This deck builds the derivative from average rates of change and secant slopes up to the limit definition. It then computes derivatives from that definition for linear, quadratic, rational, and square-root functions, reads the derivative off a graph and in real units, and settles when differentiability fails. It targets the dropped minus sign in the difference quotient, cancelling h before it is a factor, the belief that continuous means differentiable, and confusing a function value with a slope.
Subject: Calculus I · 136 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 05
From average rate of change to the slope at a single instant - and what it means when that slope refuses to exist.
Objectives
By the end of this deck you will be able to:
Warm-up
Discussion prompt
Before we open The Derivative: Definition, Meaning, and Differentiability: without looking back, what was the main idea of Limits at Infinity, End Behavior, and Asymptotes, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck is about the end behavior of functions. It explains what a limit at infinity means, then covers the divide-by-the-highest-power technique for rational functions and the three degree cases, horizontal and slant asymptotes, radicals in the negative direction, and how exponentials outrun polynomials. It targets the sign error that appears when you pull a variable out of a square root, the myth that a graph never crosses its horizontal asymptote, and the confusion between vertical-asymptote limits and limits at infinity.
Section
Part 1
Concept
Every rate you have computed so far has been an average. Over an interval, you divide the change in the output by the change in the input.
\[ \text{average rate of change on } [a,b] \;=\; \frac{f(b) - f(a)}{b - a} \]
average rate of change — The change in output divided by the change in input, measured across a whole interval. It is one number summarizing the interval - not the rate at any single instant inside it.
Counterexample
Discussion prompt
Every rate you have computed so far has been an average. Over an interval, you divide the change in the output by the change in the input.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: Average rate of change is a secant slope — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two points, one slope — that is the average rate over the interval.
Intuition
Drive 120 miles in 2 hours and your average speed is 60 miles per hour. That is the odometer talking: total distance over total time.
But you were never pinned at exactly 60. You hit 75 on the freeway and 0 at a red light. The speedometer reports something the odometer cannot: how fast you are going at one instant.
The derivative is the speedometer, built out of the odometer. That is the entire idea of this deck.
Analogy
Discussion prompt
Explain Odometer talk versus speedometer talk by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Drive 120 miles in 2 hours and your average speed is 60 miles per hour. That is the odometer talking: total distance over total time.
Concept
Figure (svg): An upward-bending curve with a dashed straight line drawn through two marked points on it
Pick two points on a graph and draw the straight line through them. That line is called a secant line.
Its slope is rise over run - which is exactly the average rate of change.
\[ m_{\text{sec}} = \frac{f(b) - f(a)}{b - a} \]
secant line — A line through two points of a curve. Its slope equals the average rate of change of the function between those two inputs.
Definition probe
Sort into buckets
Every line below is part of the definition of average rate of change or of secant line — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: average velocity of a thrown ball into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Average velocity compares two positions, so we need both of them before anything else.
Worked example
A ball is thrown straight up from the ground at 80 feet per second. Its height in feet after t seconds is:
\[ s(t) = 80t - 16t^2 \]
Find its average velocity over the first second after one second has passed - that is, from one second to two seconds.
Find the height at each end of the interval
Why: Average velocity compares two positions, so we need both of them before anything else.
\[ s(1) = 80(1) - 16(1)^2 = 80 - 16 = 64 \]
Now the second endpoint
Why: Same substitution, new time. Squaring 2 gives 4, and 16 times 4 is 64.
\[ s(2) = 80(2) - 16(2)^2 = 160 - 64 = 96 \]
Divide the change in height by the change in time
Why: This is the secant slope through the two points of the height graph - rise over run.
\[ \frac{s(2) - s(1)}{2 - 1} = \frac{96 - 64}{1} = 32 \]
State the answer with units
Why: Feet divided by seconds gives feet per second. A rate without units is not an answer.
\[ \text{average velocity} = 32 \ \text{ft/s} \]
Verify by running the average backwards
Why: Thirty-two feet per second held for one second predicts a rise of 32 feet. The ball really did go from 64 feet to 96 feet, a rise of 32 feet. The average checks out.
\[ 32 \ \tfrac{\text{ft}}{\text{s}} \cdot (1 \ \text{s}) = 32 \ \text{ft} = 96 - 64 \]
Intuition
Take any smooth curve and zoom in on one point. Zoom again. Zoom again.
The bend flattens out. At high enough magnification the curve is indistinguishable from a straight line. Earth is round, but the parking lot looks flat.
That straight line you are seeing is the tangent line, and the derivative is nothing more or less than its slope.
Picture it
Animation
Shows: Zoom in far enough and a curve looks straight — a rendered Manim animation.
Rendered with Manim.
Takeaway: Local straightness is exactly what differentiability means.
Concept
Figure (svg): A curve with a dashed secant line through two points and a solid tangent line touching the curve at the left point
A tangent line touches the curve at one point, so you cannot use rise over run on it directly - you only have one point.
So we cheat. Keep the first point fixed, and slide the second point toward it. Each position gives a secant you can measure.
As the second point closes in, the secants tilt into the tangent. The tangent slope is the limit of the secant slopes.
\[ m_{\text{tan}} = \lim_{b \to a} \frac{f(b) - f(a)}{b - a} \]
Explain it
Discussion prompt
Explain Sliding the second point toward the first to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A tangent line touches the curve at one point, so you cannot use rise over run on it directly - you only have one point.
Picture it
Animation
Shows: A secant line pivoting as its second point slides in, settling onto the tangent.
The whole definition, in one motion.
Takeaway: The second point slides in and the slope it leaves behind is the derivative. Every rule you learn later is a shortcut past this picture.
Step zero
Discussion prompt
Worked example: watching secant slopes settle down — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set up one secant slope
Answer:
Worked example
Take the squaring function and the point where the input is 1. We will compute secant slopes with the second point creeping in from both sides.
\[ f(x) = x^2, \qquad a = 1, \qquad f(1) = 1 \]
Set up one secant slope
Why: Call the gap h. The second input is one plus h, so the secant slope is the change in output over h.
\[ m_{\text{sec}} = \frac{f(1+h) - f(1)}{h} = \frac{(1+h)^2 - 1}{h} \]
Compute it for a shrinking sequence of gaps
Why: Approaching from both sides is the honest test - a two-sided limit needs both directions to agree.
| gap h | second input | secant slope |
|---|---|---|
| 1 | 2 | 3 |
| 0.5 | 1.5 | 2.5 |
| 0.1 | 1.1 | 2.1 |
| 0.01 | 1.01 | 2.01 |
| -0.01 | 0.99 | 1.99 |
| -0.1 | 0.9 | 1.9 |
| -0.5 | 0.5 | 1.5 |
Read the trend from both directions
Why: From the right the slopes fall toward 2; from the left they climb toward 2. The two sides agree, so the limit is 2.
\[ m_{\text{tan}} = 2 \]
Verify the pattern with algebra instead of arithmetic
Why: Expanding the numerator gives two h plus h squared, and h factors out cleanly. Every row of the table is exactly two plus the gap, so as the gap goes to zero the slope goes to 2. The table was not a coincidence.
\[ \frac{(1+h)^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h \;\xrightarrow[\;h \to 0\;]{}\; 2 \]
Concept
That limit has a name beyond geometry. It is the instantaneous rate of change of the function at that input.
Geometrically it is a slope. Physically it is a speed. Economically it is a marginal cost. Same number, three vocabularies.
instantaneous rate of change — The limit of average rates of change over intervals that shrink to a single point. It is the rate right now, not the rate on average.
Picture it
Animation
Shows: A secant collapsing onto a tangent on a cubic with negative slope.
The slope can be negative, and the secant says so first.
Takeaway: Nothing about the definition assumes the curve is rising. The secant reports a negative slope just as readily, before any algebra is done.
Section
Part 2
Concept
Instead of naming the second input separately, name the gap between the inputs. Call it h.
\[ \text{second input} = a + h, \qquad \text{run} = (a+h) - a = h \]
Then the secant slope has a single letter in it, and shrinking the interval just means shrinking that one letter.
difference quotient — The expression giving the secant slope over a gap of size h: the change in output, f(a+h) minus f(a), divided by h. Every derivative computation starts here.
\[ \frac{f(a+h) - f(a)}{h} \]
Picture it
Animation
Shows: The difference quotient written out and read in words.
Read the second line aloud.
Takeaway: The difference quotient is rise over run with the run shrinking to nothing. That limit is the definition; everything else is technique.
Concept
Send the gap to zero and you have the derivative. This is the definition everything else in calculus is built on.
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
Read it out loud: f prime of a is the limit, as the gap shrinks to nothing, of the average rate of change over that gap.
If that limit exists as a finite number, we say the function is differentiable at that point, and the number is its derivative there.
Intuition
The obvious move is to plug the gap in as zero. Try it and you get zero divided by zero.
\[ \frac{f(a+0) - f(a)}{0} = \frac{0}{0} \]
That is not an answer, it is a question. Zero over zero is the indeterminate form - it means the two shrinking quantities are racing, and who wins depends on the function.
So the job in every derivative-from-the-definition problem is the same: do algebra until the gap cancels out of the bottom, then let it go to zero.
Picture it
Animation
Shows: Why we cannot simply set the gap to zero — a rendered Manim animation.
Rendered with Manim.
Takeaway: The indeterminate form is the reason limits were invented.
Concept
There is a second way to write the same thing. Keep the second input named, and let it slide toward the first.
\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
This form is often faster when you want the derivative at one specific point and the algebra factors nicely.
Concept
They are not two definitions. They are the same definition in two changes of variable.
\[ h = x - a \quad \Longleftrightarrow \quad x = a + h \]
When the second input slides toward the first, the gap shrinks to zero. Same limit, different bookkeeping.
| h-form | x-approaching-a form |
|---|---|
| gap named h | second input named x |
| h goes to 0 | x goes to a |
| best for a general formula | best for one specific point |
| expand and collect | factor and cancel |
Comparison
Comparison matrix
From The two forms are one substitution apart: refill the x-approaching-a form column from what you know. The rest of the table is as it appeared.
| h-form | x-approaching-a form |
|---|---|
| gap named h | second input named x |
| h goes to 0 | x goes to a |
| best for a general formula | best for one specific point |
| expand and collect | factor and cancel |
Picture it
Animation
Shows: The two forms are one substitution apart — a rendered Manim animation.
Rendered with Manim.
Takeaway: Not two definitions. One, written from two angles.
Estimation
Predict first
Start where you already know the answer, so you can trust the machinery.
Commit before you compute: what does Worked example: the derivative of a line come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against what you already know about lines
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The graph is a straight line of slope 3, and a line is its own tangent line everywhere.
Worked example
Start where you already know the answer, so you can trust the machinery.
\[ f(x) = 3x - 5 \]
Write the difference quotient
Why: Substitute x plus h everywhere x appears in the rule, then subtract the whole original function.
\[ \frac{f(x+h) - f(x)}{h} = \frac{\left[3(x+h) - 5\right] - \left[3x - 5\right]}{h} \]
Expand and distribute the minus sign over both bracketed terms
Why: The minus in front of the second bracket hits the 3x and the negative 5. Missing that is the most common error in this whole topic.
\[ = \frac{3x + 3h - 5 - 3x + 5}{h} \]
Collect: everything without an h cancels
Why: That cancellation is not luck. The constant parts of the two heights are identical, so only the change survives.
\[ = \frac{3h}{h} = 3 \qquad (h \neq 0) \]
Take the limit
Why: There is no h left, so the limit of a constant is that constant. The derivative is the same at every input.
\[ f'(x) = 3 \]
Verify against what you already know about lines
Why: The graph is a straight line of slope 3, and a line is its own tangent line everywhere. So the tangent slope had better be 3 at every point - and it is. Notice the difference quotient never even needed a limit: it was exactly 3 for every nonzero gap.
\[ \text{slope of } y = 3x - 5 \;=\; 3 \;=\; f'(x) \]
Picture it
Animation
Shows: Each line of the worked example "the derivative of a line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The graph is a straight line of slope 3, and a line is its own tangent line everywhere. So the tangent slope had better be 3 at every point - and it is. Notice the difference quotient never even needed a limit: it was exactly 3 for every nonzero gap.
Constraint
Discussion prompt
Run Pattern: the four-step recipe with this step confiscated:
Simplify until h factors out of the whole numerator - expand, combine fractions, or multiply by a conjugate, whichever the shape demands.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Every derivative-from-the-definition problem is these same four moves, in this order.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
If step 3 leaves a term with no h in it, you made an algebra mistake. That is a built-in error detector - use it.
Missing information
Discussion prompt
Now a curve, where the slope genuinely changes from point to point.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Brackets are not decoration here; they are what keeps the minus sign honest in the next step.
Worked example
Now a curve, where the slope genuinely changes from point to point.
\[ f(x) = x^2 - 4x + 1 \]
Substitute and bracket
Why: Brackets are not decoration here; they are what keeps the minus sign honest in the next step.
\[ \frac{\left[(x+h)^2 - 4(x+h) + 1\right] - \left[x^2 - 4x + 1\right]}{h} \]
Expand the squared binomial and both products
Why: The square of x plus h is x squared plus twice the cross term plus h squared. Write all six terms before combining anything.
\[ = \frac{x^2 + 2xh + h^2 - 4x - 4h + 1 - x^2 + 4x - 1}{h} \]
Cancel every term that has no h
Why: The x squared cancels, the negative 4x cancels, the 1 cancels. Exactly three terms survive, and each of them carries an h.
\[ = \frac{2xh + h^2 - 4h}{h} \]
Factor the h out and cancel it
Why: Now h is a common factor of the whole numerator, so cancelling is legal. This is the step that kills the zero-over-zero form.
\[ = \frac{h(2x + h - 4)}{h} = 2x + h - 4 \]
Let the gap go to zero
Why: No h remains in a denominator, so direct substitution is finally legal.
\[ f'(x) = 2x - 4 \]
Verify at the vertex
Why: The parabola's vertex sits where the input is 2, and a parabola has a horizontal tangent exactly at its vertex. Our formula gives zero there, as it must. A numerical check at another point agrees too: at input 5 the formula predicts slope 6, and the secant from 5 to 5.001 has slope 6.001.
\[ f'(2) = 2(2) - 4 = 0, \qquad \frac{f(5.001) - f(5)}{0.001} = \frac{6.006001 - 6}{0.001} = 6.001 \]
Picture it
Animation
Shows: Each line of the worked example "the derivative of a parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The parabola's vertex sits where the input is 2, and a parabola has a horizontal tangent exactly at its vertex. Our formula gives zero there, as it must. A numerical check at another point agrees too: at input 5 the formula predicts slope 6, and the secant from 5 to 5.001 has slope 6.001.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The function has two terms, and only the first one gets the subtraction.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The last term should have been positive 3x.
Bracket the entire function before subtracting it, then distribute the minus across every term inside.
Why: The last term should have been positive 3x. Because it stayed negative, the two 3x pieces add instead of cancelling.
Trap
The function has two terms, and only the first one gets the subtraction.
\[ f(x) = x^2 - 3x \]
\[ f(x+h) - f(x) \;\overset{?}{=}\; (x+h)^2 - 3(x+h) - x^2 - 3x \]
Expand and collect what is left
Why: The last term should have been positive 3x. Because it stayed negative, the two 3x pieces add instead of cancelling.
\[ = 2xh + h^2 - 6x - 3h \]
Divide by h and watch a term refuse to cancel
Why: The 6x has no h attached, so dividing leaves 6x sitting over h. That is the alarm bell.
\[ \frac{2xh + h^2 - 6x - 3h}{h} = 2x + h - \frac{6x}{h} - 3 \]
The limit blows up instead of settling
Why: As the gap shrinks, the 6x over h term grows without bound for every nonzero x. A smooth parabola cannot have an infinite slope, so the algebra, not the calculus, is broken.
\[ \lim_{h \to 0}\left(2x + h - \frac{6x}{h} - 3\right) \ \text{does not exist} \]
Bracket the entire function before subtracting it, then distribute the minus across every term inside.
\[ f(x) = x^2 - 3x \]
\[ f(x+h) - f(x) = \left[(x+h)^2 - 3(x+h)\right] - \left[x^2 - 3x\right] \]
Distribute the minus: the negative 3x becomes positive 3x
Why: Subtracting a negative flips the sign. Now the two 3x pieces cancel, exactly as the constant parts should.
\[ = x^2 + 2xh + h^2 - 3x - 3h - x^2 + 3x = 2xh + h^2 - 3h \]
Every surviving term carries an h, so h factors out
Why: This is the health check from the pattern slide: no h-free leftovers means the subtraction was done correctly.
\[ \frac{h(2x + h - 3)}{h} = 2x + h - 3 \]
Let the gap go to zero
Why: A clean, finite answer. Sanity check: at input 1 it predicts slope negative 1, and the secant from 1 to 1.001 has slope negative 0.999.
\[ f'(x) = 2x - 3 \]
Notation
Annotate
From Trap: the minus sign only hits the first term — read this one piece at a time. What is each part doing?
On: \( f(x+h) - f(x) \;\overset{?}{=}\; (x+h)^2 - 3(x+h) - x^2 - 3x \)
Elimination
Eliminate the wrong options
Using the limit definition, what is the derivative of f(x) = 5x^2 - 2x?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Expanding gives f(x+h) - f(x) = 10xh + 5h^2 - 2h. Factoring out h and dividing leaves 10x + 5h - 2, and letting h go to 0 kills the 5h, leaving 10x - 2.
Check
Do it on paper before you pick. Substitute, bracket the subtraction, factor out the gap, then take the limit.
\[ f(x) = 5x^2 - 2x \]
Check your understanding
Using the limit definition, what is the derivative of f(x) = 5x^2 - 2x?
Answer: A
Why: Expanding gives f(x+h) - f(x) = 10xh + 5h^2 - 2h. Factoring out h and dividing leaves 10x + 5h - 2, and letting h go to 0 kills the 5h, leaving 10x - 2.
Step zero
Discussion prompt
Worked example: the derivative of a reciprocal — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the difference quotient
Answer:
Worked example
A fraction in the rule means a fraction inside a fraction. The fix is always the same: one common denominator.
\[ f(x) = \frac{1}{x} \]
Write the difference quotient
Why: Substituting x plus h into the reciprocal just changes what sits under the 1.
\[ \frac{f(x+h) - f(x)}{h} = \frac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h} \]
Combine the top two fractions over the common denominator
Why: You cannot cancel anything until the numerator is a single fraction. The common denominator is the product of the two bottoms.
\[ \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \]
Dividing by h is multiplying by its reciprocal
Why: Now the h in the numerator is a genuine factor, so it cancels against the h from the division.
\[ \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)} \]
Let the gap go to zero
Why: No h sits in a denominator by itself any more, so substitution is legal. The x plus h becomes x.
\[ f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2} \]
Verify numerically, and check the sign against the graph
Why: At input 2 the formula predicts a slope of negative one quarter. The secant from 2 to 2.01 has slope about negative 0.2488, closing in on negative 0.25. The sign is right too: the reciprocal curve falls everywhere it is defined, so its derivative must be negative everywhere.
\[ f'(2) = -\tfrac{1}{4} = -0.25, \qquad \frac{\frac{1}{2.01} - \frac{1}{2}}{0.01} \approx -0.2488 \]
Picture it
Animation
Shows: Each line of the worked example "the derivative of a reciprocal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At input 2 the formula predicts a slope of negative one quarter. The secant from 2 to 2.01 has slope about negative 0.2488, closing in on negative 0.25. The sign is right too: the reciprocal curve falls everywhere it is defined, so its derivative must be negative everywhere.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The gap appears twice, so it looks like the two copies should destroy each other.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It is tempting because it makes the ugly stacked fraction vanish in one stroke.
Combine into a single fraction first. Only then does a common factor exist to cancel.
Why: It is tempting because it makes the ugly stacked fraction vanish in one stroke. But cancelling is only legal on a common factor, and here the h is added to x, not multiplying anything.
Trap
The gap appears twice, so it looks like the two copies should destroy each other.
\[ \frac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h} \]
Cross the h out of the bottom and out of the sum upstairs at the same time
Why: It is tempting because it makes the ugly stacked fraction vanish in one stroke. But cancelling is only legal on a common factor, and here the h is added to x, not multiplying anything.
\[ \frac{1}{x+h} - \frac{1}{x} \;\longrightarrow\; \frac{1}{x} - \frac{1}{x} = 0 \]
The derivative comes out as zero everywhere
Why: That claims the reciprocal curve is perfectly flat at every input, which one glance at its steep drop near the origin refutes. A result that contradicts the picture is a signal to go back, not a fluke.
\[ f'(x) = 0 \quad \text{(false)} \]
Combine into a single fraction first. Only then does a common factor exist to cancel.
\[ \frac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h} \]
Put the numerator over one common denominator
Why: Subtracting the numerators gives x minus x minus h, which is exactly negative h. The gap has become a factor of the whole top.
\[ \frac{1}{x+h} - \frac{1}{x} = \frac{-h}{x(x+h)} \]
Now cancel the h legally
Why: It multiplies the entire numerator, so it may cancel against the h from the division. That is what a factor means.
\[ \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)} \;\xrightarrow[\;h \to 0\;]{}\; -\frac{1}{x^2} \]
Translation
\( \frac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Estimation
Predict first
A root in the rule calls for the conjugate - the same two terms with the sign between them flipped.
Commit before you compute: what does Worked example: the derivative of a square root come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify numerically at a friendly input
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At input 9 the formula predicts one sixth, about 0.1667.
Worked example
A root in the rule calls for the conjugate - the same two terms with the sign between them flipped.
\[ f(x) = \sqrt{x} \]
Write the difference quotient
Why: Nothing cancels yet, and no amount of staring will make it cancel. The roots have to go first.
\[ \frac{\sqrt{x+h} - \sqrt{x}}{h} \]
Multiply top and bottom by the conjugate of the numerator
Why: Multiplying by a fraction equal to 1 changes the form, not the value. The conjugate turns the numerator into a difference of squares, which erases both roots.
\[ \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \]
Collapse the numerator
Why: The product of a sum and a difference is the difference of the squares, and squaring a square root undoes it. What is left is x plus h minus x, which is just h.
\[ = \frac{(x+h) - x}{h\left(\sqrt{x+h} + \sqrt{x}\right)} = \frac{h}{h\left(\sqrt{x+h} + \sqrt{x}\right)} \]
Cancel the h and take the limit
Why: With the h gone from the denominator, substituting zero is finally legal, and the two roots become identical.
\[ = \frac{1}{\sqrt{x+h} + \sqrt{x}} \;\xrightarrow[\;h \to 0\;]{}\; \frac{1}{2\sqrt{x}} \]
Verify numerically at a friendly input
Why: At input 9 the formula predicts one sixth, about 0.1667. The secant from 9 to 9.01 has slope about 0.16662. It also matches the picture: the root curve is rising but flattening, so the derivative should be positive and shrinking as x grows.
\[ f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6} \approx 0.1667, \qquad \frac{\sqrt{9.01} - 3}{0.01} \approx 0.16662 \]
Picture it
Animation
Shows: Each line of the worked example "the derivative of a square root", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At input 9 the formula predicts one sixth, about 0.1667. The secant from 9 to 9.01 has slope about 0.16662. It also matches the picture: the root curve is rising but flattening, so the derivative should be positive and shrinking as x grows.
Pattern
Step 3 of the recipe - simplify until the gap factors out - looks different for different function shapes. There are only three moves you need.
| shape of the rule | the move | why it works |
|---|---|---|
| polynomial | expand and collect | the h-free terms cancel, leaving every term with an h |
| fraction | one common denominator | the subtraction in the numerator produces a factor of h |
| square root | multiply by the conjugate | the difference of squares erases the roots and exposes h |
In all three cases you are doing one thing: turning the gap from a term into a factor so it can legally cancel.
Trade off
Comparison matrix
From Pattern: let the shape choose the algebra: every row here is a choice with a cost. Fill the why it works column, then say which row you would actually pick and what you give up for it.
| shape of the rule | the move | why it works |
|---|---|---|
| polynomial | expand and collect | the h-free terms cancel, leaving every term with an h |
| fraction | one common denominator | the subtraction in the numerator produces a factor of h |
| square root | multiply by the conjugate | the difference of squares erases the roots and exposes h |
Prediction
Predict first
Using the limit definition, what is the derivative of f(x) = 1/(x + 3)?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: f'(x) = -1/(x + 3)^2
Why: Combining 1/(x + h + 3) - 1/(x + 3) over the common denominator gives -h divided by (x + h + 3)(x + 3). Cancelling the h and letting h go to 0 makes both factors x + 3, giving -1/(x + 3)^2.
Check
Common denominator first, then cancel. Take a minute on paper.
\[ f(x) = \frac{1}{x+3} \]
Check your understanding
Using the limit definition, what is the derivative of f(x) = 1/(x + 3)?
Answer: A
Why: Combining 1/(x + h + 3) - 1/(x + 3) over the common denominator gives -h divided by (x + h + 3)(x + 3). Cancelling the h and letting h go to 0 makes both factors x + 3, giving -1/(x + 3)^2.
Check
Recognizing the definition when it appears in disguise is worth as many exam points as computing it.
\[ f(x) = \sqrt{x}, \qquad a = 4 \]
Check your understanding
Which limit computes f'(4) for f(x) = sqrt(x) using the x-approaching-a form of the definition?
Answer: A
Why: The x-approaching-a form is the limit of f(x) minus f(a) over x minus a. With a = 4 and f(4) = 2 that is (sqrt(x) - 2)/(x - 4), which simplifies to 1/(sqrt(x) + 2) and equals 1/4.
Section
Part 3
Concept
Nothing in the definition pinned the input to one particular place. Leave it as a variable and you get a rule, not a number.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Feed the original rule an input and it hands back a height. Feed the derivative rule an input and it hands back a slope.
| input x | height f(x) = x^2 - 4x + 1 | slope f'(x) = 2x - 4 |
|---|---|---|
| 0 | 1 | -4 |
| 1 | -2 | -2 |
| 2 | -3 | 0 |
| 3 | -2 | 2 |
| 4 | 1 | 4 |
Notice the slope column passes through zero exactly where the height column bottoms out. Two different functions, tightly linked.
Pattern
Step through it
Step through The derivative is a function too one row at a time. What is driving the change, and what would the row after the last one be?
Concept
Keep these two objects separate in your head, because problems switch between them without warning.
\[ f'(a) \;=\; \text{a single number: the slope at one input} \]
\[ f' \;=\; \text{a function: the slope-reporting machine} \]
derivative function — The function whose output at each input is the slope of the tangent line to the original graph there. Its domain is every input where that limit exists.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of secant line, instantaneous rate of change, difference quotient, derivative function as The Derivative: Definition, Meaning, and Differentiability uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: A number at a point, a machine everywhere — a rendered Manim animation.
Rendered with Manim.
Takeaway: Collect the slope at every input and the result is a whole new function.
Ranking
Put in order
Put the moves of Worked example: the tangent line at a point into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The tangent line has to touch the curve, so it must pass through the point on the curve - not through the derivative's output.
Worked example
A tangent line needs two ingredients: a point and a slope. The function supplies the point, the derivative supplies the slope.
\[ f(x) = x^2 - 4x + 1, \qquad f'(x) = 2x - 4, \qquad \text{at } x = 1 \]
Get the point by evaluating the original function
Why: The tangent line has to touch the curve, so it must pass through the point on the curve - not through the derivative's output.
\[ f(1) = 1 - 4 + 1 = -2 \quad\Rightarrow\quad (1,\, -2) \]
Get the slope by evaluating the derivative
Why: This is the whole payoff of the derivative: a slope for a curve, at one point, as a plain number.
\[ f'(1) = 2(1) - 4 = -2 \]
Assemble with point-slope form
Why: Point-slope is built for exactly this input: one point and one slope, no rearranging needed.
\[ y - (-2) = -2(x - 1) \]
Simplify to slope-intercept form
Why: Distributing the negative 2 gives negative 2x plus 2, and moving the negative 2 across leaves a clean line through the origin.
\[ y = -2x \]
Verify that the line touches the parabola exactly once
Why: Setting the two rules equal gives a perfect square, so the only intersection is a doubled root at the input 1. A line meeting a parabola in a double root is precisely a tangent line. It also passes through the point: negative 2 times 1 is negative 2.
\[ x^2 - 4x + 1 = -2x \;\Longrightarrow\; x^2 - 2x + 1 = 0 \;\Longrightarrow\; (x-1)^2 = 0 \]
Picture it
Animation
Shows: Each line of the worked example "the tangent line at a point", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting the two rules equal gives a perfect square, so the only intersection is a doubled root at the input 1. A line meeting a parabola in a double root is precisely a tangent line. It also passes through the point: negative 2 times 1 is negative 2.
Concept
A function can be perfectly well behaved at an input where its derivative does not exist.
\[ f(x) = \sqrt{x}, \quad \text{domain } [0, \infty) \]
\[ f'(x) = \frac{1}{2\sqrt{x}}, \quad \text{domain } (0, \infty) \]
At an input of zero the root curve leaves the origin going straight up. There is a tangent line there - it is just vertical, and vertical lines have no slope.
Concept
Different mathematicians won different arguments, so you have to be fluent in all of these. They all mean the same thing.
| notation | read it as | handy when |
|---|---|---|
| f'(x) | f prime of x | stating general rules compactly |
| y' | y prime | the output already has a name |
| dy/dx | d y d x | you want the input and output names visible |
| df/dx | d f d x | same, with the function named f |
| d/dx [ f(x) ] | d by d x of f of x | you are giving an instruction: differentiate this |
\[ f'(x) \;=\; y' \;=\; \frac{dy}{dx} \;=\; \frac{df}{dx} \;=\; \frac{d}{dx}\left[f(x)\right] \]
Picture it
Animation
Shows: Four notations, one idea — a rendered Manim animation.
Rendered with Manim.
Takeaway: Different authors, different centuries, identical meaning.
Concept
Prime notation makes it easy to name a point: just put the input in the parentheses. Leibniz notation has no parentheses to put it in.
So you attach a vertical bar with the input written under it.
\[ \left.\frac{dy}{dx}\right|_{x=a} \;=\; f'(a) \]
Read the bar as evaluated at. Without it, the Leibniz symbol still means the whole slope function, not one number.
Intuition
The Leibniz symbol looks like a fraction because it is the fossil of one: rise over run, with both pieces shrunk to nothing.
Strictly, it is the limit of a ratio, not a ratio. You cannot split it apart and cancel the letters as though they were numbers.
But it inherits the units of a ratio exactly. If the output is measured in dollars and the input in items, the derivative is measured in dollars per item. That is why applied calculus prefers this notation.
Hypothesis
Predict first
Worked example: a cubic from the definition is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Expand the cubed binomial
Why: Each term picks up one more copy of the gap than the last. Getting the coefficients 1, 3, 3, 1 right is the only hard part of this problem.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
One more from scratch, with a cube. This one will be the running example for reading graphs later.
\[ f(x) = x^3 - 3x \]
Expand the cubed binomial
Why: Each term picks up one more copy of the gap than the last. Getting the coefficients 1, 3, 3, 1 right is the only hard part of this problem.
\[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]
Build the numerator and distribute the minus over the whole function
Why: Bracket the original rule first. The x cubed cancels and the negative 3x becomes positive 3x, cancelling as well.
\[ \left[(x+h)^3 - 3(x+h)\right] - \left[x^3 - 3x\right] = 3x^2h + 3xh^2 + h^3 - 3h \]
Factor the gap out and cancel it
Why: All four surviving terms carry at least one h, which is the sign that the subtraction was done correctly.
\[ \frac{h\left(3x^2 + 3xh + h^2 - 3\right)}{h} = 3x^2 + 3xh + h^2 - 3 \]
Let the gap go to zero
Why: Every term still containing an h dies; the two that do not survive.
\[ f'(x) = 3x^2 - 3 \]
Verify at the two inputs where the slope should vanish
Why: The formula gives zero at inputs 1 and negative 1. Testing input 1 by hand: the height there is negative 2, while just to the left and right the heights are negative 1.971 and negative 1.969, both higher. So the curve really does bottom out there with a horizontal tangent.
\[ f'(\pm 1) = 3(1) - 3 = 0, \qquad f(0.9) = -1.971 \;>\; f(1) = -2 \;<\; f(1.1) = -1.969 \]
Picture it
Animation
Shows: Each line of the worked example "a cubic from the definition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives zero at inputs 1 and negative 1. Testing input 1 by hand: the height there is negative 2, while just to the left and right the heights are negative 1.971 and negative 1.969, both higher. So the curve really does bottom out there with a horizontal tangent.
Section
Part 4
Concept
Outside a geometry class, nobody asks for the slope of a curve. They ask how fast something is changing right now.
\[ f'(a) = \text{the instantaneous rate of change of } f \text{ with respect to its input, at } x = a \]
Same limit, same computation. The only thing that changes is the story you attach to it.
Intuition
A derivative is output units divided by input units. Get that ratio right and the meaning writes itself.
| the function measures | the input measures | units of the derivative | everyday name |
|---|---|---|---|
| position in feet | time in seconds | feet per second | velocity |
| velocity in feet per second | time in seconds | feet per second, per second | acceleration |
| cost in dollars | lamps produced | dollars per lamp | marginal cost |
| bacteria in a dish | time in hours | bacteria per hour | growth rate |
| water in a tank, in liters | depth in centimeters | liters per centimeter | how much a centimeter of depth holds |
If your answer's units do not read as something per something, you have not computed a rate.
Comparison
Comparison matrix
From The units tell you what it is: refill the everyday name column from what you know. The rest of the table is as it appeared.
| the function measures | the input measures | units of the derivative | everyday name |
|---|---|---|---|
| position in feet | time in seconds | feet per second | velocity |
| velocity in feet per second | time in seconds | feet per second, per second | acceleration |
| cost in dollars | lamps produced | dollars per lamp | marginal cost |
| bacteria in a dish | time in hours | bacteria per hour | growth rate |
| water in a tank, in liters | depth in centimeters | liters per centimeter | how much a centimeter of depth holds |
Picture it
Animation
Shows: The units tell you what it is — a rendered Manim animation.
Rendered with Manim.
Takeaway: A derivative always inherits output-units over input-units.
Step zero
Discussion prompt
Worked example: from position to velocity — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the difference quotient in the time variable
Answer:
Worked example
Back to the thrown ball. Earlier we got its average velocity over a whole second. Now we want its speedometer reading at one instant.
\[ s(t) = 80t - 16t^2 \quad \text{(height in feet, } t \text{ in seconds)} \]
Write the difference quotient in the time variable
Why: Nothing about the definition cares that the letter is t instead of x. Substitute t plus h everywhere t appears.
\[ \frac{\left[80(t+h) - 16(t+h)^2\right] - \left[80t - 16t^2\right]}{h} \]
Expand, then cancel the h-free terms
Why: The 80t cancels and the 16t squared cancels, exactly as the pattern promised. Three terms survive, all carrying an h.
\[ = \frac{80h - 32th - 16h^2}{h} \]
Cancel the gap and take the limit
Why: Dividing each term by h leaves one term still containing h, and that one dies in the limit.
\[ = 80 - 32t - 16h \;\xrightarrow[\;h \to 0\;]{}\; v(t) = 80 - 32t \]
Evaluate at the instant asked for, with units
Why: One second after release the ball is still climbing, so a positive velocity is what we expect.
\[ v(1) = 80 - 32(1) = 48 \ \text{ft/s} \]
Verify against a very short average
Why: Over the hundredth of a second after t equals 1 the ball rises from 64 feet to 64.4784 feet, an average velocity of 47.84 feet per second. That is closing in on 48, exactly as an instantaneous rate should be approached by short averages.
\[ \frac{s(1.01) - s(1)}{0.01} = \frac{64.4784 - 64}{0.01} = 47.84 \ \text{ft/s} \]
Picture it
Animation
Shows: Each line of the worked example "from position to velocity", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Over the hundredth of a second after t equals 1 the ball rises from 64 feet to 64.4784 feet, an average velocity of 47.84 feet per second. That is closing in on 48, exactly as an instantaneous rate should be approached by short averages.
Concept
Velocity carries a direction; speed does not. The sign of the derivative is the direction.
\[ v(t) = 80 - 32t \]
| time t (seconds) | velocity (ft/s) | what the ball is doing |
|---|---|---|
| 1 | 48 | rising, and fast |
| 2 | 16 | still rising, but slowing |
| 2.5 | 0 | at the top, momentarily motionless |
| 3 | -16 | falling |
| 4 | -48 | falling fast |
The derivative hitting zero is what identifies the peak. Speed is the size of the velocity with the sign thrown away, so the ball's speed at four seconds is 48 feet per second even though its velocity is negative.
Trade off
Comparison matrix
From The sign of a rate is information: every row here is a choice with a cost. Fill the what the ball is doing column, then say which row you would actually pick and what you give up for it.
| time t (seconds) | velocity (ft/s) | what the ball is doing |
|---|---|---|
| 1 | 48 | rising, and fast |
| 2 | 16 | still rising, but slowing |
| 2.5 | 0 | at the top, momentarily motionless |
| 3 | -16 | falling |
| 4 | -48 | falling fast |
Missing information
Discussion prompt
A workshop's total cost in dollars to produce a batch of lamps depends on how many lamps are in the batch.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The fixed 5000 will cancel immediately, which is the algebra's way of saying overhead does not change with one more lamp.
Worked example
A workshop's total cost in dollars to produce a batch of lamps depends on how many lamps are in the batch.
\[ C(q) = 5000 + 12q + 0.01q^2 \]
How fast is cost rising when 200 lamps are already planned?
Build the difference quotient
Why: The fixed 5000 will cancel immediately, which is the algebra's way of saying overhead does not change with one more lamp.
\[ \frac{\left[5000 + 12(q+h) + 0.01(q+h)^2\right] - \left[5000 + 12q + 0.01q^2\right]}{h} \]
Expand and cancel
Why: The 5000 cancels, the 12q cancels, and the 0.01 q squared cancels. Everything left has an h.
\[ = \frac{12h + 0.02qh + 0.01h^2}{h} = 12 + 0.02q + 0.01h \]
Let the gap go to zero
Why: The last term is the only one still holding an h, so it vanishes and leaves the marginal cost function.
\[ C'(q) = 12 + 0.02q \]
Evaluate at 200 lamps and attach units
Why: Dollars divided by lamps gives dollars per lamp. This is the marginal cost at that production level.
\[ C'(200) = 12 + 0.02(200) = 16 \ \text{dollars per lamp} \]
Verify by actually costing out the next lamp
Why: The true cost of the 201st lamp is the total at 201 minus the total at 200, which is 16 dollars and one cent. The derivative predicted 16 dollars. It is an excellent approximation of the cost of one more unit, which is exactly what economists mean by marginal.
\[ C(201) - C(200) = 7816.01 - 7800 = 16.01 \ \text{dollars} \]
Picture it
Animation
Shows: Each line of the worked example "marginal cost", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The true cost of the 201st lamp is the total at 201 minus the total at 200, which is 16 dollars and one cent. The derivative predicted 16 dollars. It is an excellent approximation of the cost of one more unit, which is exactly what economists mean by marginal.
Concept
Because the derivative is a rate per unit of input, multiplying it by one unit of input estimates the effect of one more.
\[ C'(q) \cdot 1 \;\approx\; C(q+1) - C(q) \]
marginal cost — The derivative of the total-cost function. It approximates the additional cost of producing one more unit at the current production level, and it is measured in dollars per unit.
The same trick names marginal revenue, marginal profit, and every other marginal quantity you will meet.
Commit first
Predict first
What does C'(200) = 16 tell the workshop owner?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: At a batch size of 200 lamps, total cost is rising at about 16 dollars per additional lamp, so the 201st lamp costs roughly 16 dollars.
Why: The derivative has units of dollars per lamp, so it is a rate, not a total. Costing the next lamp exactly gives C(201) - C(200) = 16.01 dollars, which the derivative's 16 predicts almost perfectly.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
A workshop's total cost in dollars for a batch of lamps, and the marginal cost we just computed:
\[ C(q) = 5000 + 12q + 0.01q^2, \qquad C'(200) = 16 \]
Check your understanding
What does C'(200) = 16 tell the workshop owner?
Answer: A
Why: The derivative has units of dollars per lamp, so it is a rate, not a total. Costing the next lamp exactly gives C(201) - C(200) = 16.01 dollars, which the derivative's 16 predicts almost perfectly.
Section
Part 5
Concept
Figure (svg): The graph of x cubed minus three x: it rises to a local high point at input negative one, falls to a local low point at input one, then rises again. Both turning points are marked.
You do not need a formula to say a lot about the derivative. The picture is enough.
Where the curve climbs as you read left to right, every tangent tilts upward, so the derivative is positive there.
Where the curve falls, every tangent tilts downward, so the derivative is negative there.
On this cubic: positive out on the left, negative through the dip in the middle, positive again on the right.
Picture it
Animation
Shows: The sign of the derivative says which way — a rendered Manim animation.
Rendered with Manim.
Takeaway: Rising where the slope is positive, falling where it is negative.
Concept
Between climbing and falling there has to be an instant of neither. At that instant the tangent line is flat.
\[ \text{horizontal tangent at } x = a \quad \Longleftrightarrow \quad f'(a) = 0 \]
On the graph, the dashed level segments sit at the two turning points. Those are precisely where the derivative crosses zero.
This is the seed of every optimization problem you will ever solve: to find a peak or a valley, hunt for where the derivative is zero.
Concept
Sign tells you the direction. Size tells you how dramatic the change is.
| what the graph looks like | what the derivative is doing |
|---|---|
| nearly vertical climb | large and positive |
| gentle climb | small and positive |
| momentarily level | zero |
| gentle drop | small and negative |
| steep drop | large and negative |
So the graph of the derivative is not a mystery object. It is a report card on the original graph's steepness, input by input.
Comparison
Comparison matrix
From Steepness is the size of the derivative: refill the what the derivative is doing column from what you know. The rest of the table is as it appeared.
| what the graph looks like | what the derivative is doing |
|---|---|
| nearly vertical climb | large and positive |
| gentle climb | small and positive |
| momentarily level | zero |
| gentle drop | small and negative |
| steep drop | large and negative |
Estimation
Predict first
Suppose you were handed only the graph of the cubic - no formula. Describe its derivative.
Commit before you compute: what does Worked example: describing the derivative from the picture… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the derivative we actually computed
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Earlier, straight from the definition, we got three x squared minus three.
Worked example
Suppose you were handed only the graph of the cubic - no formula. Describe its derivative.
\[ f(x) = x^3 - 3x \quad \text{(formula hidden for now)} \]
Mark the level moments first
Why: The curve turns at the local high point and the local low point, and at both the tangent is horizontal. Those become the zeros of the derivative.
\[ f'(-1) = 0, \qquad f'(1) = 0 \]
Assign a sign to each interval between them
Why: Left of the high point the curve climbs, so the derivative is positive. Between the two turning points it falls, so the derivative is negative. Right of the low point it climbs again, so positive.
| interval of inputs | the curve is | so the derivative is |
|---|---|---|
| left of -1 | climbing | positive |
| between -1 and 1 | falling | negative |
| right of 1 | climbing | positive |
Judge the sizes
Why: The steepest falling happens midway between the turning points, right at the origin, so the derivative reaches its most negative value there. Far out on either side the curve is nearly vertical, so the derivative is large and positive at both ends.
\[ f'(0) \text{ is the minimum of } f' \]
Assemble the shape
Why: A function that is positive outside two roots, negative between them, and bottoms out midway is an upward-opening parabola with roots at negative 1 and 1.
\[ f'(x) \text{ looks like } a(x-1)(x+1) \text{ with } a > 0 \]
Verify against the derivative we actually computed
Why: Earlier, straight from the definition, we got three x squared minus three. Factoring it gives three times the quantity x minus one times x plus one - an upward parabola with roots at exactly negative 1 and 1, and value negative 3 at the origin. The picture predicted every feature.
\[ f'(x) = 3x^2 - 3 = 3(x-1)(x+1), \qquad f'(0) = -3 \]
Picture it
Animation
Shows: Each line of the worked example "describing the derivative from the picture alone", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Earlier, straight from the definition, we got three x squared minus three. Factoring it gives three times the quantity x minus one times x plus one - an upward parabola with roots at exactly negative 1 and 1, and value negative 3 at the origin. The picture predicted every feature.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The cubic reaches its local high point at the input negative 1, and the question asks for the derivative there.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The point is sitting at a height of 2, and that is the only number the picture seems to be offering.
Ask what the tangent line is doing there, not where the point is sitting.
Why: The point is sitting at a height of 2, and that is the only number the picture seems to be offering. So the answer must be 2.
Trap
The cubic reaches its local high point at the input negative 1, and the question asks for the derivative there.
\[ \text{find } f'(-1) \]
Read the number off the vertical axis and hand it in
Why: The point is sitting at a height of 2, and that is the only number the picture seems to be offering. So the answer must be 2.
\[ f'(-1) \overset{?}{=} 2 \]
But that answer contradicts the picture
Why: A slope of 2 would mean the curve is climbing briskly through that point. The graph shows it levelling off at a peak. Height and slope are different questions about the same point, and this answered the wrong one.
Ask what the tangent line is doing there, not where the point is sitting.
\[ \text{find } f'(-1) \]
Lay a straightedge along the curve at the peak
Why: At a local high point the curve is momentarily level, so the tangent line is horizontal - and a horizontal line has slope zero.
\[ f'(-1) = 0 \]
Confirm with the formula, and keep both facts
Why: The computed derivative is three x squared minus three, which is zero at negative 1. The height there is 2 and the slope there is 0. Both are true, and they answer different questions.
\[ f(-1) = 2 \quad \text{and} \quad f'(-1) = 3(-1)^2 - 3 = 0 \]
Notation
Annotate
From Trap: reading the height when the question asked for the… — read this one piece at a time. What is each part doing?
On: \( f(-1) = 2 \quad \text{and} \quad f'(-1) = 3(-1)^2 - 3 = 0 \)
Check
Use the picture, the factored derivative, or both.
\[ f(x) = x^3 - 3x, \qquad f'(x) = 3x^2 - 3 \]
Check your understanding
For f(x) = x^3 - 3x, on which inputs is f'(x) negative?
Answer: A
Why: Factoring gives f'(x) = 3(x - 1)(x + 1), which is negative exactly between its two roots. That matches the picture: the curve falls from its local high point at -1 to its local low point at 1.
Section
Part 6
Concept
The definition is a limit, and limits are allowed to not exist. When that happens, the derivative simply is not there.
differentiable at a point — A function is differentiable at an input when the limit defining the derivative there exists as a finite number. If the limit fails, or runs off to infinity, the function is not differentiable there.
\[ f \text{ differentiable at } a \iff \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \ \text{exists and is finite} \]
A function is called differentiable on an interval when it is differentiable at every input in that interval.
Concept
There is one implication between these two ideas, and it runs in exactly one direction.
\[ f \text{ differentiable at } a \;\Longrightarrow\; f \text{ continuous at } a \]
In words: a graph that has a tangent line at a point cannot be broken at that point. Having a slope is a strictly stronger property than being connected.
The useful way to use it is backwards: if a function is not continuous somewhere, it cannot possibly be differentiable there. No further work needed.
Picture it
Animation
Shows: The implication from differentiable to continuous, and its failure in reverse.
The arrow points one way only.
Takeaway: Differentiability implies continuity, never the reverse. Absolute value at zero is continuous with no tangent at all.
Concept
The proof is one line of algebra plus the product law for limits. Continuity at a point means the change in output shrinks to nothing as the input closes in.
\[ f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a) \qquad (x \neq a) \]
Multiplying and dividing by the same nonzero quantity changes nothing - but it makes the difference quotient appear.
\[ \lim_{x \to a}\left[f(x) - f(a)\right] = f'(a) \cdot 0 = 0 \]
The first factor settles down to a finite number precisely because the function is differentiable, and the second factor goes to zero. A finite number times zero is zero, so the output difference vanishes - which is exactly continuity.
\[ \lim_{x \to a} f(x) = f(a) \]
Concept
Reverse the arrow and the statement dies. Continuity does not buy you a derivative.
\[ f \text{ continuous at } a \;\not\Longrightarrow\; f \text{ differentiable at } a \]
One counterexample is enough to kill it, and the standard one is the absolute-value function at the origin: unbroken, but with a sharp corner.
\[ f(x) = |x| \ \text{ is continuous at } 0 \ \text{ but not differentiable at } 0 \]
Explain it
Discussion prompt
Explain The converse is false to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Reverse the arrow and the statement dies. Continuity does not buy you a derivative.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The absolute-value graph is a single unbroken V. You can draw the whole thing without lifting your pencil.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: No break means no problem, and the V clearly bottoms out at the origin - so the slope there should be zero, the way it is at the bottom of a parabola.
Continuity forbids a break. It does not forbid a sudden turn.
Why: No break means no problem, and the V clearly bottoms out at the origin - so the slope there should be zero, the way it is at the bottom of a parabola.
Trap
The absolute-value graph is a single unbroken V. You can draw the whole thing without lifting your pencil.
\[ f(x) = |x| \]
Conclude that a connected graph must have a slope everywhere
Why: No break means no problem, and the V clearly bottoms out at the origin - so the slope there should be zero, the way it is at the bottom of a parabola.
\[ f'(0) \overset{?}{=} 0 \]
But no single line hugs the graph at the origin
Why: Just to the right the graph runs along a line of slope 1. Just to the left it runs along a line of slope negative 1. There is no one tangent line, so there is no derivative. Continuity was never enough.
Continuity forbids a break. It does not forbid a sudden turn.
\[ f(x) = |x| \]
Test the definition from each side separately
Why: For a positive gap the absolute value leaves the gap alone; for a negative gap it flips the sign. The two one-sided difference quotients are constants, and they are different constants.
\[ \lim_{h \to 0^{+}} \frac{|h|}{h} = 1, \qquad \lim_{h \to 0^{-}} \frac{|h|}{h} = -1 \]
The two-sided limit fails, so the derivative does not exist
Why: A two-sided limit exists only when both one-sided limits agree. Here they disagree by a jump of 2, so the derivative at the origin does not exist.
\[ f'(0) \ \text{does not exist} \]
Keep the implication pointed the right way
Why: Differentiable always implies continuous. Continuous never implies differentiable. Memorize the direction, not just the words.
\[ \text{differentiable} \Rightarrow \text{continuous}, \qquad \text{continuous} \nRightarrow \text{differentiable} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Step zero
Discussion prompt
Worked example: the corner at the origin — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the difference quotient
Answer:
Worked example
Figure (svg): The V-shaped graph of the absolute value function with a sharp corner marked at the origin, where the two straight arms meet.
Run the definition on the absolute-value function at the origin, carefully, one side at a time.
\[ f(x) = |x|, \qquad a = 0, \qquad f(0) = 0 \]
Write the difference quotient
Why: Since the function value at the origin is zero, the numerator is just the absolute value of the gap.
\[ \frac{f(0+h) - f(0)}{h} = \frac{|h| - 0}{h} = \frac{|h|}{h} \]
Approach from the right, where the gap is positive
Why: A positive number is its own absolute value, so the quotient is the gap divided by itself. It equals 1 for every positive gap, no matter how small.
\[ h > 0 \;\Rightarrow\; \frac{|h|}{h} = \frac{h}{h} = 1 \]
Approach from the left, where the gap is negative
Why: The absolute value of a negative number is its opposite, so the quotient is negative gap over gap, which is negative 1 for every negative gap.
\[ h < 0 \;\Rightarrow\; \frac{|h|}{h} = \frac{-h}{h} = -1 \]
Compare the two sides
Why: The one-sided limits both exist, but they are not equal. A two-sided limit requires agreement, so this limit does not exist and neither does the derivative.
\[ 1 \neq -1 \;\Longrightarrow\; f'(0) \ \text{does not exist} \]
Verify with a table of actual gaps
Why: No matter how close the gap gets to zero, the quotient never settles: it is stuck at 1 on one side and negative 1 on the other. Shrinking the gap does not help, which is the signature of a corner rather than a rounding error.
| gap h | absolute value of h | difference quotient |
|---|---|---|
| 0.1 | 0.1 | 1 |
| 0.001 | 0.001 | 1 |
| 0.000001 | 0.000001 | 1 |
| -0.000001 | 0.000001 | -1 |
| -0.001 | 0.001 | -1 |
| -0.1 | 0.1 | -1 |
Concept
Figure (svg): Three small sketches side by side: a sharp V-shaped corner, a cusp where two branches meet vertically, and a jump discontinuity with a filled dot on the left piece and an open dot on the right piece.
There are exactly three shapes that kill a derivative, and the pictures are worth memorizing.
| failure | what the graph does | what the definition does |
|---|---|---|
| corner | two straight arms meet at an angle | the one-sided limits exist but disagree |
| cusp or vertical tangent | the curve turns straight up | the quotient grows without bound |
| discontinuity | the graph breaks, jumps, or has a hole | the limit cannot exist at all |
The third one is free: no continuity means no derivative, by the implication we just proved. The first two are the ones that fool people, because the graph is perfectly connected.
Picture it
Animation
Shows: Three ways differentiability fails — a rendered Manim animation.
Rendered with Manim.
Takeaway: All three survive the zoom test as a visible kink, not a line.
Picture it
Animation
Shows: The absolute value graph with its corner at the origin.
Two candidate slopes, no way to choose.
Takeaway: At a corner the left slope and the right slope disagree, so there is no single number for the derivative to be.
Intuition
Here is a test you can run in your head on any graph, with no algebra at all.
Zoom in on the point. Keep zooming. If the graph eventually looks like a single straight, non-vertical line, the function is differentiable there and that line is the tangent.
A corner stays a corner no matter how far you zoom - the angle never flattens. A cusp gets more vertical, not more flat. A break stays broken.
Differentiable is just a formal word for locally straight.
Analogy
Discussion prompt
Explain The zoom test by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Zoom in on the point. Keep zooming. If the graph eventually looks like a single straight, non-vertical line, the function is differentiable there and that line is the tangent.
Ranking
Put in order
Put the moves of Worked example: a cusp into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Dividing by the gap subtracts 1 from its exponent.
Worked example
The cube root squared makes a curve that comes to a sharp vertical point at the origin. Its two arms are both smooth, but they arrive going straight up.
\[ f(x) = x^{2/3}, \qquad a = 0, \qquad f(0) = 0 \]
Write the difference quotient and simplify the exponent
Why: Dividing by the gap subtracts 1 from its exponent. Two thirds minus one is negative one third.
\[ \frac{f(0+h) - f(0)}{h} = \frac{h^{2/3}}{h} = h^{-1/3} = \frac{1}{\sqrt[3]{h}} \]
Approach from the right
Why: The cube root of a tiny positive number is a small positive number, and one divided by something tiny is enormous. The quotient grows without bound.
\[ \lim_{h \to 0^{+}} \frac{1}{\sqrt[3]{h}} = +\infty \]
Approach from the left
Why: The cube root of a negative number is negative, so the quotient is enormous and negative. It runs off the other way.
\[ \lim_{h \to 0^{-}} \frac{1}{\sqrt[3]{h}} = -\infty \]
State the conclusion
Why: Neither one-sided limit is a finite number, and they head in opposite directions. There is no tangent line with a slope, so the derivative does not exist at the origin.
\[ f'(0) \ \text{does not exist (cusp)} \]
Verify with actual gaps
Why: As the gap shrinks by a factor of a thousand, the quotient grows by a factor of ten each time - and the sign flips depending on which side you come from. That runaway growth is exactly what a vertical arrival looks like in the algebra.
| gap h | cube root of h | difference quotient |
|---|---|---|
| 0.001 | 0.1 | 10 |
| 0.000001 | 0.01 | 100 |
| 0.000000001 | 0.001 | 1000 |
| -0.001 | -0.1 | -10 |
| -0.000001 | -0.01 | -100 |
Picture it
Animation
Shows: Each line of the worked example "a cusp", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: As the gap shrinks by a factor of a thousand, the quotient grows by a factor of ten each time - and the sign flips depending on which side you come from. That runaway growth is exactly what a vertical arrival looks like in the algebra.
Concept
Both involve the graph turning straight up, and neither has a derivative. They differ in whether the two sides run off the same way.
\[ g(x) = x^{1/3}: \qquad \frac{h^{1/3}}{h} = h^{-2/3} \;\longrightarrow\; +\infty \ \text{ from both sides} \]
Because both sides run off the same way, the cube-root curve has an honest vertical tangent line at the origin - it just has no slope, since vertical lines do not have one.
The cusp from the last slide has the two sides running off in opposite directions, so there is not even a vertical line to call the tangent. Either way, no derivative.
Counterexample
Discussion prompt
Both involve the graph turning straight up, and neither has a derivative. They differ in whether the two sides run off the same way.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Because both sides run off the same way, the cube-root curve has an honest vertical tangent line at the origin - it just has no slope, since vertical lines do not have one.
Ranking
Put in order
These are the steps of Pattern: is it differentiable here?, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Run these in order and stop at the first failure.
For a piecewise rule, step 2 is the whole game: match the heights first to get continuity, then match the one-sided slopes to get differentiability.
Edge cases
Discussion prompt
Pattern: is it differentiable here? works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Run these in order and stop at the first failure.
Prediction
Predict first
Which statement about g(x) = |x - 3| + 2 at the input x = 3 is true?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: g is continuous at 3 but not differentiable there
Why: The graph is an unbroken V with its corner at the point (3, 2), so g is continuous there. But the difference quotient tends to 1 from the right and -1 from the left, so no single slope exists and g is not differentiable at 3.
Check
Picture the graph first: it is the usual V, shifted right and up.
\[ g(x) = |x - 3| + 2 \]
Check your understanding
Which statement about g(x) = |x - 3| + 2 at the input x = 3 is true?
Answer: A
Why: The graph is an unbroken V with its corner at the point (3, 2), so g is continuous there. But the difference quotient tends to 1 from the right and -1 from the left, so no single slope exists and g is not differentiable at 3.
Elimination
Eliminate the wrong options
The limit as h approaches 0 of ((2 + h)^3 - 8)/h is a derivative in disguise. What is its value?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: This is the h-form definition for the cubing function at the input 2, since 2 cubed is 8. Expanding gives (12h + 6h^2 + h^3)/h = 12 + 6h + h^2, which tends to 12 as h goes to 0.
Check
Exams love to hand you a bare limit and ask for its value. Match it to the h-form definition before you touch any algebra.
\[ \lim_{h \to 0} \frac{(2+h)^3 - 8}{h} \]
Check your understanding
The limit as h approaches 0 of ((2 + h)^3 - 8)/h is a derivative in disguise. What is its value?
Answer: A
Why: This is the h-form definition for the cubing function at the input 2, since 2 cubed is 8. Expanding gives (12h + 6h^2 + h^3)/h = 12 + 6h + h^2, which tends to 12 as h goes to 0.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — From Average to Instantaneous · The Definition of the Derivative · The Derivative as a Function · What a Derivative Means · Reading the Derivative off a Graph · When the Derivative Fails to Exist. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
| function | derivative, straight from the definition |
|---|---|
| 3x - 5 | 3 |
| x^2 - 4x + 1 | 2x - 4 |
| x^3 - 3x | 3x^2 - 3 |
| 1/x | -1/x^2 |
| sqrt(x) | 1/(2 sqrt(x)) |
Next up: the power, constant, sum, and difference rules - shortcuts that reproduce every one of those answers in a single line, so you never have to run the definition again unless a problem asks you to.
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