The tangent line as the limit of secant lines, the two equivalent forms of the difference quotient, computing a derivative at a point directly from the definition, writing the equation of a tangent line, velocity as an instantaneous rate of change, and estimating a derivative from a table of values.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Defining the Derivative
Objectives
Five outcomes. The first is a definition Section 2.1 spent a whole section motivating and Chapter 2 spent four sections making legitimate.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 188-202 — the section these objectives are drawn from
Warm-up
Section 2.1 computed secant slopes approaching a tangent slope but had no way to finish. Chapter 2 supplied exactly what was missing.
Discussion prompt
The secant slope through the points at 1 and at x on the squaring function simplifies to x plus 1, for every x except 1. What is the tangent slope at 1, and which chapter's machinery justifies the answer?
Hint: It is a limit, and Section 2.3 says how to evaluate it.
Answer:
\[ \frac{x^2 - 1}{x - 1} = x + 1 \;(x \ne 1) \;\Longrightarrow\; \lim_{x \to 1} = 2 \]
The tangent slope is 2, and the justification is the limit — which Chapter 2 defined, made computable with the limit laws, and finally made rigorous with epsilon and delta.
So the whole apparatus of Chapter 2 exists to make one number well defined. This section gives that number a name, a notation, and a definition that applies to every function at once.
Concept
The derivative of a function at a point is the limit of the difference quotient as the second point closes in. It is the slope of the tangent line, the instantaneous rate of change, and — when the function is a position — the velocity. All three are the same number.
the derivative at a point — The derivative of f at a is the limit as h approaches zero of f of a plus h minus f of a, all over h, provided that limit exists. It is written f prime of a.
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
The quotient is undefined at h equal to zero, which is exactly why Section 2.2 insisted a limit ignores the point it approaches. Without that insistence the definition would be empty.
Figure (svg): The difference quotient drawn on a curve, with the rise, the run h, and the secant that becomes the tangent
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 188-191
Section
Section 1
Concept
The difference quotient can be written with the second input as a plus h, or simply as x. The two forms are the same quantity: substituting x equal to a plus h turns one into the other, and h approaching zero is exactly x approaching a.
difference quotient — The slope of the secant line through the points at a and at a second nearby input. Written as f of a plus h minus f of a over h, or equivalently as f of x minus f of a over x minus a.
\[ f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} = \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \]
In practice the h form is usually easier to compute with, because the algebra collapses cleanly, while the x form is easier to read as an average rate of change. Choose whichever shortens the work.
Figure (svg): The two equivalent forms of the difference quotient, with the substitution linking them
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 188-193 — the tangent line and the derivative
Picture it
The rise, the run, and the secant that becomes the tangent.
Figure (svg): The difference quotient drawn on a curve, with the rise, the run h, and the secant that becomes the tangent
The run is h and the rise is the change in output across it. Shrinking h swings the dashed secant onto the solid tangent, and the limit is what that swing settles at.
Worked example
Example 3.1. Expand, cancel, then let h go.
\[ \text{For } f(x) = x^2, \text{ find } f'(3) \text{ from the definition.} \]
Write the quotient with a equal to 3
Why: The definition, with the numbers put in.
\[ \frac{(3 + h) ^{2} - 9}{h} \]
Expand the square
Why: Nine plus 6h plus h squared.
\[ \frac{9 + 6 h + h ^{2} - 9}{h} \]
Simplify: the constants cancel
Why: This always happens, and it is what makes h a factor.
\[ \frac{6 h + h ^{2}}{h} \]
Factor out h and cancel
Why: Legal because h is not zero.
\[ 6 + h \]
Take the limit
Why: As h approaches 0.
\[ f'(3) = 6 \]
Figure (svg): The solution to Worked example a derivative from the definition shown as a ladder of expressions, one row per legal move
\[ f'(3) = 6 \]
Verify: check against the secant slopes and the general pattern
Why: At h equal to 0.001 the quotient is 6.001, and at negative 0.001 it is 5.999 — closing on 6 from both sides. The same computation at a general a gives 2a plus h, whose limit is 2a; at a equal to 3 that is 6. The cancellation of the constant terms in step three is the structural reason this always works: the numerator is guaranteed to have h as a factor, because it vanishes when h does.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 191-192
Matching
The two forms, and when each is shorter.
Match the pairs
Why: Every one of these is a technique from Section 2.3, reappearing in a single setting. The derivative is not a new kind of computation — it is the indeterminate quotient, always of the same shape, which is why one chapter of limit technique suffices for the whole of Chapter 3.
Worked example
Checkpoint 3.1. Factoring replaces expanding.
\[ \text{Find } f'(3) \text{ for } f(x) = x^2 \text{ using the } x \text{ form.} \]
Write the quotient with the second input as x
Why: The alternative form.
\[ \frac{x ^{2} - 9}{x - 3} \]
Factor the numerator
Why: A difference of squares.
\[ (x - 3) (x + 3) / (x - 3) \]
Cancel
Why: Legal because x is not 3.
\[ x + 3 \]
Take the limit as x approaches 3
Why: Substitute into the simplified form.
\[ f'(3) = 6 \]
Figure (svg): The solution to Worked example the same derivative in the x form shown as a ladder of expressions, one row per legal move
\[ f'(3) = \lim_{x \to 3}(x+3) = 6 \]
Verify: compare the work in the two forms
Why: Both give 6, as they must. The h form required expanding a square; the x form required factoring a difference of squares. For a power, factoring is often quicker; for a root or a reciprocal, the h form usually is. Since they are the same limit, the choice is purely tactical and it is worth trying the other one when the algebra gets ugly.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 192-193
Trap
\[ \frac{(3+h)^2 - 9}{h} \quad \text{at } h = 0 \]
Put h equal to zero straight into the quotient
Why: The student evaluates before simplifying.
\[ = \frac{0}{0} \quad \text{(and the student concludes no derivative exists)} \]
The quotient is undefined at h equal to zero by design, and always will be. Concluding failure from that is concluding failure from the definition itself.
\[ \frac{6h + h^2}{h} = 6 + h \;(h \ne 0) \;\Longrightarrow\; \lim_{h \to 0} = 6 \]
Simplify while h is still non-zero, THEN take the limit
Why: The cancellation is legal at every h the limit uses.
Zero over zero is the expected form here, not a warning sign. Every derivative computation in this course produces it and then removes it, which is exactly the pattern Section 2.3 catalogued — and the reason that section came first.
Fill the middle
The derivative of the squaring function at 3, after expanding.
Fill in the blanks
\frac6 + h___ = ___ \;(h \ne 0)
Why: Factoring out h and cancelling leaves 6 plus h, whose limit as h approaches 0 is 6. The cancellation requires h to be non-zero, which every input the limit uses satisfies.
Ranking
A derivative from the definition.
Put in order
Why: Step c is the one that reveals the structure: the constants must cancel, because the numerator vanishes when h does. If they do not cancel, an algebra slip has occurred and step d will be impossible — which makes step c a useful checkpoint.
Prediction
Commit before reasoning.
Predict first
Why does the numerator of a difference quotient always have h as a factor?
Correct: Because the numerator is zero when h is zero, so h divides it.
\[ N(h) = f(a+h) - f(a), \quad N(0) = 0 \;\Longrightarrow\; h \mid N(h) \]
Why: The numerator is f of a plus h minus f of a, which is exactly zero at h equal to zero. For a polynomial the Factor Theorem then guarantees h is a factor; for other functions the same cancellation appears after rationalising or combining. This is why every derivative computation follows the same arc — produce zero over zero, remove the shared h, evaluate — and why it works for functions far beyond polynomials.
Section
Section 2
Concept
For a square root the quotient is cleared by rationalising the numerator; for a reciprocal it is cleared by combining the fractions. In both cases the h cancels and the limit is then a substitution.
clearing the quotient — The algebraic preparation that exposes the factor of h in a difference quotient. Which preparation is needed depends on the function: expanding for powers, rationalising for roots, combining for reciprocals.
\[ \frac{\sqrt{a+h}-\sqrt{a}}{h} \cdot \frac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}} = \frac{1}{\sqrt{a+h}+\sqrt{a}} \]
Each of these is a technique from Section 2.3 applied to a quotient of a fixed shape. That the same three or four moves suffice for every function in the course is what makes the general rules of Section 3.3 possible.
Figure (svg): The two equivalent forms of the difference quotient, with the substitution linking them
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 193-196 — derivatives from the definition
Picture it
The h form and the x form, and the substitution between them.
Figure (svg): The two equivalent forms of the difference quotient, with the substitution linking them
For a root the h form and rationalising is usually shortest; for a reciprocal the h form and combining is. Recognising the shape before starting saves a page of algebra.
Worked example
Example 3.3. Rationalise the numerator.
\[ \text{For } f(x) = \sqrt{x}, \text{ find } f'(4). \]
Write the quotient
Why: With a equal to 4.
\[ \frac{\sqrt{4 + h} - 2}{h} \]
Multiply by the conjugate over itself
Why: This is multiplying by 1.
\[ \times(\sqrt{4 + h} + 2) / (\sqrt{4 + h} + 2) \]
Simplify the numerator
Why: A difference of squares: 4 plus h minus 4.
\[ \frac{h}{h(\sqrt{4 + h} + 2)} \]
Cancel h
Why: Legal because h is not zero.
\[ \frac{1}{\sqrt{4 + h} + 2} \]
Take the limit
Why: The root approaches 2.
\[ \frac{1}{4} \]
Figure (svg): The solution to Worked example the derivative of a square root shown as a ladder of expressions, one row per legal move
\[ f'(4) = \tfrac{1}{4} \]
Verify: check numerically and against the general pattern
Why: At h equal to 0.001 the quotient is about 0.24998, closing on 0.25. Running the same computation at a general a gives one over the root of a plus h plus the root of a, whose limit is one over twice the root of a — at a equal to 4 that is one quarter. Section 2.3 computed exactly this limit as an exercise in rationalising; here it turns out to have been a derivative all along.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 194-195
Fill the middle
The square root derivative from the worked example, after multiplying by the conjugate.
Fill in the blanks
\frac1/4___+2\right)} = \frac______+2} \;\longrightarrow\; ___
Why: As h approaches 0 the root approaches 2, so the denominator approaches 4 and the whole expression approaches one quarter. The conjugate manufactured the h in the numerator that cancelled the h in the denominator.
Worked example
Checkpoint 3.3. Combine the fractions first.
\[ \text{For } f(x) = \frac{1}{x}, \text{ find } f'(2). \]
Write the quotient
Why: With a equal to 2.
\[ \frac{\frac{1}{2 + h} - \frac{1}{2}}{h} \]
Combine the inner fractions
Why: Common denominator 2 times 2 plus h.
\[ \frac{2 - (2 + h)}{2(2 + h)} \]
Simplify the numerator
Why: The 2s cancel, leaving negative h.
\[ -\frac{h}{2(2 + h)} \]
Divide by h, that is cancel it
Why: Legal because h is not zero.
\[ -\frac{1}{2(2 + h)} \]
Take the limit
Why: As h approaches 0.
\[ -\frac{1}{4} \]
Figure (svg): The solution to Worked example the derivative of a reciprocal shown as a ladder of expressions, one row per legal move
\[ f'(2) = -\tfrac{1}{4} \]
Verify: check the sign, which is the whole content here
Why: The reciprocal function is decreasing on the positive numbers, so its derivative must be negative — and negative one quarter is. Numerically, at h equal to 0.001 the quotient is about negative 0.24994. The general computation gives negative one over a squared, which at 2 is negative one quarter. Losing the minus sign when the numerator's 2s cancel is the standard error, and the decreasing graph is the check that catches it.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 195-196
Error analysis
A student computes a derivative from the definition.
Annotate
On: \( \frac{6h + h^2}{h} = 6 + h^2 \)
Here the final answer survives — both 6 plus h and 6 plus h squared tend to 6 — which is exactly what makes the error dangerous. Factor the h out explicitly rather than cancelling term by term, and the mistake becomes impossible.
Sorting
Look at the shape of the function.
Sort into buckets
Sort each derivative-from-definition problem.
Three preparations cover every derivative computed from the definition in this course. Section 3.3 will replace all of them with rules, but the rules are proved by exactly these computations.
Two truths and a lie
All three are about computing from the definition.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and believing it would make every derivative fail. The form 0/0 is what a difference quotient always produces; it is the starting point, not a verdict. The derivative exists exactly when the limit of the simplified quotient exists, which is a question the algebra answers only after the h has been cancelled.
Prediction
Commit before reasoning.
Predict first
Why must the derivative of 1/x be negative for every positive x?
Correct: Because the function is decreasing there, so every tangent slopes downward.
\[ f'(a) = -\frac{1}{a^2} < 0 \quad \text{for every } a \ne 0 \]
Why: A larger input gives a smaller output on the positive half of the reciprocal's graph, so every secant and every tangent has a negative slope. The sign is forced by the shape of the graph, not produced by the algebra — which makes it an excellent check. If the algebra ever produces a positive answer here, a sign has been lost, and this reasoning finds it faster than rechecking the computation. Chapter 4 will turn this observation into a theorem: the sign of the derivative determines whether a function rises or falls.
Section
Section 3
Concept
A tangent line needs two pieces of information: a point it passes through and a slope. The point comes from evaluating the function; the slope comes from the derivative. Point-slope form then assembles them.
equation of the tangent line — The tangent to the graph of f at the input a is the line through the point a comma f of a with slope f prime of a, written in point-slope form as y minus f of a equals f prime of a times x minus a.
\[ y - f(a) = f'(a)(x - a) \]
The commonest error is using the derivative as the y-coordinate of the point. Keeping the two numbers labelled — f of a for height, f prime of a for slope — prevents it entirely.
Figure (svg): A curve with its tangent line at a point, and the point-slope equation beside it
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 191-194 — the tangent line
Picture it
The tangent to the squaring function at 2.
Figure (svg): A curve with its tangent line at a point, and the point-slope equation beside it
Here f of 2 and f prime of 2 are both 4, which is a coincidence of this example and a trap in general. Computing them as separate quantities is what keeps the coincidence from becoming a habit.
Worked example
Example 3.4. Two evaluations, then point-slope form.
\[ \text{Find the tangent to } f(x) = x^2 - 4x + 6 \text{ at } x = 1. \]
Evaluate the function for the point
Why: One minus 4 plus 6.
\[ f(1) = 3,\text{ point } (1, 3) \]
Compute the derivative from the definition
Why: Expand and cancel.
\[ \frac{(1 + h) ^{2} - 4(1 + h) + 6 - 3}{h} \]
Simplify the numerator
Why: The constants cancel, leaving negative 2h plus h squared.
\[ \frac{-2 h + h ^{2}}{h} = -2 + h \]
Take the limit for the slope
Why: As h approaches 0.
\[ f'(1) = -2 \]
Assemble in point-slope form
Why: Point and slope together.
\[ y - 3 = -2(x - 1) \]
Figure (svg): A curve with its tangent line at a point, and the point-slope equation beside it
\[ y - 3 = -2(x-1) \;\Longrightarrow\; y = -2x + 5 \]
Verify: confirm the line touches rather than crosses
Why: Setting the parabola equal to the line gives x squared minus 4x plus 6 equal to negative 2x plus 5, so x squared minus 2x plus 1 equals 0, which is x minus 1 all squared. The repeated root at 1 confirms tangency — a secant would give two distinct roots. Also, the vertex of this parabola is at x equal to 2, and 1 is to the left of it where the parabola is falling, so a negative slope is exactly right.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 193-194
Fill the middle
The parabola from the worked example, with its point and slope computed.
Fill in the blanks
y - 3 = -2(x-1) \;\Longrightarrow\; y = -2x + 5
Why: Distributing gives negative 2x plus 2, and adding 3 gives negative 2x plus 5. Substituting x equal to 1 gives 3, which is f of 1 — the check that the line really passes through the point of tangency.
Worked example
Checkpoint 3.4. The same two numbers, harder algebra.
\[ \text{Find the tangent to } f(x) = \sqrt{x} \text{ at } x = 9. \]
Evaluate for the point
Why: The root of 9.
\[ f(9) = 3,\text{ point } (9, 3) \]
Use the general root derivative
Why: One over twice the root, from the earlier idea.
\[ f'(x) = \frac{1}{2 \sqrt{x}} \]
Evaluate the slope
Why: One over 2 times 3.
\[ f'(9) = \frac{1}{6} \]
Assemble
Why: Point-slope form.
\[ y - 3 = (\frac{1}{6}) (x - 9) \]
Simplify
Why: Distribute and collect.
\[ y = \frac{x}{6} + \frac{3}{2} \]
Figure (svg): The solution to Worked example a tangent to a root function shown as a ladder of expressions, one row per legal move
\[ y = \tfrac{x}{6} + \tfrac{3}{2} \]
Verify: check the line approximates the curve nearby
Why: At x equal to 10 the line gives 10 over 6 plus 1.5, which is about 3.1667, while the actual root of 10 is about 3.1623 — agreeing to two decimal places one unit away. That a tangent line closely approximates the curve near the point of contact is not a coincidence; it is the whole basis of linear approximation in Section 4.2, and it is a fast sanity check on any tangent line.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 194-194
Trap
\[ f(x) = x^2 - 4x + 6 \text{ at } x = 1: \; f'(1) = -2 \]
Write the tangent through the point (1, -2)
Why: The student uses the derivative as the y-coordinate.
\[ y + 2 = -2(x-1) \quad \text{(wrong point)} \]
The point on the curve is at height f of 1, which is 3, not negative 2. The resulting line does not touch the curve at all.
\[ \text{point } (1, f(1)) = (1, 3), \quad \text{slope } f'(1) = -2 \]
Compute the two numbers separately and label them
Why: f of a is a height on the curve; f prime of a is a slope.
The check is immediate: the tangent must pass through a point ON the curve, so substituting a into both the curve and the line must give the same height. Any tangent line failing that test is wrong before anything else is examined.
Matching
Two numbers, two jobs.
Match the pairs
Why: Keeping the first two visibly separate is the whole discipline here. When they happen to be equal, as at x equal to 2 on the squaring function, it is a coincidence of that example — and treating it as a pattern produces wrong tangent lines everywhere else.
Two truths and a lie
All three are about tangent lines.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it repeats an over-reading from Section 2.1. Tangency is a local condition about matching direction at the point; the line is free to meet the curve again elsewhere. The tangent to a cubic at one point commonly crosses it again, and at an inflection point the tangent crosses the curve at the very point it touches.
Prediction
Commit before reasoning.
Predict first
If s(t) is a position in metres and t is in seconds, what are the units of s'(t)?
Correct: Metres per second.
\[ [f'] = \frac{[\text{output}]}{[\text{input}]} \]
Why: A derivative is a limit of quotients of an output change by an input change, so its units are always output units divided by input units — exactly as slope was in Section 1.2. This makes units a genuine error check: a derivative reported in metres has confused a value with a rate. It also predicts what a second derivative measures, since differentiating again divides by seconds once more and gives metres per second squared, an acceleration.
Section
Section 4
Concept
The derivative of a position function is the velocity at that instant. More generally, the derivative of any quantity with respect to any variable is the instantaneous rate at which the first changes as the second does.
instantaneous rate of change — The derivative of a quantity with respect to a variable, interpreted as the rate at which that quantity is changing at a single value of the variable. Its units are the quantity's units divided by the variable's.
\[ v(t) = s'(t) = \lim_{h \to 0}\frac{s(t+h)-s(t)}{h} \]
Section 2.1 could describe instantaneous velocity only as a limit of averages, with no way to compute it. The derivative is that limit, now equipped with a definition, a notation and a method.
Figure (svg): A position graph with its tangent, showing instantaneous velocity as a derivative
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 196-200 — velocities and rates of change
Picture it
A falling body's position graph with its tangent.
Figure (svg): A position graph with its tangent, showing instantaneous velocity as a derivative
The tangent's slope at each instant is the velocity then. Where the curve steepens the body is moving faster, which is the reading a speedometer gives directly.
Worked example
Example 3.6. Differentiate, then evaluate.
\[ \text{A ball falls } s(t) = 16t^2 \text{ feet. Find its velocity at } t = 1.5 \text{ s.} \]
Write the difference quotient
Why: At a general time t.
\[ \frac{16(t + h) ^{2} - 16 t ^{2}}{h} \]
Expand and simplify
Why: The constant terms cancel.
Cancel h
Why: Legal because h is not zero.
\[ 32 t + 16 h \]
Take the limit for the velocity function
Why: As h approaches 0.
\[ v(t) = 32 t \]
Evaluate at the given time
Why: Thirty-two times 1.5.
\[ v(1.5) = 48 \text{ft} / s \]
Figure (svg): The solution to Worked example velocity from a position function shown as a ladder of expressions, one row per legal move
\[ v(t) = 32t, \qquad v(1.5) = 48 \text{ ft/s} \]
Verify: compare with the averages from Section 2.1
Why: That section computed average velocities over shrinking intervals from t equal to 1 and found them closing on 32 feet per second — and the formula gives v of 1 equal to 32 exactly. The units also check: 32t has units of feet per second squared times seconds, which is feet per second. Notice that differentiating once produced a function, not just a number, which is the subject of Section 3.2.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 197-198
Sorting
Ask whether an interval or a moment is named.
Sort into buckets
Sort each quantity.
Every instantaneous item on this list is the same number viewed three ways: a derivative, a speedometer reading, and a tangent slope. That these are one quantity is the central claim of the section.
Worked example
Checkpoint 3.6. Same mathematics, different units.
\[ \text{Revenue is } R(x) = 60x - 0.02x^2 \text{ dollars for } x \text{ units. Find } R'(500) \text{ and interpret it.} \]
Write the quotient at a general x
Why: The definition.
\[ \frac{R(x + h) - R(x)}{h} \]
Expand and simplify
Why: The constant terms cancel as always.
\[ \frac{60 h - 0.04 x h - 0.02 h ^{2}}{h} \]
Cancel h
Why: Legal because h is not zero.
\[ 60 - 0.04 x - 0.02 h \]
Take the limit
Why: As h approaches 0.
\[ R'(x) = 60 - 0.04 x \]
Evaluate and interpret
Why: Sixty minus 20.
\[ R'(500) = 40\text{ dollars per unit} \]
Figure (svg): The solution to Worked example a rate outside physics shown as a ladder of expressions, one row per legal move
\[ R'(x) = 60 - 0.04x, \qquad R'(500) = 40 \]
Verify: check the interpretation against a direct computation
Why: Selling one more unit changes revenue from R of 500 to R of 501, a difference of about 39.98 dollars — very close to the derivative's 40. That is exactly what the derivative predicts: it is the rate per unit, so it approximates the change from one additional unit. Economists call this the marginal revenue, and the small discrepancy is why the derivative is an approximation to the next unit's revenue rather than its exact value.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 199-200
Error analysis
A student is asked for the velocity of a falling body at t equal to 1 second.
Annotate
On: \( \frac{s(2) - s(1)}{2 - 1} = \frac{64 - 16}{1} = 48 \text{ ft/s} \)
An average rate uses a fixed interval; an instantaneous rate takes the limit as the interval shrinks to nothing. Whenever a question names a single moment rather than a span, the answer is a derivative.
Fill the middle
The falling body from the worked example, after cancelling h.
Fill in the blanks
\lim_32t(32t + 16h) = ___
Why: The term carrying h vanishes in the limit, leaving 32t as the velocity function. Evaluating it at any instant gives the speedometer reading then, with units of feet per second.
Matching
Output units over input units, every time.
Match the pairs
Why: The last row is a derivative of a derivative, and dividing by seconds a second time produces the squared unit — which is why acceleration is measured in metres per second squared. Units are a reliable check on any applied derivative, and they also reveal what a higher derivative must mean.
Prediction
Commit before reasoning.
Predict first
If R'(500) = 40 dollars per unit, what does that tell a manager?
Correct: Selling one more unit adds about 40 dollars.
\[ R(501) - R(500) = 39.98 \approx 40 = R'(500) \]
Why: A derivative is a rate per unit of input, so multiplying it by a one-unit change estimates the resulting output change. The actual change here is 39.98 dollars, so the estimate is very good but not exact — and that gap is precisely what Section 4.2's linear approximation quantifies. Total revenue at 500 units is 25,000 dollars and average revenue is 50 dollars per unit, so all three quantities differ and only the derivative answers the marginal question.
Section
Section 5
Concept
When no formula is available, a derivative can be estimated from a table by computing difference quotients over shrinking intervals on both sides. And a derivative need not exist: a corner, a vertical tangent or a discontinuity all defeat the limit.
differentiability — A function is differentiable at a point when the limit defining the derivative exists there. Differentiability requires continuity, but continuity is not enough — a corner is continuous and has no derivative.
\[ f \text{ differentiable at } a \;\Longrightarrow\; f \text{ continuous at } a \]
The implication runs one way only. Every differentiable function is continuous, since the quotient could not have a finite limit otherwise; but the absolute value is continuous everywhere and has no derivative at the origin.
Figure (svg): Two points where a derivative fails: a corner and a vertical tangent
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 200-202 — estimating derivatives and differentiability
Picture it
A corner and a vertical tangent, both on continuous graphs.
Figure (svg): Two points where a derivative fails: a corner and a vertical tangent
At the corner the one-sided quotients settle on negative 1 and 1, so the limit fails by a jump. At the vertical tangent they grow without bound. Both graphs are unbroken, which is exactly the point.
Worked example
Example 3.8. Both sides, shrinking intervals.
\[ \text{Given } f(1.9) = 3.61, f(1.99) = 3.9601, f(2) = 4, f(2.01) = 4.0401, \text{ estimate } f'(2). \]
Compute the quotient from the left
Why: Using the value at 1.99.
\[ \frac{4 - 3.9601}{0.01} = 3.99 \]
Compute a wider one from the left
Why: Using 1.9.
\[ \frac{4 - 3.61}{0.1} = 3.9 \]
Compute from the right
Why: Using 2.01.
\[ \frac{4.0401 - 4}{0.01} = 4.01 \]
Compare the two sides
Why: They straddle a common value.
\[ 3.99\text{ and } 4.01 \]
Read the estimate
Why: The trend from both sides.
\[ \text{about } 4 \]
Figure (svg): A table of difference quotients from both sides, closing on the derivative
\[ f'(2) \approx 4 \]
Verify: recognise the function and check
Why: The tabulated values are squares, so this is the squaring function, whose derivative at 2 is 4 exactly. The two-sided estimates 3.99 and 4.01 straddle it symmetrically, which is characteristic of a smooth function. Averaging the two gives 4.00, and that symmetric-difference trick is usually more accurate than either one-sided estimate — worth knowing when only data is available.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 201-202
Sorting
Check continuity first, then whether a single direction exists.
Sort into buckets
Sort each function at the point named.
The first three are all continuous at 0 and only one is differentiable, which is the whole point of the distinction. The reciprocal fails for a stronger reason — it is not continuous — and that failure is detectable without computing anything.
Worked example
Checkpoint 3.8. Continuous, and still not differentiable.
\[ \text{Show that } f(x) = |x| \text{ has no derivative at } 0. \]
Write the difference quotient at 0
Why: The value at 0 is 0.
\[ | h | / h \]
Evaluate for positive h
Why: The absolute value is h itself.
\[ = 1 \]
Evaluate for negative h
Why: The absolute value flips the sign.
\[ = -1 \]
Compare the one-sided limits
Why: One and negative 1.
Conclude
Why: The two-sided limit fails.
\[ \text{not differentiable at } 0 \]
Figure (svg): The solution to Worked example where the derivative fails shown as a ladder of expressions, one row per legal move
\[ \lim_{h \to 0^-}\frac{|h|}{h} = -1 \ne 1 = \lim_{h \to 0^+}\frac{|h|}{h} \]
Verify: confirm the function is nevertheless continuous
Why: The limit of the absolute value at 0 is 0, which equals its value there, so it is perfectly continuous — the graph is unbroken. What fails is the DERIVATIVE, because the graph changes direction abruptly and has no single tangent. This is the standard example proving that continuity does not imply differentiability, and it is the exact quotient Section 2.3 used as its jump example.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 202-202
Trap
\[ f(x) = |x| \text{ is continuous at } 0 \]
Conclude the derivative exists there
Why: The student treats continuity as sufficient.
\[ \text{so } f'(0) \text{ exists} \quad \text{(wrong)} \]
The graph has a sharp corner. Approaching from the left the slope is negative 1 and from the right it is 1, so no single tangent direction exists.
\[ \text{differentiable} \Rightarrow \text{continuous}, \quad \text{but not conversely} \]
Read the implication in the correct direction only
Why: A derivative requires more than an unbroken graph: it requires a single well-defined direction.
The correct implication is genuinely useful in the other form: a function that is DISCONTINUOUS at a point certainly has no derivative there, which rules out differentiability at every jump and every asymptote without any computation. Section 3.2 catalogues the three failure modes in full.
Fill the middle
The absolute value at the origin, approached from the left.
Fill in the blanks
\lim_-1\frac______ = ___
Why: For negative h the absolute value is negative h, so the quotient is negative 1. From the right it is 1, and the disagreement is exactly why no derivative exists at the corner.
Two truths and a lie
All three are about differentiability.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A vertical tangent means the difference quotient grows without bound, so the limit does not exist and the derivative is undefined — even though the curve has a perfectly definite direction there. The cube root at the origin is the standard example: the graph is smooth and unbroken, and the slope is infinite rather than a number.
Prediction
Commit before reasoning.
Predict first
A shipping cost function is continuous but has a corner at 10 kg. What can you not do there?
Correct: State a single marginal cost per kilogram at exactly 10 kg.
\[ \text{left rate} \ne \text{right rate} \;\Longrightarrow\; \text{no single marginal cost} \]
Why: The cost itself is perfectly well defined, and so is its limit — the function is continuous, so both exist and agree. What fails is the RATE: approaching from below gives one cost per kilogram and from above gives another, so there is no single marginal cost at the threshold. In practice you would quote the two one-sided rates instead, which is exactly the honest answer a corner permits. This is the same distinction as Section 2.4's transfer example about billing thresholds.
Comparison
Fill the blanks. Every row is the same limit.
Comparison matrix
| Reading | What it measures | Units |
|---|---|---|
| Tangent slope | the curve's direction at a point | output units per input unit |
| Instantaneous rate | how fast the output changes per unit input | output units per input unit |
| Velocity | how fast a position changes | metres per second |
| Marginal quantity | the effect of one more unit | dollars per unit |
These are not four related quantities but one quantity with four names, which is why a single definition serves geometry, physics and economics at once.
Pattern
Given a function and a point, find the derivative from the definition.
For a tangent line, do this once for the slope and evaluate the original function once for the point. They are two separate computations, and confusing them is the commonest error in the section.
Stewart, Calculus: Early Transcendentals 8e, §2.7 Derivatives and Rates of Change §2.7, pp. 140-151
Check
From the definition. Expand and cancel.
Check your understanding
For f(x) = x^2, what is f'(3)?
Answer: A
Why: The quotient simplifies to 6 + h, whose limit is 6.
Check
Tangent lines. Two separate numbers.
Check your understanding
For f(x) = x^2 - 4x + 6 with f(1) = 3 and f'(1) = -2, what is the tangent at x = 1?
Answer: A
Why: Point-slope from (1, 3) with slope -2 gives y - 3 = -2(x-1), so y = -2x + 5.
Check
Differentiability. Continuity is not enough.
Check your understanding
Is f(x) = |x| differentiable at 0?
Answer: A
Why: The quotient is -1 for negative h and 1 for positive h, so the limit does not exist.
Real world
A drug's concentration in the bloodstream, in milligrams per litre, is measured every fifteen minutes after a dose. The readings at 45, 60 and 75 minutes are 8.2, 7.1 and 6.3.
Discussion prompt
Estimate the rate at which the concentration is changing at 60 minutes, give its units, and say why a clinician would want it rather than the concentration itself.
Hint: This is a derivative estimated from a table, with both sides available.
Answer:
Compute the difference quotients on each side of 60 minutes:
\[ \text{left: } \frac{7.1 - 8.2}{60-45} = -0.0733, \qquad \text{right: } \frac{6.3 - 7.1}{75-60} = -0.0533 \]
Averaging the two gives about negative 0.063 milligrams per litre per minute — the symmetric estimate, which is generally better than either side alone. Equivalently, about 3.8 mg/L per hour, falling.
The units are the giveaway for what has been computed: milligrams per litre per minute is a rate, not a level. The negative sign says the drug is clearing.
A clinician wants the rate because it predicts when the concentration will fall below the therapeutic threshold. A single reading of 7.1 says nothing about how long it will stay useful; a rate of negative 0.063 per minute says roughly 17 minutes until it reaches 6.0, which is when the next dose must be scheduled.
The honest caveat is the one from Section 2.1: with readings every fifteen minutes, this is an estimate over a fairly wide interval, and the concentration curve is bending throughout it. The two one-sided estimates differ by about 30 percent, which is a direct measure of how much the rate itself is changing — and that difference is the second derivative making itself felt.
Commit first
Answer, then rate your confidence honestly.
Predict first
What is the relationship between continuity and differentiability?
Correct: Differentiable implies continuous, but not conversely.
\[ \text{differentiable} \Rightarrow \text{continuous}; \quad |x| \text{ at } 0 \text{ blocks the converse} \]
Why: If the difference quotient has a finite limit, the numerator must vanish as h does, which forces the function to be continuous. The converse fails at any corner: the absolute value is continuous at the origin and has one-sided slopes of negative 1 and 1, so no derivative. The useful direction in practice is the contrapositive — a function discontinuous at a point certainly has no derivative there, which rules out differentiability at every jump and asymptote instantly.
Explain it
They ask why the definition bothers with a limit when you could just make h very small.
Discussion prompt
In four sentences or fewer, explain why h must go to zero as a limit rather than being set to a small number.
Hint: Ask what answer a specific small h gives.
Answer:
Ask them to pick a small h — say a thousandth — for the squaring function at 3. They get 6.001, which is close to 6 but not 6, and picking a smaller h gives a different answer again. Every choice gives a different number, so no choice is THE answer.
Setting h to zero exactly is worse: the quotient becomes zero over zero and gives nothing at all. The limit is the only thing that picks out a single number, and it does so precisely because it uses every small h at once rather than any particular one.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the algebra, identify the shape first — power, root or reciprocal — and use the matching preparation. For tangent lines, write the two numbers on separate lines and label them height and slope. For interpretation, always attach output units over input units and read the sign. For existence, check continuity first, then whether the two one-sided quotients agree and stay bounded. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a curve with a point marked, a second point h to its right, and the secant between them, labelling the rise, the run and the quotient; then draw the tangent and write the definition beneath. Below that, write both forms of the definition side by side and the substitution linking them. In the middle of the page, compute three derivatives from the definition in full: the squaring function at 3, the square root at 4, and the reciprocal at 2, showing which preparation each needed. Beside each, note the sign of the answer and whether the graph rising or falling agrees. Then take the parabola x squared minus 4x plus 6 at x equal to 1, compute f of 1 and f prime of 1 as two clearly separate numbers, and write the tangent line. At the bottom, draw the absolute value and the cube root, mark the origin on each, and write one sentence per graph saying which part of the definition fails there. In a margin, write the units of a derivative in general form.
If any of your three derivative computations did not pass through a stage where the constant terms cancelled, check it — that cancellation is guaranteed by the structure of the quotient, and its absence means an algebra slip rather than an unusual function.
Recap
Five things, and the first is the definition the whole of Chapter 2 was built to make legitimate.
| If you see | Then |
|---|---|
| A binomial power | Expand; the constants cancel |
| A difference of roots | Rationalise the numerator |
| A difference of fractions | Combine over a common denominator |
| A request for a tangent line | Compute f(a) and f'(a) separately |
| A single moment named | The answer is a derivative |
| A corner in the graph | No derivative, though the function is continuous |
| A discontinuity | No derivative, and no computation needed to say so |
Section 3.2 stops computing derivatives one point at a time. Treating the derivative as a function of the point produces a new function, and reading its graph from the original's is the skill that makes all of Chapter 4 possible.
OpenStax Calculus Volume 1, §3.1 Defining the Derivative §3.1, pp. 188-202 — everything on these slides traces back here
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