixed point.
A spreadsheet does the same thing without a picture: one column of $h$, one of $f(a+h)$, one of the gradient. Both are showing the same limit.
Try it at a corner. On $y = |x|$ at $x = 0$, dragging in from the right gives $1$ and from the left gives $-1$. The two sides disagree, so there is no tangent and no derivative there.
Fill the blank: as the second point approaches the first, the secant converges on the .
Graphing software shows the secant rotating onto the tangent as the second point approaches; a spreadsheet shows the same limit as a column of numbers. At a corner the two sides give different values, so no derivative exists.
Pause, copy the drag-the-point method, the spreadsheet columns, and the $y = |x|$ corner result, into your book.
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When the numerical estimate breaks down
core concept
We just saw that smaller $h$ gives a better estimate. That raises a question: is smaller always better? This card answers it → no, past a point the arithmetic itself fails.
Mathematically, smaller $h$ is always better. On a machine it is not, because $f(a+h)$ and $f(a)$ become nearly equal and subtracting them destroys most of the significant figures.
For $f(x) = x^2$ at $a = 2$, an estimate at $h = 10^{-8}$ is excellent. At $h = 10^{-14}$ a spreadsheet may return something visibly wrong, or even $0$.
So a numerical derivative has a sweet spot: small enough that the secant is close to the tangent, large enough that the subtraction still carries digits.
This is a fact about calculators, not about calculus. The limit is exact; the machine estimating it is not. Knowing where a tool stops being trustworthy is part of using it.
Which statement about numerical derivatives is FALSE?
Shrinking $h$ improves the estimate mathematically but eventually destroys it numerically, because subtracting two nearly equal values loses significant figures. A good numerical derivative uses a moderately small $h$, not the smallest available.
Pause, copy the sweet-spot idea and the note that this is a limitation of the machine, not of the limit, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · A TABLE OF SECANT GRADIENTS
For $f(x) = x^2$, tabulate the secant gradient at $a = 2$ for $h = 1, 0.5, 0.1, 0.01$, and state the limit.
Estimate $f'(3)$ for $f(x) = x^3$ using $h = 0.001$, and compare with the exact value.
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$f(3.001) = 27.027009$ and $f(3) = 27$
Evaluate both to enough decimal places.
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$\dfrac{27.027009 - 27}{0.001} = 27.009$
The numerical estimate.
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Exact: $f'(x) = 3x^2$ so $f'(3) = 27$; the estimate is high by $0.009$
The error is roughly proportional to $h$.
PROBLEM 3 · A TWO-SIDED CHECK AT A CORNER
For $f(x) = |x|$ at $a = 0$, estimate the derivative using $h = 0.001$ and $h = -0.001$.
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$h = 0.001$: $\dfrac{0.001 - 0}{0.001} = 1$
Approaching from the right.
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$h = -0.001$: $\dfrac{0.001 - 0}{-0.001} = -1$
Approaching from the left.
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The two sides disagree, so $f'(0)$ does not exist
A corner has no single tangent.
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Quick-fire practice
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Quick-fire practice
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For $f(x) = x^2$ at $a = 5$, find the secant gradient with $h = 0.1$.
A table of secant gradients reads $7.3, 7.03, 7.003$. What is the derivative?
Why does an extremely small $h$ make a spreadsheet estimate worse?
A two-sided estimate gives $2$ from the right and $-2$ from the left. What does that tell you?
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Revisit your three chord gradients
Run the quick drill and copy the summary into your book.
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Revisit your three chord gradients
At the start you found the chord gradients for $f(x) = x^2$ from $x = 2$ to $x = 3$, $2.5$ and $2.1$. State the number they were approaching, and explain how the algebra $4 + h$ accounts for every value you calculated.
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Multiple choice
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Multiple choice
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Short answer
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ApplyBand 33 marks
Q1. For $f(x) = x^2$, find the secant gradients at $a = 3$ for $h = 1$, $h = 0.1$ and $h = 0.01$, and state the value they are converging on. (3 marks)
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ApplyBand 43 marks
Q2. Show algebraically that the secant gradient for $f(x) = x^2$ at $x = a$ simplifies to $2a + h$, and explain what this tells you about the table you would get for any $a$. (3 marks)
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UnderstandBand 53 marks
Q3. A student says the best numerical estimate of a derivative always comes from the smallest $h$ their calculator allows. Explain why this is wrong, and describe what actually happens. (3 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $\dfrac{5.1^2 - 25}{0.1} = \dfrac{26.01 - 25}{0.1} = 10.1$. Practice 2: $7$. Practice 3: $f(a+h)$ and $f(a)$ become nearly equal, so subtracting them loses significant figures. Practice 4: the two sides disagree, so the derivative does not exist there; the graph has a corner.
Q2 (3 marks): $\dfrac{(a+h)^2 - a^2}{h} = \dfrac{a^2 + 2ah + h^2 - a^2}{h}$ [1]. $= \dfrac{2ah + h^2}{h} = 2a + h$ [1]. So for any $a$ the table is just $2a$ plus the current $h$, which is why the values fall towards $2a$ in step with $h$ [1].
Q3 (3 marks): Mathematically a smaller $h$ does give a better approximation, so the reasoning is right about the limit [1]. On a machine, $f(a+h)$ and $f(a)$ become nearly equal and the subtraction cancels most of the significant figures, so the stored result carries very few correct digits [1]. The estimate therefore improves as $h$ shrinks, reaches a sweet spot, and then degrades, sometimes collapsing to $0$ [1].
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Boss battle · Secant Sprint
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Build secant tables, read their limits, and judge when a numerical estimate can be trusted. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.