The derivative as a function in its own right, sketching the graph of f prime from the graph of f, the Lagrange and Leibniz notations, the theorem that differentiability implies continuity and its useful contrapositive, the three ways a derivative fails, and higher-order derivatives with their interpretation.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
The Derivative as a Function
Objectives
Five outcomes. The second is a graphical skill, and it is the one that makes Chapter 4 readable rather than mechanical.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 203-215 — the section these objectives are drawn from
Warm-up
Section 3.1 computed the derivative of the squaring function at 3, getting 6. It could equally have been computed at 5, or at negative 2.
Discussion prompt
Run the definition for the squaring function at a general input a rather than at a specific number. What comes out?
Hint: Expand, cancel and take the limit exactly as before, but keep a as a letter.
Answer:
\[ \frac{(a+h)^2 - a^2}{h} = \frac{2ah + h^2}{h} = 2a + h \;\longrightarrow\; 2a \]
The answer is 2a — not a number but a rule, giving the slope at every input at once. Computing at 3 was a special case of it.
So the derivative is not merely a number attached to a point; it is a function derived from the original. That shift is what this section is about, and it is what makes it possible to ask questions about where a function rises, turns and bends.
Concept
Letting the point vary turns the derivative into a new function whose input is a point and whose output is the slope there. Its domain is every input at which the original is differentiable, which may be smaller than the original's domain.
the derivative function — The function f prime whose value at x is the limit of the difference quotient at x, defined at every input where that limit exists. Its domain is a subset of the original function's domain.
\[ f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \]
The domain can genuinely shrink. The absolute value is defined everywhere and its derivative is defined everywhere except the origin, so f prime lives on a strictly smaller set than f does.
Figure (svg): A cubic above its derivative, with the turning points of one aligned to the zeros of the other
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 203-206
Section
Section 1
Concept
The derivative function is found by running the difference quotient with a general input instead of a specific one. The algebra is identical to Section 3.1; only the bookkeeping changes.
differentiating — The process of producing the derivative function from a given function. A function that has a derivative at every point of an interval is called differentiable on that interval.
\[ f(x) = x^2 \;\Longrightarrow\; f'(x) = 2x \]
Doing it once with a letter replaces doing it forever with numbers, which is the same economy that produced the quadratic formula in algebra. Section 3.3 takes the idea further and does it once per family.
Figure (svg): The rules for sketching a derivative from a graph, each stated as a correspondence
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 203-207 — derivative functions
Picture it
The correspondence, stated five ways.
Figure (svg): The rules for sketching a derivative from a graph, each stated as a correspondence
Every row is the same sentence: the derivative's height records the original's steepness, with the sign recording direction. Holding that one sentence makes all five rows automatic.
Worked example
Example 3.9. Expand, cancel, and keep x as a letter.
\[ \text{Find } f'(x) \text{ for } f(x) = x^3. \]
Write the quotient at a general x
Why: The definition.
\[ \frac{(x + h) ^{3} - x ^{3}}{h} \]
Expand the cube
Why: x cubed plus 3x squared h plus 3x h squared plus h cubed.
\[ \frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} \]
Note the x cubed terms cancel
Why: As they always must.
Factor out h and cancel
Why: Legal because h is not zero.
\[ 3 x ^{2} + 3 x h + h ^{2} \]
Take the limit
Why: Every term with h vanishes.
\[ f'(x) = 3 x ^{2} \]
Figure (svg): The solution to Worked example the derivative of a cubic shown as a ladder of expressions, one row per legal move
\[ f'(x) = 3x^2 \]
Verify: check at a point against the graph, and note the pattern
Why: At x equal to 2 the derivative is 12, and the cubic is indeed steeply rising there. At x equal to 0 it is 0, matching the horizontal tangent the cubic has at the origin. And the answer is never negative, which agrees with the cubic being increasing everywhere. The pattern so far — x squared gives 2x, x cubed gives 3x squared — is the power rule appearing, and Section 3.3 will prove it in general.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 205-206
Matching
Each computed from the definition.
Match the pairs
Why: The first two show the power pattern emerging. The last two are worth noting for their domains: both derivatives exclude a point the original may or may not have included, and the reciprocal's derivative is negative everywhere, matching a decreasing graph on each branch.
Worked example
Checkpoint 3.9. The derivative can live on less ground.
\[ \text{Find } f'(x) \text{ for } f(x) = \sqrt{x} \text{ and compare the domains.} \]
Write the quotient and rationalise
Why: Multiply by the conjugate over itself.
\[ \frac{1}{\sqrt{x + h} + \sqrt{x}} \]
Take the limit
Why: Both roots approach the root of x.
\[ f'(x) = \frac{1}{2 \sqrt{x}} \]
State the original's domain
Why: The radicand must be non-negative.
\[ f\text{ on } [0, \infty] \]
State the derivative's domain
Why: The denominator must also be non-zero.
\[ \text{f' on } (0, \infty) \]
Figure (svg): The solution to Worked example a derivative with a smaller domain shown as a ladder of expressions, one row per legal move
\[ f'(x) = \frac{1}{2\sqrt{x}}, \quad D_{f'} = (0, \infty) \]
Verify: check what happens at the excluded endpoint
Why: At x equal to 0 the formula would divide by zero, and the graph confirms why: the square root has a vertical tangent at the origin, so the difference quotient grows without bound rather than settling. The function is defined and continuous at 0, and its derivative is not — a concrete case of the domain shrinking. Every derivative domain must be checked separately rather than inherited.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 206-207
Trap
\[ f(x) = \sqrt{x} \text{ on } [0, \infty) \;\Longrightarrow\; f'(x) = \frac{1}{2\sqrt{x}} \text{ on } [0,\infty) \]
Carry the original's domain across to the derivative
Why: The student assumes the two match.
\[ f'(0) = \frac{1}{0} \quad \text{(undefined)} \]
The formula divides by zero at the endpoint, and the graph has a vertical tangent there, so the derivative genuinely does not exist at 0.
\[ D_f = [0,\infty), \qquad D_{f'} = (0,\infty) \]
Determine the derivative's domain from the derivative
Why: It is where the LIMIT exists, which may be a strictly smaller set.
The absolute value is the sharper example: it is differentiable on every real number except one, so its derivative's domain has a point removed from the middle rather than from an end. Checking the derivative's own domain is not pedantry — Chapter 4 will hunt for extreme values at exactly the points where a derivative fails to exist.
Fill the middle
The cubic's difference quotient, after cancelling h.
Fill in the blanks
\lim_3x^2\left(3x^2 + 3xh + h^2\right) = ___
Why: The two terms containing h vanish, leaving 3x squared. This is the derivative function, giving the slope at every input at once rather than at one point.
Sorting
Compare where f is defined with where f prime is.
Sort into buckets
Sort each function.
Polynomials are the well-behaved family: their derivatives are defined wherever they are, which is everywhere. Roots and absolute values each lose exactly one point, and those lost points are precisely where Chapter 4 will look for extreme values.
Prediction
Commit before reasoning.
Predict first
What is different about computing f'(x) rather than f'(3)?
Correct: Nothing in the method: the point stays a letter, so the output is a rule rather than a number.
\[ f'(3) = 6 \text{ is one value of } f'(x) = 2x \]
Why: The same quotient, the same cancellation, the same limit as h approaches zero — only the bookkeeping differs. What is gained is enormous, though: one computation now answers the question at every input, and the result is an object that can itself be graphed, differentiated again, or set equal to zero. That last move is what Chapter 4 is built on.
Section
Section 2
Concept
Read the original graph's slope at each input and plot that value. Where the curve rises the derivative is positive, where it falls the derivative is negative, and where it has a horizontal tangent the derivative is zero.
graphical differentiation — Constructing the graph of f prime by reading slopes off the graph of f: sign from the direction of travel, magnitude from the steepness, and zeros at horizontal tangents.
\[ f \text{ increasing} \iff f' > 0; \qquad f \text{ decreasing} \iff f' < 0 \]
Stacking the two graphs vertically with the same horizontal scale is what makes the reading reliable. Every feature of one then sits directly above or below the corresponding feature of the other.
Figure (svg): A cubic above its derivative, with the turning points of one aligned to the zeros of the other
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 207-210 — graphing a derivative
Picture it
A cubic stacked above its derivative.
Figure (svg): A cubic above its derivative, with the turning points of one aligned to the zeros of the other
The maximum and the minimum of the cubic sit directly above the two zeros of the parabola beneath. That vertical alignment is the whole technique, and it works in both directions.
Worked example
Example 3.11. Read the sign, then the size.
\[ \text{Sketch } f' \text{ for } f(x) = x^3 - 3x \text{ from its graph.} \]
Find where the tangent is horizontal
Why: The turning points, at negative 1 and 1.
\[ \text{f' } = 0\text{ at } x = -1\text{ and } 1 \]
Read the sign on the far left
Why: The curve is rising steeply.
\[ \text{f' } > 0\text{ for } x < -1 \]
Read the sign in the middle
Why: The curve falls between the turning points.
\[ \text{f' } < 0\text{ on } (-1, 1) \]
Read the sign on the right
Why: Rising again, and steepening.
\[ \text{f' } > 0\text{ for } x > 1 \]
Assemble
Why: Positive, zero, negative, zero, positive: an upward parabola.
\[ \text{f' is } a\text{ parabola with roots at } +- 1 \]
Figure (svg): A cubic above its derivative, with the turning points of one aligned to the zeros of the other
\[ f'(x) = 3x^2 - 3 \]
Verify: compute it and compare with the sketch
Why: Differentiating gives 3x squared minus 3, which is an upward parabola with zeros at plus and minus 1 — exactly what the graphical reading produced. The value at 0 is negative 3, and the cubic is indeed falling most steeply at the origin. The two routes agreeing is the point: the sketch can be made without the algebra, and Chapter 4 will often have only the sketch.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 208-209
Matching
The correspondence, item by item.
Match the pairs
Why: The third row is the one that surprises people: a straight line has the same slope everywhere, so its derivative is a horizontal line, not a slanted one. Confusing the graph of f with the graph of f prime is the standard error, and the constant-derivative case is the fastest way to catch it.
Worked example
Checkpoint 3.11. The correspondence runs both ways.
\[ \text{Given that } f' \text{ is positive on } (-\infty, 2), \text{ zero at } 2, \text{ and negative after, describe } f. \]
Translate the positive stretch
Why: A positive derivative means a rising function.
\[ f\text{ increases up to } 2 \]
Translate the zero
Why: A zero derivative means a horizontal tangent.
\[ f\text{ has } a\text{ horizontal tangent at } 2 \]
Translate the negative stretch
Why: A negative derivative means a falling function.
\[ f\text{ decreases after } 2 \]
Combine
Why: Rising, then turning, then falling.
\[ f\text{ has } a\text{ maximum at } x = 2 \]
Figure (svg): The solution to Worked example reading f from f prime shown as a ladder of expressions, one row per legal move
\[ f \text{ has a maximum at } x = 2 \]
Verify: check that the sign change, not the zero, identifies the maximum
Why: A zero derivative alone does not give a maximum: the cubic x cubed has a zero derivative at the origin and no turning point at all, because the derivative does not change sign there. What identifies the maximum is the derivative going from positive to negative. This distinction is exactly the first derivative test, stated properly in Section 4.5, and getting it right here prevents the commonest error there.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 209-210
Error analysis
A student analyses a function whose derivative vanishes at the origin.
Annotate
On: \( f(x) = x^3: \; f'(0) = 0 \;\Longrightarrow\; f \text{ has a maximum or minimum at } 0 \)
A zero derivative marks a horizontal tangent and nothing more. Whether it is a maximum, a minimum or neither is decided by what the sign does on either side, which is why sketching f prime is more informative than solving f prime equals zero.
Fill the middle
The cubic from the worked example, whose turning points sit above the derivative's roots.
Fill in the blanks
f(x) = x^3 - 3x \text\pm 1 x = -1 \text___ x = 1 \;\Longrightarrow\; f'(x) = 0 \text___ x = ___
Why: The turning points of f are exactly the zeros of f prime, because a turning point is where the tangent is horizontal. Stacking the graphs makes this vertical alignment visible at a glance.
Sorting
Read the original's behaviour and translate.
Sort into buckets
Sort each description of f.
The two falling cases share a bucket because sign and magnitude are separate readings: the sign says which side of the axis, the steepness says how far. Doing them as two questions rather than one makes the sketch far more reliable.
Prediction
Commit before reasoning.
Predict first
If f is the line y = 3x - 5, what does the graph of f prime look like?
Correct: The horizontal line y equals 3.
\[ f(x) = 3x - 5 \;\Longrightarrow\; f'(x) = 3 \text{ for every } x \]
Why: The slope of a straight line is the same at every input, so the derivative function is constant — and the constant is the slope, 3. Its graph is horizontal, not slanted. This is the cleanest test of whether the distinction between f and f prime has landed: the two graphs here look nothing alike, and a student who draws a slanted line for f prime is still graphing f. The intercept has no effect at all, which is why the derivative cannot recover it.
Section
Section 3
Concept
Lagrange notation writes f prime and is compact, which suits stating rules. Leibniz notation writes dy by dx and names both variables, which suits rates of change and, later, the chain rule and substitution.
Leibniz notation — The derivative of y with respect to x written as dy over dx. It is a single symbol, not a fraction, but it displays which variable is the input and which the output — and it behaves like a fraction often enough to be a genuine aid.
\[ f'(x) = \frac{dy}{dx} = \frac{d}{dx}\left[f(x)\right] = y' \]
To indicate a derivative at a specific point in Leibniz notation, a vertical bar with the input is used. That is more cumbersome than f prime of a, which is exactly why both systems survive.
Figure (svg): The four common notations for a derivative, with what each is good for
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 210-212 — derivative notation
Picture it
The notations and what each is good for.
Figure (svg): The four common notations for a derivative, with what each is good for
Leibniz notation carries the units with it: a derivative of metres with respect to seconds visibly has units of metres per second. That is why every applied rate problem in Chapter 4 uses it.
Worked example
Example 3.12. The same statement, four ways.
\[ \text{Write the derivative of } y = x^3 \text{ at } x = 2 \text{ in each notation.} \]
Lagrange, as a function
Why: Prime on the function name.
\[ f'(x) = 3 x ^{2} \]
Lagrange, at the point
Why: Evaluate.
\[ f'(2) = 12 \]
Leibniz, as a function
Why: Derivative of y with respect to x.
\[ \,dy / \,dx = 3 x ^{2} \]
Leibniz, at the point
Why: A vertical bar carries the input.
\[ \,dy / \,dx\text{ at } x = 2\text{ equals } 12 \]
Figure (svg): The solution to Worked example translating between notations shown as a ladder of expressions, one row per legal move
\[ f'(2) = \left.\frac{dy}{dx}\right|_{x=2} = 12 \]
Verify: note which notation is shorter for which job
Why: For stating the rule, Lagrange is shorter. For evaluating at a point, Lagrange is much shorter — the Leibniz bar is cumbersome. But for a rate problem where the variables are volume and time rather than y and x, Leibniz is far clearer: dV by dt says what is changing with respect to what, while V prime leaves the input to be inferred. Both survive because neither wins outright.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 211-211
Matching
Read each aloud.
Match the pairs
Why: The last is an operator rather than a value — it names the operation being applied, which is why it is the natural way to state a rule. The third shows Leibniz notation doing its real work: two named variables and, for free, the units of the answer.
Worked example
Checkpoint 3.12. Where Leibniz earns its keep.
\[ \text{A balloon's volume } V \text{ in cm}^3 \text{ changes with time } t \text{ in seconds. Interpret } \frac{dV}{dt} = 12. \]
Read the notation aloud
Why: The derivative of volume with respect to time.
Read the units off the notation
Why: Output units over input units.
\[ \text{cm} ^{3}\text{ per second} \]
Read the sign
Why: Positive means the output is increasing.
State the interpretation
Why: Together.
\[ \text{growing at } 12 \text{cm} ^{3}\text{ per second} \]
Figure (svg): The solution to Worked example notation carrying the units shown as a ladder of expressions, one row per legal move
\[ \frac{dV}{dt} = 12 \text{ cm}^3\text{/s} \]
Verify: compare with what V prime would have conveyed
Why: Writing V prime equals 12 leaves the input variable unstated — it could be a rate with respect to time, or with respect to the radius, and those are completely different quantities. The Leibniz form makes it unambiguous and hands you the units for free. This is why Section 4.1's related rates problems, which juggle several variables at once, are written in Leibniz notation throughout.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 212-212
Trap
\[ \frac{dy}{dx} = 3x^2 \]
Cancel the d's
Why: The student treats the symbol as a genuine quotient.
\[ \frac{y}{x} = 3x^2 \quad \text{(meaningless)} \]
The d is not a quantity multiplying y, so there is nothing to cancel. The whole symbol is a single piece of notation for one limit.
\[ \frac{dy}{dx} \text{ is one symbol denoting } \lim_{h \to 0}\frac{\Delta y}{\Delta x} \]
Read it as a name for a limit, not as a ratio
Why: It was designed to LOOK like the quotient it is the limit of, which is a genuine aid and also the source of the trap.
The notation does behave like a fraction in several important places — the chain rule in Section 3.6 and substitution in Section 5.5 both exploit that, and Leibniz designed it so. But those are theorems that must be proved, not consequences of the symbol's shape.
Fill the middle
A balloon whose volume in cubic centimetres changes with time in seconds.
Fill in the blanks
\fracsecond___ = 12 \;\Longrightarrow\; \text___^3 \text___ ___
Why: The units are output over input, which the notation displays directly: cubic centimetres per second. This is Leibniz notation's practical advantage and the reason applied problems are written in it.
Two truths and a lie
All three are about notation.
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. The symbol is a single unit of notation denoting a limit; the d is not a factor. It was deliberately shaped to resemble the quotient of small changes it is the limit of, and it does behave like a fraction in the chain rule and in substitution — but those are proved theorems, not permissions granted by the notation's appearance.
Prediction
Commit before reasoning.
Predict first
Why has neither notation replaced the other in three centuries?
Correct: Each is better at a different job.
\[ \text{stating a rule: } (x^n)' = nx^{n-1}; \quad \text{tracking a rate: } \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Why: Lagrange's prime is compact, which is what a table of rules needs — the power rule is unreadable in Leibniz form repeated fifty times. Leibniz's form names both variables, which is what a related rates problem with four interlinked quantities needs, and it hands you the units. Neither is more rigorous; they denote the same limit. The historical rivalry is real but is not why both survived, and a working mathematician switches between them within a single page.
Section
Section 4
Concept
Every differentiable function is continuous, because the difference quotient could not have a finite limit otherwise. The converse fails, and there are exactly three ways: a corner, a vertical tangent, and a discontinuity.
the three failures — A derivative fails to exist at a corner, where the one-sided quotients differ; at a vertical tangent or cusp, where the quotient grows without bound; and at any discontinuity, where the function is not even continuous.
\[ f \text{ differentiable at } a \;\Longrightarrow\; f \text{ continuous at } a \]
The contrapositive is the version that saves work. A function discontinuous at a point cannot be differentiable there, so every jump and every asymptote is ruled out by inspection, with no quotient computed.
Figure (svg): The three ways a derivative fails: a corner, a cusp with a vertical tangent, and a discontinuity
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 212-214 — differentiability and continuity
Picture it
A corner, a cusp and a jump.
Figure (svg): The three ways a derivative fails: a corner, a cusp with a vertical tangent, and a discontinuity
Only the third is caught by continuity. The first two have perfectly unbroken graphs and no derivative, which is why differentiability is a strictly stronger condition and worth a separate name.
Worked example
Example 3.14. The proof is three lines and worth seeing.
\[ \text{Show that if } f'(a) \text{ exists then } f \text{ is continuous at } a. \]
Write the change in output as a product
Why: Multiply and divide by the change in input.
\[ f(x) - f(a) = (\frac{f(x) - f(a)}{x - a}) (x - a) \]
Take the limit of each factor
Why: The first is the derivative; the second goes to zero.
\[ \to f'(a) \times 0 \]
Evaluate the product
Why: A finite number times zero.
\[ = 0 \]
Interpret
Why: The output change vanishes as the input approaches a.
\[ \lim f(x) = f(a) \]
Figure (svg): The solution to Worked example proving differentiability implies continuity shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to a}f(x) = f(a) \]
Verify: identify exactly where differentiability was used
Why: It was used in step two, to say the first factor has a FINITE limit. Had the quotient been unbounded, the product of an unbounded quantity with a vanishing one would be indeterminate and nothing would follow — which is precisely the situation at a vertical tangent, where the function is continuous anyway but the argument does not deliver it. So the proof genuinely needs the hypothesis, and the multiply-and-divide trick in step one is the whole idea.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 213-213
Sorting
Check continuity, then the one-sided quotients.
Sort into buckets
Sort each function at the origin.
Only the last bucket is detectable without touching the difference quotient. The first two are the reason differentiability needs a separate definition from continuity, and both live on perfectly unbroken graphs.
Worked example
Checkpoint 3.14. Which of the three is it?
\[ \text{Why is } f(x) = x^{2/3} \text{ not differentiable at } 0? \]
Check continuity first
Why: The function is defined and its limit at 0 is 0.
\[ \text{continuous at } 0 \]
Write the difference quotient
Why: With the value at 0 being 0.
\[ h ^{\frac{2}{3}} / h = h ^{-\frac{1}{3}} \]
Examine the behaviour as h shrinks
Why: A negative power grows without bound.
\[ | h ^{-\frac{1}{3}} | \to \infty \]
Check the two sides
Why: The sign differs, and both are unbounded.
Classify
Why: Unbounded with opposite signs.
Figure (svg): The solution to Worked example classifying a failure shown as a ladder of expressions, one row per legal move
\[ \lim_{h \to 0}h^{-1/3} \text{ does not exist} \]
Verify: distinguish a cusp from a plain vertical tangent
Why: The cube root of x also has an unbounded quotient at 0, but there both sides go to POSITIVE infinity, so the tangent is a single vertical line. Here the two sides go to opposite infinities, so the curve comes to a sharp point — a cusp. Both fail to be differentiable, and the distinction matters for sketching: a vertical tangent is smooth, a cusp is not.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 214-214
Trap
\[ f \text{ is continuous at } a \]
Conclude the derivative exists there
Why: The student reverses the theorem.
\[ \text{so } f'(a) \text{ exists} \quad \text{(wrong)} \]
The absolute value, the cube root and the two-thirds power are all continuous at the origin and none of them is differentiable there.
\[ \text{differentiable} \Rightarrow \text{continuous}, \quad \text{and the useful form is the contrapositive} \]
Use it as: NOT continuous implies NOT differentiable
Why: That direction is valid and eliminates work.
The contrapositive is genuinely labour-saving. At any jump, any asymptote and any removable hole, the derivative fails immediately with no quotient to compute. What continuity cannot do is certify a derivative, and the three continuous counterexamples are worth keeping in mind for exactly that reason.
Fill the middle
The multiply-and-divide step in the proof that differentiability implies continuity.
Fill in the blanks
f(x) - f(a) = \frac0___\cdot(x-a) \;\longrightarrow\; f'(a)\cdot 0 = ___
Why: The product tends to zero, so the output change vanishes and the limit equals the value — which is continuity. The step needs f prime of a to be FINITE, which is exactly where the hypothesis is used.
Two truths and a lie
All three are about the relationship.
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. The cube root has a vertical tangent at the origin and is perfectly continuous — its graph is unbroken and passes through the point. What fails is only the derivative, because the slope grows without bound rather than settling. Continuity and differentiability fail independently, which is the section's central point.
Prediction
Commit before reasoning.
Predict first
You are told a function has a jump at x = 4. What follows immediately?
Correct: It has no derivative at 4 — the contrapositive settles it instantly.
\[ \text{not continuous at } 4 \;\Longrightarrow\; \text{not differentiable at } 4 \]
Why: Differentiability implies continuity, so failing continuity forces failing differentiability. No difference quotient need be written. This is the direction of the theorem that does practical work: it lets you rule out differentiability at every jump, asymptote and hole by inspection. The other direction certifies nothing, which is why the three continuous counterexamples matter so much.
Section
Section 5
Concept
The derivative is a function, so it can be differentiated again. The second derivative measures the rate at which the rate is changing, which for a position function is the acceleration and for a graph is the bending.
second derivative — The derivative of the derivative, written f double prime or d squared y by dx squared. It measures how fast the first derivative is changing, and its sign records whether the graph bends upward or downward.
\[ f''(x) = \frac{d}{dx}\left[f'(x)\right] = \frac{d^2 y}{dx^2} \]
The Leibniz notation for the second derivative puts the exponent in two different places, which looks odd until you see it as d applied twice to y, over dx multiplied by itself. It is a notation, not an algebraic identity.
Figure (svg): Position, velocity and acceleration stacked, each the derivative of the one above
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 214-215 — higher-order derivatives
Picture it
Three graphs, each the derivative of the one above.
Figure (svg): Position, velocity and acceleration stacked, each the derivative of the one above
At the marked instant the velocity is zero and the acceleration is positive, which means the position is at a minimum. Reading two derivatives together like this is the second derivative test of Section 4.5.
Worked example
Example 3.16. Differentiate repeatedly.
\[ \text{For } f(x) = 2x^3 - 5x, \text{ find } f', f'', f''' \text{ and } f^{(4)}. \]
Differentiate once
Why: Using the pattern from earlier: x cubed gives 3x squared.
\[ f'(x) = 6 x ^{2} - 5 \]
Differentiate again
Why: The constant contributes nothing.
\[ f''(x) = 12 x \]
Differentiate a third time
Why: The coefficient survives.
\[ f'''(x) = 12 \]
Differentiate once more
Why: A constant has zero derivative.
\[ f ^{4}(x) = 0 \]
Note the pattern
Why: Every further derivative vanishes.
\[ \text{all higher ones are } 0 \]
Figure (svg): The solution to Worked example successive derivatives shown as a ladder of expressions, one row per legal move
\[ f' = 6x^2-5, \; f'' = 12x, \; f''' = 12, \; f^{(4)} = 0 \]
Verify: check the pattern against the degree
Why: Each differentiation lowers the degree by one, so a cubic runs out after four steps — which is exactly what happened. In general the nth derivative of a degree-n polynomial is a constant and everything beyond it is zero. That is a useful structural check: if a fourth derivative of a cubic came out non-zero, something has gone wrong earlier.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 215-215
Matching
Each one describes the one above.
Match the pairs
Why: Each row describes the row above it, which is the whole idea of a higher derivative. The last row is the geometric reading that Section 4.5 will turn into a test for maxima and minima, and it is why the second derivative is worth computing even when there is no physical motion involved.
Worked example
Checkpoint 3.16. Two derivatives read together.
\[ \text{A particle has } s(t) = t^3 - 3t^2 + 2. \text{ Find } v \text{ and } a \text{ at } t = 1 \text{ and interpret.} \]
Differentiate for the velocity
Why: The rate of position change.
\[ v(t) = 3 t ^{2} - 6 t \]
Evaluate at the instant
Why: Three minus 6.
\[ v(1) = -3 \]
Differentiate again for the acceleration
Why: The rate of velocity change.
\[ a(t) = 6 t - 6 \]
Evaluate
Why: Six minus 6.
\[ a(1) = 0 \]
Interpret both together
Why: Moving backwards, momentarily at constant speed.
\[ \text{velocity } -3,\text{ not changing} \]
Figure (svg): The solution to Worked example interpreting the second derivative shown as a ladder of expressions, one row per legal move
\[ v(1) = -3, \qquad a(1) = 0 \]
Verify: check the units and what a zero acceleration means
Why: Velocity carries units of distance per time and acceleration distance per time squared, since differentiating divides by time a second time. A zero acceleration does NOT mean the particle has stopped — its velocity is negative 3, so it is moving briskly backwards; what is momentarily zero is the CHANGE in velocity. Confusing zero acceleration with zero velocity is the standard error, and reading the two numbers as answers to different questions prevents it.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 215-215
Error analysis
A student analyses a particle at an instant where the acceleration vanishes.
Annotate
On: \( a(1) = 0 \;\Longrightarrow\; \text{the particle is at rest at } t = 1 \)
At rest means the velocity is zero, which is a statement about the FIRST derivative. Each derivative answers a question about the one below it, and reading down the chain carefully is what keeps them apart.
Fill the middle
The cubic from the worked example, differentiated once already.
Fill in the blanks
f'(x) = 6x^2 - 5 \;\Longrightarrow\; f''(x) = 12x
Why: Differentiating 6x squared gives 12x, and the constant negative 5 contributes nothing. Each differentiation lowers the degree by one, so a cubic reaches zero after four steps.
Sorting
Read what the question is asking about.
Sort into buckets
Sort each question about a moving particle.
Items b and d both concern the first derivative but ask different things of it: one reads its sign, the other its zeros. That the same function answers several questions depending on what you read off it is the pattern the whole of Chapter 4 exploits.
Prediction
Commit before reasoning.
Predict first
How many non-zero derivatives does a polynomial of degree 5 have?
Correct: Five — the fifth derivative is a constant and the sixth is zero.
\[ \deg f = 5 \;\Longrightarrow\; f^{(5)} \text{ constant}, \; f^{(6)} = 0 \]
Why: Each differentiation lowers the degree by exactly one, so degree 5 becomes 4, then 3, 2, 1, and finally 0 — a constant — at the fifth step. The sixth derivative is zero and so is everything beyond. The coefficients affect the values but not the count, since the leading term's degree is what drives the process. This is a useful structural check on any repeated differentiation, and it is also why polynomials are the easiest functions to work with in Chapter 4.
Comparison
Fill the blanks. Each column answers a different question about the same graph.
Comparison matrix
| What you want to know | Which function | What to read |
|---|---|---|
| Where the graph is | f | its value |
| Whether it rises or falls | f' | the sign |
| Where it turns | f' | where the sign changes |
| How it bends | f'' | the sign |
The third row is the one to be careful with: a turning point needs the derivative's sign to CHANGE, not merely to vanish. The cubic at the origin is the counterexample worth remembering.
Pattern
Given the graph of a function, produce the graph of its derivative.
Doing sign and magnitude as two separate readings is what makes this reliable. Trying to judge both at once is where most inaccurate sketches come from.
Stewart, Calculus: Early Transcendentals 8e, §2.8 The Derivative as a Function §2.8, pp. 152-164
Check
The derivative function. Keep the point as a letter.
Check your understanding
For f(x) = x^3, what is f'(x)?
Answer: A
Why: The quotient simplifies to 3x^2 + 3xh + h^2, whose limit as h approaches 0 is 3x^2.
Check
Reading a graph. Straight lines have constant slope.
Check your understanding
If f is the line y = 3x - 5, what is the graph of f prime?
Answer: A
Why: A line has the same slope everywhere, so its derivative is the constant 3.
Check
Differentiability. Continuity is necessary, not sufficient.
Check your understanding
A function has a jump at x = 4. What can you conclude about f'(4)?
Answer: A
Why: Differentiability implies continuity, so a discontinuity rules out a derivative immediately.
Real world
A news report states that the rate of inflation is still positive but has been falling for three months. A second report says a spacecraft's engines have shut down, so its acceleration is now zero.
Discussion prompt
For the first, say what is true of the price level, its first derivative and its second derivative. For the second, say whether the spacecraft has stopped, and why the two situations are the same mathematics.
Hint: Each statement is about a different rung of the derivative ladder.
Answer:
Inflation. The price level is P. Inflation is the rate at which prices rise, which is P prime. So 'inflation is positive' says P prime is above zero — prices are still going up. 'Inflation is falling' says P prime is decreasing, which means P double prime is negative.
\[ P' > 0 \text{ and } P'' < 0: \quad \text{prices rising, but less quickly each month} \]
So prices are not falling. A great many people read 'falling inflation' as 'falling prices', and the error is exactly one rung on the ladder — mistaking a statement about the second derivative for one about the first.
The spacecraft. Zero acceleration means s double prime is zero. Its velocity s prime is unchanged, not zero — with no engine and no friction it coasts at whatever speed it had. It has certainly not stopped.
The two are the same mathematics and the same error. In each case a statement about the rate of change of a rate has been misread as a statement about the quantity itself. The ladder is what keeps them apart: f is the amount, f prime is how fast it changes, f double prime is how fast that change is changing — and each rung is a genuinely different claim about the world.
Commit first
Answer, then rate your confidence honestly.
Predict first
The derivative of a function is zero at x = 2. What follows?
Correct: It has a horizontal tangent at 2, and nothing more without further checking.
\[ x^3: \; f'(0) = 0 \text{ but } f' > 0 \text{ on both sides, so no turning point} \]
Why: A zero derivative marks a horizontal tangent, which a maximum, a minimum and a mere flattening all have. The cubic has a zero derivative at the origin and no turning point, because its derivative is positive on both sides and never changes sign. The value of the function at 2 is a different quantity entirely, and a constant function would need the derivative to vanish on a whole interval rather than at one point. Deciding which case you are in requires the sign of the derivative on each side, which is the first derivative test.
Explain it
They keep drawing the graph of f when asked for the graph of f prime, and cannot see the difference.
Discussion prompt
In four sentences or fewer, give them a test case that makes the distinction unmissable.
Hint: Pick a function whose two graphs look nothing alike.
Answer:
Give them the straight line y equals 3x minus 5 and ask for its derivative's graph. The slope is 3 at every single point, so the derivative is the constant 3 and its graph is a horizontal line — nothing like the slanted original.
Then point out that the intercept, negative 5, has vanished entirely: the derivative cannot see it, because moving the whole line up or down does not change any slope. If their sketch of f prime is slanted, or if it remembers the intercept, they have drawn f again.
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 sketching, mark the horizontal tangents first and do sign and magnitude as two separate readings. For keeping them apart, test yourself on a straight line, where the two graphs look nothing alike. For the failures, check continuity first — it settles the third case instantly — then compare the one-sided quotients. For second derivatives, say aloud which rung each answers: amount, rate, rate of the rate. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Divide the page into two stacked strips with the same horizontal scale. In the upper strip draw the cubic x cubed minus 3x, marking its two turning points; in the lower strip draw its derivative, lining the derivative's zeros up directly beneath those turning points. Beside the pair, write the five correspondence rules — rising, falling, horizontal, steep, straight — each as an arrow from a feature of f to a feature of f prime. Below, compute the derivative of x cubed from the definition in full, keeping x as a letter, then differentiate the result twice more and note where the chain runs out. On the right, draw the absolute value, the cube root and a jump function, mark the origin on each, and label which of the three failures each shows and whether it is continuous. At the bottom, write the four notations for a derivative and one sentence saying what each is best at. In a margin, write the sentence relating differentiability and continuity, with an arrow showing the one direction that is valid.
If the zeros of your lower graph do not sit directly beneath the turning points of your upper one, the two strips are misaligned — redraw with a shared vertical guide line, because that alignment is the entire content of the picture.
Recap
Five things, and the second is the graphical skill the whole of Chapter 4 assumes.
| If you see | Then |
|---|---|
| f rising on an interval | f' is positive there |
| A turning point of f | f' changes sign there |
| A zero of f' with no sign change | A flattening, not a turning point |
| A straight stretch of f | f' is constant there |
| A corner in f | No derivative, though f is continuous |
| A discontinuity in f | No derivative, by the contrapositive |
| A question about how a rate is changing | The second derivative |
Section 3.3 stops computing difference quotients altogether. One rule per family — powers, sums, products and quotients — replaces the definition, and every one of them is proved by exactly the computations of these two sections.
OpenStax Calculus Volume 1, §3.2 The Derivative as a Function §3.2, pp. 203-215 — everything on these slides traces back here
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