This deck explains why the derivative of a product is not the product of the derivatives, then gives the product rule with its expanding-rectangle picture and the quotient rule, paying close attention to the order of the terms in its numerator. It covers when to simplify instead of grinding, products of three factors, and combining the rules, and finishes with rate-of-change applications such as marginal revenue and average cost. It targets the product-of-derivatives error, the reversed numerator, the unsquared denominator, and reaching for the quotient rule when plain division is easier.
Subject: Calculus I · 135 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I · Deck 07
Two functions multiplied. Two functions divided. Neither one behaves the way you would guess.
Objectives
By the end of this deck you can:
The rules themselves are two lines long. The whole deck is really about not making the four errors that cost the most points on an exam.
Warm-up
Discussion prompt
Before we open The Product and Quotient Rules: without looking back, what was the main idea of Power, Constant, Sum, and Difference Rules, 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:
The shortcut rules that replace the limit definition: constant, power, constant-multiple, sum and difference, plus the natural exponential. Covers rewriting radicals and reciprocals as powers before differentiating, higher-order derivatives, tangent lines, and horizontal tangents.
Section
Section 1
Concept
The sum rule is generous. The derivative of a sum really is the sum of the derivatives.
\[ \frac{d}{dx}\left[f(x)+g(x)\right] = f'(x)+g'(x) \]
So it feels obvious that a product should behave the same way.
\[ \frac{d}{dx}\left[f(x)\,g(x)\right] \overset{?}{=} f'(x)\,g'(x) \]
That guess is false. Do not take my word for it. Watch it fail on a product you can already differentiate two other ways.
Counterexample
Discussion prompt
The sum rule is generous. The derivative of a sum really is the sum of the derivatives.
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: The guess that feels right, and is not — a rendered Manim animation.
Rendered with Manim.
Takeaway: One counterexample settles it permanently.
Ranking
Put in order
Put the moves of Test the guess on a product you already know 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 two factors multiply out to x squared, and the power rule handles that instantly.
Worked example
Take the simplest product there is: a number times itself.
\[ f(x) = x \cdot x = x^{2} \]
Differentiate the real function
Why: The two factors multiply out to x squared, and the power rule handles that instantly.
\[ f'(x) = 2x \]
Now run the tempting guess
Why: Each factor has derivative 1, so the product of the derivatives is 1 times 1.
\[ f'(x) \overset{?}{=} (1)(1) = 1 \]
Compare the two answers at one point
Why: A single point is enough to kill a false rule. At x = 3 the truth is 6 and the guess says 1.
\[ 2(3) = 6 \qquad \text{versus} \qquad 1 \]
Verify with nearby values which answer the graph agrees with
Why: Average slopes from x = 3 to a point just to the right march toward 6, not toward 1. The guess is dead.
| h | f(3 + h) | average slope over h |
|---|---|---|
| 0.1 | 9.61 | 6.1 |
| 0.01 | 9.0601 | 6.01 |
| 0.001 | 9.006001 | 6.001 |
Trap
Differentiate each factor and multiply the results.
\[ f(x) = x \cdot x^{2} \]
Multiply the two derivatives
Why: Derivative of the first factor is 1; derivative of the second is 2x. Multiplying them feels like the sum rule with different punctuation.
\[ f'(x) \overset{?}{=} (1)(2x) = 2x \]
Evaluate at x = 2
Why: The shortcut predicts a slope of 4 there.
\[ f'(2) \overset{?}{=} 4 \]
Multiply the factors out first, then use the power rule you already trust.
\[ f(x) = x \cdot x^{2} = x^{3} \]
Differentiate x cubed
Why: The power rule is not in dispute here, so this answer is the reference truth.
\[ f'(x) = 3x^{2} \]
Evaluate at x = 2 and confirm with real slopes
Why: Average slopes from x = 2 close in on 12, exactly three times the shortcut's answer. The shortcut is not slightly off, it is wrong.
\[ f'(2) = 3(2)^{2} = 12 \]
| x | x cubed | average slope from 2 |
|---|---|---|
| 2.1 | 9.261 | 12.61 |
| 2.01 | 8.120601 | 12.0601 |
| 2.001 | 8.012006001 | 12.006001 |
Intuition
Picture a rectangle whose width is one function and whose height is the other. Their product is its area.
Figure (svg): A rectangle of width u and height v, with a thin vertical strip of width du added on the right, a thin horizontal strip of height dv added on top, and a tiny corner square where the two strips meet.
Now let both sides grow a little. The new area arrives in two strips: the height times the extra width, plus the width times the extra height.
There is also a tiny corner square where the two strips overlap, but it is small times small. Shrink the change and that corner vanishes faster than everything else.
Two strips, so two terms. That is the whole product rule, and it is why one term could never be enough.
Analogy
Discussion prompt
Explain The expanding rectangle 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:
Picture a rectangle whose width is one function and whose height is the other. Their product is its area.
Picture it
Animation
Shows: The expanding rectangle — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two strips grow at once, and each contributes its own term.
Concept
Here is the rule those two strips were describing.
\[ \frac{d}{dx}\left[f(x)\,g(x)\right] = f'(x)\,g(x) + f(x)\,g'(x) \]
product rule — Differentiate one factor at a time, leaving the other one alone, then add the two results.
Notice what each term keeps. In every term, exactly one factor has been differentiated and the other is untouched. If a term has both derivatives in it, or neither, it does not belong.
Explain it
Discussion prompt
Explain The product rule 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:
Notice what each term keeps. In every term, exactly one factor has been differentiated and the other is untouched. If a term has both derivatives in it, or neither, it does not belong.
Picture it
Animation
Shows: The product rule with its spoken form underneath.
The sentence survives exam pressure. The symbols do not.
Takeaway: First prime second, plus first second prime. Saying it aloud is what keeps it intact when the algebra gets busy.
Concept
Out loud, every time, until it is automatic: first times the derivative of the second, plus second times the derivative of the first.
\[ (fg)' = f\,g' + g\,f' \]
Some books write the other term first. Both chants produce the same two terms, so pick one and never switch. Consistency is what stops you from dropping a term under time pressure.
Concept
Addition does not care about order, so these are the same answer.
\[ f'g + fg' \;=\; fg' + f'g \]
It also does not matter which factor you call the first one. The product rule is symmetric.
\[ \frac{d}{dx}\left[g f\right] = g'f + gf' = f'g + fg' \]
Hold on to that freedom. The quotient rule, coming later, has none of it, and that difference is where most lost points live.
Concept
When the two factors are named as variables, the rule reads like the rectangle picture.
\[ \frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx} \]
Read it as: width times the change in height, plus height times the change in width.
In physics and economics you will meet this form far more often than the prime form, so get comfortable reading both.
Picture it
Animation
Shows: The same rule in Leibniz notation — a rendered Manim animation.
Rendered with Manim.
Takeaway: Same statement, different century.
Concept
A quick sketch of the proof, so the rule is not a magic spell. Start with the change in the product over a small step.
\[ \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \]
Add and subtract the same middle quantity in the numerator. Adding zero changes nothing, but it splits the mess into two clean pieces.
\[ f(x+h)g(x+h) \;-\; f(x+h)g(x) \;+\; f(x+h)g(x) \;-\; f(x)g(x) \]
Group and factor, and each piece is one function times a difference quotient of the other.
\[ f(x+h)\,\frac{g(x+h)-g(x)}{h} \;+\; g(x)\,\frac{f(x+h)-f(x)}{h} \]
Let the step shrink to zero and you get the two strips from the rectangle picture.
\[ f(x)\,g'(x) + g(x)\,f'(x) \]
Step zero
Discussion prompt
Product rule on two polynomials — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the two factors and differentiate each one separately
Answer:
Worked example
A first drill, chosen so we can check the answer a completely different way.
\[ f(x) = (3x^{2}+1)(x^{3}-5x) \]
Name the two factors and differentiate each one separately
Why: Writing the four pieces down before assembling them is what keeps a term from going missing.
\[ u = 3x^{2}+1,\quad u' = 6x,\qquad v = x^{3}-5x,\quad v' = 3x^{2}-5 \]
Assemble the two product-rule terms
Why: Derivative of the first times the second, plus the first times the derivative of the second.
\[ f'(x) = 6x\left(x^{3}-5x\right) + \left(3x^{2}+1\right)\left(3x^{2}-5\right) \]
Expand each term
Why: Multiply out before combining, so no like terms get merged by accident.
\[ = 6x^{4} - 30x^{2} \;+\; \left(9x^{4} - 15x^{2} + 3x^{2} - 5\right) \]
Combine like terms
Why: Six plus nine gives fifteen fourth-power terms; minus thirty minus twelve gives minus forty-two square terms.
\[ f'(x) = 15x^{4} - 42x^{2} - 5 \]
Verify by expanding the original product and using the power rule
Why: The original multiplies out to three x to the fifth minus fourteen x cubed minus five x, whose power-rule derivative is the same fifteen, minus forty-two, minus five. Two independent routes, one answer.
\[ f(x) = 3x^{5} - 14x^{3} - 5x \;\Longrightarrow\; f'(x) = 15x^{4} - 42x^{2} - 5 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Product rule on two polynomials", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original multiplies out to three x to the fifth minus fourteen x cubed minus five x, whose power-rule derivative is the same fifteen, minus forty-two, minus five. Two independent routes, one answer.
Section
Section 2
Concept
In the last example we could have skipped the product rule and just multiplied out. That was on purpose, so the check was airtight.
Most products refuse to be multiplied out. There is nothing to expand here.
\[ x^{2}e^{x}, \qquad \sqrt{x}\,(x+4), \qquad \frac{e^{x}}{x^{2}+1} \]
That is why the rule exists. It is not a shortcut for polynomials, it is the only route once a factor is an exponential, a root, or an unnamed function from a table.
Picture it
Animation
Shows: When expanding is not an option — a rendered Manim animation.
Rendered with Manim.
Takeaway: Expanding works for polynomials and almost nothing else.
Estimation
Predict first
Recall the rule from the last deck: the natural exponential is its own derivative.
Commit before you compute: what does A polynomial times an exponential 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 the two predicted flat spots against real function values
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 exponential is never zero, so the factored derivative is zero only at x = 0 and x = -2.
Worked example
Recall the rule from the last deck: the natural exponential is its own derivative.
\[ f(x) = x^{2}e^{x}, \qquad \frac{d}{dx}\left[e^{x}\right] = e^{x} \]
Name the factors and their derivatives
Why: The exponential factor is the easy one; its derivative is itself.
\[ u = x^{2},\quad u' = 2x,\qquad v = e^{x},\quad v' = e^{x} \]
Apply the product rule
Why: Derivative of the first times the second, plus the first times the derivative of the second.
\[ f'(x) = 2x\,e^{x} + x^{2}e^{x} \]
Factor out what both terms share
Why: Both terms carry an exponential and at least one x. Factoring makes the zeros visible, which is what most exam questions ask about next.
\[ f'(x) = x\,e^{x}\left(x+2\right) \]
Verify the two predicted flat spots against real function values
Why: The exponential is never zero, so the factored derivative is zero only at x = 0 and x = -2. Values around x = -2 rise then fall, which is exactly the local maximum a zero derivative there predicts.
| x | value of x squared times e to the x | what it shows |
|---|---|---|
| -3 | about 0.448 | lower |
| -2 | about 0.541 | highest of the three, so flat here |
| -1 | about 0.368 | lower |
| 0 | 0 | a minimum, so flat here too |
Picture it
Animation
Shows: Each line of the worked example "A polynomial times an exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The exponential is never zero, so the factored derivative is zero only at x = 0 and x = -2. Values around x = -2 rise then fall, which is exactly the local maximum a zero derivative there predicts.
Missing information
Discussion prompt
Roots go into the product rule as fractional powers, exactly as in the last deck.
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 power rule on the one-half power subtracts one from the exponent, giving negative one-half.
Worked example
Roots go into the product rule as fractional powers, exactly as in the last deck.
\[ f(x) = \sqrt{x}\,(x+4) = x^{1/2}(x+4) \]
Name the factors and differentiate each
Why: The power rule on the one-half power subtracts one from the exponent, giving negative one-half.
\[ u = x^{1/2},\quad u' = \tfrac{1}{2}x^{-1/2},\qquad v = x+4,\quad v' = 1 \]
Apply the product rule
Why: One term differentiates the root and keeps the binomial; the other keeps the root and differentiates the binomial.
\[ f'(x) = \tfrac{1}{2}x^{-1/2}(x+4) + x^{1/2}(1) \]
Put everything over the common denominator
Why: Write the negative power as a root in the denominator, then give the second term the same denominator so the numerators can combine.
\[ f'(x) = \frac{x+4}{2\sqrt{x}} + \frac{2x}{2\sqrt{x}} = \frac{3x+4}{2\sqrt{x}} \]
Verify by distributing first and using the power rule instead
Why: Distributing gives the three-halves power plus four times the one-half power, whose derivative is three-halves root x plus two over root x. Over the common denominator that is the same three x plus four over two root x.
\[ f(x) = x^{3/2}+4x^{1/2} \Longrightarrow f'(x) = \tfrac{3}{2}x^{1/2}+2x^{-1/2} = \frac{3x+4}{2\sqrt{x}} \;\checkmark \]
Intuition
Almost every dropped term traces back to doing the labeling in your head.
Write the four pieces on the page before you assemble anything: the first factor, its derivative, the second factor, its derivative.
| piece | what goes here |
|---|---|
| first factor | copy it exactly |
| derivative of the first | differentiate it alone |
| second factor | copy it exactly |
| derivative of the second | differentiate it alone |
Then the rule is just reading two of those boxes, then the other two. Bookkeeping beats memory every single time.
Picture it
Animation
Shows: Label first, differentiate second — a rendered Manim animation.
Rendered with Manim.
Takeaway: Bookkeeping is what keeps a two-rule problem honest.
Concept
Group two of the three factors and apply the rule twice, and the pattern that falls out is easy to remember.
\[ (fgh)' = f'gh + fg'h + fgh' \]
One term per factor. In each term exactly one factor is differentiated and the rest are copied down untouched.
Same shape for four factors: four terms. If your term count does not match your factor count, something is missing.
Picture it
Animation
Shows: Three factors: the rule extends — a rendered Manim animation.
Rendered with Manim.
Takeaway: The pattern continues for any number of factors.
Step zero
Discussion prompt
Product rule with three factors — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate the first factor, copy the other two
Answer:
Worked example
\[ f(x) = x(x+1)(x+2) \]
Three factors, so expect three terms.
Differentiate the first factor, copy the other two
Why: Derivative of x is 1, so this term is just the product of the two survivors.
\[ (1)(x+1)(x+2) = x^{2}+3x+2 \]
Differentiate the middle factor, copy the outer two
Why: Derivative of x plus one is 1, leaving x times the quantity x plus two.
\[ x(1)(x+2) = x^{2}+2x \]
Differentiate the last factor, copy the first two
Why: Derivative of x plus two is 1, leaving x times the quantity x plus one.
\[ x(x+1)(1) = x^{2}+x \]
Add the three terms
Why: Three square terms, and the linear parts add to six x.
\[ f'(x) = 3x^{2}+6x+2 \]
Verify by expanding the original and differentiating
Why: The original expands to x cubed plus three x squared plus two x, whose power-rule derivative is three x squared plus six x plus two. Same answer.
\[ f(x) = x^{3}+3x^{2}+2x \Longrightarrow f'(x) = 3x^{2}+6x+2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Product rule with three factors", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original expands to x cubed plus three x squared plus two x, whose power-rule derivative is three x squared plus six x plus two. Same answer.
Ranking
Put in order
These are the steps of How to run the product rule, 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
The reusable recipe, good for two factors or ten.
Step 3 is the whole rule. Steps 2 and 4 are what keep it from going wrong at speed.
Elimination
Eliminate the wrong options
What is the derivative of the function shown?
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: The product rule gives 2x times (x - 5) plus (x squared + 3) times 1, which expands to 2x squared - 10x + x squared + 3, and that combines to 3x squared - 10x + 3. Expanding the original to x cubed - 5x squared + 3x - 15 and differentiating confirms it.
Check
Work it on paper before you pick. Label the factors first.
\[ f(x) = (x^{2}+3)(x-5) \]
Check your understanding
What is the derivative of the function shown?
Answer: A
Why: The product rule gives 2x times (x - 5) plus (x squared + 3) times 1, which expands to 2x squared - 10x + x squared + 3, and that combines to 3x squared - 10x + 3. Expanding the original to x cubed - 5x squared + 3x - 15 and differentiating confirms it.
Check
You do not need formulas for the two functions. Values at the one point are enough.
| function | value at x = 2 | derivative at x = 2 |
|---|---|---|
| f | 3 | -1 |
| g | 5 | 4 |
Check your understanding
Using the table, what is the derivative of the product f times g at x = 2?
Answer: A
Why: The product rule gives f prime times g plus f times g prime, which is (-1)(5) + (3)(4) = -5 + 12 = 7 at x = 2. Both terms are needed and both are added.
Comparison
Comparison matrix
From Check yourself: a product rule from a table: refill the value at x = 2 column from what you know. The rest of the table is as it appeared.
| function | value at x = 2 | derivative at x = 2 |
|---|---|---|
| f | 3 | -1 |
| g | 5 | 4 |
Section
Section 3
Concept
Before learning a new rule, kill the shortcut again. Here is a quotient you can simplify away.
\[ f(x) = \frac{x^{3}}{x} = x^{2} \quad\Longrightarrow\quad f'(x) = 2x \]
Now try dividing the derivatives.
\[ \frac{d}{dx}\left[\frac{x^{3}}{x}\right] \overset{?}{=} \frac{3x^{2}}{1} = 3x^{2} \]
At x equal to 1 the truth is 2 and the shortcut says 3. Quotients need their own rule too, and this one is uglier because subtraction gets involved.
Intuition
Think of a fraction as a share: numerator is what you have, denominator is how many ways it is split.
Growing the top pushes the fraction up. Growing the bottom pushes the fraction down. The two effects fight.
That fight is the minus sign in the quotient rule. It is not decoration; it is the denominator's contribution pulling the other way.
And because the two effects pull in opposite directions, swapping them does not give a slightly different answer. It gives the exact negative of the right one.
Concept
\[ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\left[g(x)\right]^{2}} \]
quotient rule — Bottom times the derivative of the top, minus top times the derivative of the bottom, all divided by the bottom squared.
Three things must be right: the order of the two products, the minus between them, and the square on the denominator. Miss any one and the answer is wrong.
It is valid wherever the denominator is not zero, which is exactly where the original fraction was defined anyway.
Sorting
Sort into buckets
These are the pieces of The Product and Quotient Rules, out of order. Put each one back under the part of the lesson it belongs to.
Picture it
Animation
Shows: The quotient rule with the subtraction highlighted.
The minus sign is what makes order matter.
Takeaway: Because the numerator subtracts, swapping the two terms flips every sign downstream. The order is not a convention you may vary.
Concept
Say it with the rhythm and it sticks: low d-high, minus high d-low, over the square of what is below.
\[ \left(\frac{\text{hi}}{\text{lo}}\right)' = \frac{\text{lo}\cdot d(\text{hi}) - \text{hi}\cdot d(\text{lo})}{\text{lo}^{2}} \]
The chant starts at the bottom, and so does the correct numerator. That is not an accident, it is the whole point of the mnemonic.
Whatever chant you use, write the denominator squared first, before you fill in the numerator. Nobody forgets a square they already wrote.
Concept
You do not have to take it on faith. It is the product rule wearing a disguise. Call the quotient by a name.
\[ q(x) = \frac{f(x)}{g(x)} \quad\Longrightarrow\quad f(x) = q(x)\,g(x) \]
Differentiate that product.
\[ f' = q'g + qg' \]
Solve for the derivative you want, then put the quotient back in for its name.
\[ q' = \frac{f' - qg'}{g} = \frac{f' - \dfrac{f}{g}g'}{g} \]
Clear the inner fraction by multiplying top and bottom by the denominator, and the square appears on its own.
\[ q' = \frac{f'g - fg'}{g^{2}} \]
Missing information
Discussion prompt
The cleanest possible first quotient, and one we can check a second way.
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:
Same discipline as the product rule: get all four pieces on the page before assembling.
Worked example
\[ f(x) = \frac{2x+1}{x-3} \]
The cleanest possible first quotient, and one we can check a second way.
Label top and bottom and differentiate each
Why: Same discipline as the product rule: get all four pieces on the page before assembling.
\[ \text{hi} = 2x+1,\ d(\text{hi}) = 2,\qquad \text{lo} = x-3,\ d(\text{lo}) = 1 \]
Write the denominator squared first, then fill in the numerator
Why: Writing the square immediately removes the most common quotient-rule error before it can happen.
\[ f'(x) = \frac{2(x-3) - (2x+1)(1)}{(x-3)^{2}} \]
Distribute the minus sign across the entire second product
Why: The minus applies to both terms of two x plus one, not just the first. This is where sign errors are born.
\[ = \frac{2x - 6 - 2x - 1}{(x-3)^{2}} \]
Combine the numerator
Why: The x terms cancel completely, leaving a constant on top.
\[ f'(x) = \frac{-7}{(x-3)^{2}} \]
Verify by rewriting the original as a constant plus a simple reciprocal
Why: Splitting off the whole part gives two plus seven over x minus three, whose derivative by the power rule is negative seven over the square. Identical answer, and the negative sign is confirmed by the fact that this function decreases on both sides of its asymptote.
\[ f(x) = \frac{2(x-3)+7}{x-3} = 2 + 7(x-3)^{-1} \Longrightarrow f'(x) = -7(x-3)^{-2} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Quotient rule on a linear over linear", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Splitting off the whole part gives two plus seven over x minus three, whose derivative by the power rule is negative seven over the square. Identical answer, and the negative sign is confirmed by the fact that this function decreases on both sides of its asymptote.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Starting the numerator with the top instead of the bottom.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The two products look symmetric, so the order feels harmless.
Bottom times derivative of top comes first. Always.
Why: The two products look symmetric, so the order feels harmless.
Trap
Starting the numerator with the top instead of the bottom.
\[ f(x) = \frac{2x+1}{x-3} \]
Write top times derivative of bottom, minus bottom times derivative of top
Why: The two products look symmetric, so the order feels harmless.
\[ f'(x) \overset{?}{=} \frac{(2x+1)(1) - 2(x-3)}{(x-3)^{2}} \]
Simplify and get the wrong sign everywhere
Why: Two x plus one minus two x plus six is positive seven. This claims the function is increasing near x = 4, when it is actually falling steeply.
\[ f'(x) \overset{?}{=} \frac{7}{(x-3)^{2}} \]
Bottom times derivative of top comes first. Always.
\[ f(x) = \frac{2x+1}{x-3} \]
Write bottom times derivative of top, minus top times derivative of bottom
Why: The chant starts low, so the numerator starts with the low function.
\[ f'(x) = \frac{2(x-3) - (2x+1)(1)}{(x-3)^{2}} = \frac{-7}{(x-3)^{2}} \]
Confirm the sign against real values
Why: At x = 4 the function is 9 and at x = 4.1 it is about 8.36, so the function is decreasing and the derivative must be negative. Only the correct order gives that.
| x | value of the function | moving which way |
|---|---|---|
| 4 | 9 | start |
| 4.1 | about 8.364 | down |
| 4.5 | about 6.667 | down |
Pattern
Step through it
Step through Trap: reversing the order in the numerator one row at a time. What is driving the change, and what would the row after the last one be?
Concept
Subtraction is the only operation in these rules that cares about order, and the quotient rule is the only place it appears.
\[ \frac{fg' - f'g}{g^{2}} = -\,\frac{f'g - fg'}{g^{2}} \]
So a reversed numerator does not give an answer that is a little off. It gives the exact negative of the truth: every increase becomes a decrease, every maximum becomes a minimum.
That is why a sign sanity-check is worth ten seconds. Pick a convenient x, see whether the function is rising or falling there, and confirm your derivative agrees.
Estimation
Predict first
This one cannot be simplified away, so the rule is the only route.
Commit before you compute: what does Quotient rule with a square in the denominator 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 with a nearby-value slope and a symmetry check
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 formula predicts a slope of one half at x = 1, and real values give about 0.4975 over a step of one hundredth.
Worked example
\[ f(x) = \frac{x^{2}}{x^{2}+1} \]
This one cannot be simplified away, so the rule is the only route.
Label the pieces
Why: Top and bottom differ by one, which is exactly the kind of near-match that produces heavy cancellation later.
\[ \text{hi} = x^{2},\ d(\text{hi}) = 2x,\qquad \text{lo} = x^{2}+1,\ d(\text{lo}) = 2x \]
Assemble, denominator squared first
Why: The bottom squared is the quantity x squared plus one, squared. Do not expand it; you almost never need it expanded.
\[ f'(x) = \frac{2x\left(x^{2}+1\right) - x^{2}(2x)}{\left(x^{2}+1\right)^{2}} \]
Expand only the numerator
Why: Two x cubed plus two x, minus two x cubed. The cubic terms cancel.
\[ = \frac{2x^{3}+2x-2x^{3}}{\left(x^{2}+1\right)^{2}} = \frac{2x}{\left(x^{2}+1\right)^{2}} \]
Verify with a nearby-value slope and a symmetry check
Why: The formula predicts a slope of one half at x = 1, and real values give about 0.4975 over a step of one hundredth. The original is an even function, so its derivative must be odd, and two x over an even square is indeed odd.
| x | value of the function | average slope from x = 1 |
|---|---|---|
| 1 | 0.5 | - |
| 1.01 | about 0.504975 | about 0.4975 |
| formula | 2x over the square | 0.5 at x = 1 |
Picture it
Animation
Shows: Each line of the worked example "Quotient rule with a square in the denominator", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts a slope of one half at x = 1, and real values give about 0.4975 over a step of one hundredth. The original is an even function, so its derivative must be odd, and two x over an even square is indeed odd.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Copying the denominator down as it was.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The numerator work is fine, so the error hides in the one place nobody re-reads.
The denominator is squared. Write the square before the numerator.
Why: The numerator work is fine, so the error hides in the one place nobody re-reads.
Trap
Copying the denominator down as it was.
\[ f(x) = \frac{x^{2}}{x^{2}+1} \]
Assemble the numerator correctly, then reuse the original denominator
Why: The numerator work is fine, so the error hides in the one place nobody re-reads.
\[ f'(x) \overset{?}{=} \frac{2x}{x^{2}+1} \]
Test it at x = 1
Why: This claims a slope of 1 at x = 1. But the function itself can never exceed 1, and it is already at one half, so a slope of 1 would push it past its ceiling almost immediately. Impossible.
\[ f'(1) \overset{?}{=} \frac{2}{2} = 1 \]
The denominator is squared. Write the square before the numerator.
\[ f(x) = \frac{x^{2}}{x^{2}+1} \]
Keep the square on the bottom
Why: The rule divides by the bottom squared, which came from clearing the inner fraction in the derivation.
\[ f'(x) = \frac{2x}{\left(x^{2}+1\right)^{2}} \]
Test the same point against real values
Why: The correct formula predicts one half at x = 1, and the nearby-value slope is about 0.4975. The unsquared version overshoots by a factor of two.
| what | slope predicted at x = 1 | real nearby slope |
|---|---|---|
| unsquared version | 1 | about 0.4975 |
| correct version | 0.5 | about 0.4975 |
Trade off
Comparison matrix
From Trap: forgetting to square the denominator: every row here is a choice with a cost. Fill the slope predicted at x = 1 column, then say which row you would actually pick and what you give up for it.
| what | slope predicted at x = 1 | real nearby slope |
|---|---|---|
| unsquared version | 1 | about 0.4975 |
| correct version | 0.5 | about 0.4975 |
Concept
A square is never negative, so the bottom of a quotient-rule answer contributes nothing to the sign.
\[ \left[g(x)\right]^{2} > 0 \quad \text{wherever } g(x) \neq 0 \]
So the derivative is positive exactly where the numerator is positive. Increasing and decreasing questions reduce to reading the top.
That is a genuine time saver later, in curve sketching. It also gives you a quick way to spot a reversed numerator: if the sign contradicts the picture, you swapped the order.
Picture it
Animation
Shows: The squared denominator is a free sign check — a rendered Manim animation.
Rendered with Manim.
Takeaway: If your sign looks wrong, it is the numerator that is wrong.
Ranking
Put in order
Put the moves of A tangent line to a quotient 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. A tangent line needs a point as well as a slope, and the point comes from the original function, not its derivative.
Worked example
Find the tangent line at the point where x equals 1.
\[ f(x) = \frac{x}{x+1} \]
Find the point on the curve
Why: A tangent line needs a point as well as a slope, and the point comes from the original function, not its derivative.
\[ f(1) = \frac{1}{2} \]
Differentiate with the quotient rule
Why: Bottom times derivative of top is x plus one; top times derivative of bottom is x. Their difference is 1.
\[ f'(x) = \frac{(1)(x+1) - x(1)}{(x+1)^{2}} = \frac{1}{(x+1)^{2}} \]
Evaluate the derivative at the point
Why: The derivative is a function; the slope of this particular tangent is its value at x equal to 1.
\[ f'(1) = \frac{1}{4} \]
Write the line in point-slope form and simplify
Why: Point-slope is safest: the point and the slope go straight in with no rearranging.
\[ y - \tfrac{1}{2} = \tfrac{1}{4}(x-1) \quad\Longrightarrow\quad y = \tfrac{1}{4}x + \tfrac{1}{4} \]
Verify the line touches the curve and hugs it nearby
Why: At x = 1 the line gives one quarter plus one quarter, which is one half, matching the curve exactly. At x = 1.1 the curve is about 0.5238 and the line gives 0.525, so they agree to three decimals just as a tangent should.
| x | curve value | tangent line value |
|---|---|---|
| 1 | 0.5 | 0.5 |
| 1.1 | about 0.5238 | 0.525 |
| 0.9 | about 0.4737 | 0.475 |
Picture it
Animation
Shows: Each line of the worked example "A tangent line to a quotient", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x = 1 the line gives one quarter plus one quarter, which is one half, matching the curve exactly. At x = 1.1 the curve is about 0.5238 and the line gives 0.525, so they agree to three decimals just as a tangent should.
Pattern
Prediction
Predict first
Which expression is the derivative of the function shown?
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: (-x^2 - 4x + 1) / (x^2 + 1)^2
Why: Bottom times derivative of top is x squared plus 1; top times derivative of bottom is 2x squared plus 4x. Subtracting gives x squared + 1 - 2x squared - 4x, which is -x squared - 4x + 1, over the bottom squared.
Check
Do it on paper. Write the squared denominator first.
\[ f(x) = \frac{x+2}{x^{2}+1} \]
Check your understanding
Which expression is the derivative of the function shown?
Answer: A
Why: Bottom times derivative of top is x squared plus 1; top times derivative of bottom is 2x squared plus 4x. Subtracting gives x squared + 1 - 2x squared - 4x, which is -x squared - 4x + 1, over the bottom squared.
Check
Same two functions as before, same single point.
| function | value at x = 2 | derivative at x = 2 |
|---|---|---|
| f | 3 | -1 |
| g | 5 | 4 |
Check your understanding
Using the table, what is the derivative of f divided by g at x = 2?
Answer: A
Why: The quotient rule gives (f prime times g minus f times g prime) over g squared, which is ((-1)(5) - (3)(4)) divided by 25, that is (-5 - 12)/25 = -17/25 at x = 2.
Comparison
Comparison matrix
From Check yourself: a quotient from a table: refill the derivative at x = 2 column from what you know. The rest of the table is as it appeared.
| function | value at x = 2 | derivative at x = 2 |
|---|---|---|
| f | 3 | -1 |
| g | 5 | 4 |
Section
Section 4
Concept
A fraction bar is not a command to use the quotient rule. It is a division problem, and sometimes the division is trivial.
When the denominator is one term, split the fraction into pieces and use the power rule.
\[ \frac{a+b+c}{m} = \frac{a}{m} + \frac{b}{m} + \frac{c}{m} \]
That is only legal when the denominator is the single term. You may never split a sum in the denominator that way.
\[ \frac{m}{a+b} \neq \frac{m}{a} + \frac{m}{b} \]
Picture it
Animation
Shows: A quotient simplified before differentiating.
Simplify first whenever the algebra allows.
Takeaway: If the division simplifies, do it before differentiating. The quotient rule is for quotients you cannot get rid of.
Hypothesis
Predict first
Simplify first, then differentiate 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: Divide every term in the numerator by the single term below
Why: Each division is just subtracting one from the exponent, which is arithmetic rather than calculus.
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
\[ f(x) = \frac{x^{3}+4x^{2}-x}{x}, \qquad x \neq 0 \]
The quotient rule would work here. It is also three times the labor.
Divide every term in the numerator by the single term below
Why: Each division is just subtracting one from the exponent, which is arithmetic rather than calculus.
\[ f(x) = x^{2} + 4x - 1 \]
Differentiate with the power rule
Why: A quadratic differentiates in one line and there is no denominator left to get wrong.
\[ f'(x) = 2x + 4 \]
Now run the quotient rule as a cross-check
Why: Bottom times derivative of top is x times the quantity three x squared plus eight x minus one; top times derivative of bottom is the numerator itself.
\[ f'(x) = \frac{x\left(3x^{2}+8x-1\right) - \left(x^{3}+4x^{2}-x\right)(1)}{x^{2}} \]
Verify the two routes agree
Why: The numerator collapses to two x cubed plus four x squared, and dividing by x squared gives two x plus four, exactly what the one-line method produced. Same answer, five times the work.
\[ = \frac{2x^{3}+4x^{2}}{x^{2}} = 2x+4 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Simplify first, then differentiate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The numerator collapses to two x cubed plus four x squared, and dividing by x squared gives two x plus four, exactly what the one-line method produced. Same answer, five times the work.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Seeing a fraction bar and reaching for a rule, then taking a shortcut inside it.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This is the quotient version of the product-of-derivatives error, and it is most tempting exactly when the fraction looks simple.
The bottom is one term, so divide through and forget the rule entirely.
Why: This is the quotient version of the product-of-derivatives error, and it is most tempting exactly when the fraction looks simple.
Trap
Seeing a fraction bar and reaching for a rule, then taking a shortcut inside it.
\[ f(x) = \frac{6x^{5}-4x^{3}}{2x^{2}} \]
Differentiate the top and the bottom and divide
Why: This is the quotient version of the product-of-derivatives error, and it is most tempting exactly when the fraction looks simple.
\[ f'(x) \overset{?}{=} \frac{30x^{4}-12x^{2}}{4x} = 7.5x^{3}-3x \]
Test at x = 1
Why: The shortcut predicts a slope of 4.5 there. Real nearby values give about 7.09, so this is not a rounding disagreement.
\[ f'(1) \overset{?}{=} 4.5 \]
The bottom is one term, so divide through and forget the rule entirely.
\[ f(x) = \frac{6x^{5}}{2x^{2}} - \frac{4x^{3}}{2x^{2}} = 3x^{3}-2x \]
Differentiate the simplified form
Why: Two power-rule terms, no fraction, nothing to square.
\[ f'(x) = 9x^{2}-2 \]
Confirm at x = 1 with real values
Why: The correct formula predicts 7, and the average slope from x = 1 to x = 1.01 is about 7.09, closing in on 7. The full quotient rule, done properly, also returns nine x squared minus two.
| x | value of the function | average slope from x = 1 |
|---|---|---|
| 1 | 1 | - |
| 1.01 | about 1.070903 | about 7.09 |
| 1.001 | about 1.007009 | about 7.009 |
Pattern
Step through it
Step through Trap: grabbing the quotient rule instead of dividing one row at a time. What is driving the change, and what would the row after the last one be?
Concept
A second escape hatch: any single-term denominator can move upstairs as a negative power.
\[ \frac{1}{x^{n}} = x^{-n} \]
That turns a quotient into a sum of powers, or into a product, and both of those you can already handle.
Use it whenever the denominator is a single power of x. It is not worth it when the denominator is a sum, because a negative power of a sum needs the chain rule from the next deck.
Picture it
Animation
Shows: Negative exponents beat the quotient rule — a rendered Manim animation.
Rendered with Manim.
Takeaway: Reach for the quotient rule only when nothing simpler applies.
Step zero
Discussion prompt
Rewriting with negative exponents — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split the fraction and reduce each piece
Answer:
Worked example
\[ f(x) = \frac{x^{2}+1}{x^{3}} \]
Single-term denominator, so split it up rather than grinding.
Split the fraction and reduce each piece
Why: The first piece loses three from its exponent, leaving a negative first power; the second becomes a negative third power.
\[ f(x) = \frac{x^{2}}{x^{3}} + \frac{1}{x^{3}} = x^{-1} + x^{-3} \]
Differentiate term by term with the power rule
Why: Subtract one from each negative exponent and multiply by the old exponent, keeping the signs straight.
\[ f'(x) = -x^{-2} - 3x^{-4} \]
Rewrite as a single fraction if the answer must be presented that way
Why: Common denominator is the fourth power, so the first term contributes a square on top.
\[ f'(x) = -\frac{1}{x^{2}} - \frac{3}{x^{4}} = \frac{-x^{2}-3}{x^{4}} \]
Verify by running the quotient rule on the original
Why: Bottom times derivative of top is two x to the fourth; top times derivative of bottom is three x to the fourth plus three x squared. Subtracting and dividing by x to the sixth gives negative x squared minus three, over x to the fourth. Same answer.
\[ \frac{2x\cdot x^{3} - \left(x^{2}+1\right)3x^{2}}{x^{6}} = \frac{-x^{4}-3x^{2}}{x^{6}} = \frac{-x^{2}-3}{x^{4}} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Rewriting with negative exponents", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Bottom times derivative of top is two x to the fourth; top times derivative of bottom is three x to the fourth plus three x squared. Subtracting and dividing by x to the sixth gives negative x squared minus three, over x to the fourth. Same answer.
Constraint
Discussion prompt
Run Choosing your method with this step confiscated:
Are both factors short polynomials? Multiplying out is usually faster than the product rule.
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
Spend five seconds here and you will often save two minutes.
Both routes give the same derivative when both are legal. The rules are never wrong, only slower, so if you are unsure, use the rule and keep moving.
Edge cases
Discussion prompt
Choosing your method 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:
Spend five seconds here and you will often save two minutes.
Commit first
Predict first
Which line is both the fastest legal route and the correct derivative?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Divide first to get (1/3)x^3 - 2x, then differentiate: x^2 - 2
Why: The denominator is the single term 3x, so dividing through gives one third x cubed minus 2x. The power rule then gives x squared minus 2. Running the full quotient rule returns the same x squared minus 2, with far more algebra.
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
Look at the denominator before you decide anything.
\[ f(x) = \frac{x^{4}-6x^{2}}{3x} \]
Check your understanding
Which line is both the fastest legal route and the correct derivative?
Answer: A
Why: The denominator is the single term 3x, so dividing through gives one third x cubed minus 2x. The power rule then gives x squared minus 2. Running the full quotient rule returns the same x squared minus 2, with far more algebra.
Section
Section 5
Concept
Real problems nest. The top of a fraction is often itself a product.
\[ f(x) = \frac{x(x+2)}{x-1} \]
The outer structure decides which rule you start with. This is a fraction, so the quotient rule runs the outside.
Then, when the quotient rule asks for the derivative of the top, you answer that smaller question with the product rule. Outside first, inside second.
Pattern
Predict first
The table runs: 2.9 | about 7.4789 | just left · 3 | 7.5 | the point · 3.1 | about 7.5286 | just right
In Product rule inside the quotient rule, given the rows so far: what is the next one — the row where x is slope across?
Correct: slope across | about 0.248 | formula says 0.25
| x | value of the function | note |
|---|---|---|
| 2.9 | about 7.4789 | just left |
| 3 | 7.5 | the point |
| 3.1 | about 7.5286 | just right |
| slope across | about 0.248 | formula says 0.25 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. One times the quantity x plus two, plus x times one.
Worked example
\[ f(x) = \frac{x(x+2)}{x-1} \]
Answer the inner question before assembling the outer one.
Differentiate the top with the product rule
Why: One times the quantity x plus two, plus x times one. Two terms because there are two factors.
\[ \text{hi} = x(x+2),\qquad d(\text{hi}) = (1)(x+2) + x(1) = 2x+2 \]
Assemble the quotient rule, denominator squared first
Why: Bottom times derivative of top, minus top times derivative of bottom, over the bottom squared.
\[ f'(x) = \frac{(2x+2)(x-1) - x(x+2)(1)}{(x-1)^{2}} \]
Expand the numerator carefully
Why: The first product is two x squared minus two; the second is x squared plus two x. Subtract the whole second product, not just its first term.
\[ = \frac{\left(2x^{2}-2\right) - \left(x^{2}+2x\right)}{(x-1)^{2}} \]
Combine like terms
Why: Two x squared minus x squared leaves one square term; there is no x term to cancel the minus two x.
\[ f'(x) = \frac{x^{2}-2x-2}{(x-1)^{2}} \]
Verify the value at x = 3 against real nearby slopes
Why: The formula gives nine minus six minus two, over four, which is one quarter. Real values on either side of x = 3 give an average slope of about 0.248, so the answer holds.
| x | value of the function | note |
|---|---|---|
| 2.9 | about 7.4789 | just left |
| 3 | 7.5 | the point |
| 3.1 | about 7.5286 | just right |
| slope across | about 0.248 | formula says 0.25 |
Picture it
Animation
Shows: Each line of the worked example "Product rule inside the quotient rule", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives nine minus six minus two, over four, which is one quarter. Real values on either side of x = 3 give an average slope of about 0.248, so the answer holds.
Concept
When rules nest, the failure mode is never the calculus. It is losing track of which piece you were computing.
Answer the inner derivative on its own line, box it, and only then carry it into the outer rule.
| question | answer it separately |
|---|---|
| what is the top | the product x times the quantity x plus two |
| what is the derivative of the top | two x plus two, by the product rule |
| what is the bottom | x minus one |
| what is the derivative of the bottom | one |
Four boxes, then one assembly. It is slower for the first three problems and faster forever after that.
Trade off
Comparison matrix
From Bookkeeping keeps combined problems honest: every row here is a choice with a cost. Fill the answer it separately column, then say which row you would actually pick and what you give up for it.
| question | answer it separately |
|---|---|
| what is the top | the product x times the quantity x plus two |
| what is the derivative of the top | two x plus two, by the product rule |
| what is the bottom | x minus one |
| what is the derivative of the bottom | one |
Intuition
A store sells lemonade. Money comes in as price times cups sold.
Drop the price and you sell more cups, but each cup brings in less. Two quantities are moving, and they move in opposite directions.
The product rule is the accounting for exactly that situation: one term for the effect of selling more, one term for the effect of charging less.
Anyone who says revenue rises just because sales rose has kept only one of the two terms. That is the product-of-derivatives error in business clothes.
Concept
In a demand model the price you can charge depends on how many units you want to move.
\[ R(x) = x \cdot p(x) \]
marginal revenue — The derivative of revenue with respect to quantity. It estimates the extra money brought in by selling one more unit.
Because revenue is literally a product, its derivative is literally a product rule. The two terms have names: more units sold, and a lower price on every unit.
Explain it
Discussion prompt
Explain Revenue is price times quantity 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:
In a demand model the price you can charge depends on how many units you want to move.
Picture it
Animation
Shows: Revenue is price times quantity — a rendered Manim animation.
Rendered with Manim.
Takeaway: Marginal revenue is a product rule wearing an economics hat.
Estimation
Predict first
A shop can sell x units at the price given below, in dollars per unit. Find the marginal revenue at 20 units.
Commit before you compute: what does Marginal revenue from a demand equation 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 by expanding the revenue and by an actual one-unit step
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. Expanding gives sixty x minus one half x squared, whose derivative is sixty minus x, matching.
Worked example
A shop can sell x units at the price given below, in dollars per unit. Find the marginal revenue at 20 units.
\[ p(x) = 60 - 0.5x, \qquad R(x) = x\,p(x) \]
Label the two factors and differentiate each
Why: Quantity is the first factor and price is the second. The price is falling, so its derivative is negative.
\[ u = x,\ u' = 1,\qquad v = 60-0.5x,\ v' = -0.5 \]
Apply the product rule
Why: First term is the price itself, the money from the extra unit. Second term is the loss from charging every existing unit a little less.
\[ R'(x) = (1)(60-0.5x) + x(-0.5) = 60 - x \]
Evaluate at the quantity asked for
Why: Marginal revenue at 20 units is 60 minus 20, in dollars per unit.
\[ R'(20) = 40 \ \text{dollars per unit} \]
Verify by expanding the revenue and by an actual one-unit step
Why: Expanding gives sixty x minus one half x squared, whose derivative is sixty minus x, matching. And real revenue rises from 1000 dollars at 20 units to 1039.50 at 21 units, an increase of 39.50, which is what a marginal revenue of 40 dollars per unit predicts.
| units | price per unit | revenue |
|---|---|---|
| 20 | 50 | 1000 |
| 21 | 49.50 | 1039.50 |
| change | down 0.50 | up 39.50 |
Picture it
Animation
Shows: Each line of the worked example "Marginal revenue from a demand equation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Expanding gives sixty x minus one half x squared, whose derivative is sixty minus x, matching. And real revenue rises from 1000 dollars at 20 units to 1039.50 at 21 units, an increase of 39.50, which is what a marginal revenue of 40 dollars per unit predicts.
Concept
The units come straight from the ratio the derivative is built out of: dollars on top, units on the bottom.
\[ R'(x) = 60 - x \quad \left[\text{dollars per unit}\right] \]
So at 20 units, the twenty-first unit is worth about 40 dollars. That is a decision number, not just a symbol.
Notice where the marginal revenue hits zero: at 60 units. Past that, selling more actually brings in less money, because the price cut outweighs the extra volume. The second product-rule term has taken over.
Analogy
Discussion prompt
Explain Reading marginal revenue, with units 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:
The units come straight from the ratio the derivative is built out of: dollars on top, units on the bottom.
Concept
Total cost divided by how many you made is the cost per item.
\[ A(x) = \frac{C(x)}{x} \]
average cost — Total cost divided by quantity. Its derivative tells you whether making one more unit pulls the per-item cost up or down.
This is a genuine quotient with a fixed cost buried in the numerator, so it does not simplify away. The quotient rule earns its keep here.
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 product rule, quotient rule, marginal revenue, average cost as The Product and Quotient Rules uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: Average cost is a quotient — a rendered Manim animation.
Rendered with Manim.
Takeaway: Average cost bottoms out exactly where marginal cost meets it.
Step zero
Discussion prompt
Minimizing average cost — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Label the pieces of the quotient
Answer:
Worked example
Total cost in dollars for making x units, including a fixed cost of 200 dollars. Find the production level with the lowest cost per unit.
\[ C(x) = 0.5x^{2}+2x+200, \qquad A(x) = \frac{C(x)}{x} \]
Label the pieces of the quotient
Why: The top is the total cost and the bottom is just x, whose derivative is one.
\[ \text{hi} = 0.5x^{2}+2x+200,\ d(\text{hi}) = x+2,\qquad \text{lo} = x,\ d(\text{lo}) = 1 \]
Apply the quotient rule
Why: Bottom times derivative of top, minus top times derivative of bottom, over x squared.
\[ A'(x) = \frac{x(x+2) - \left(0.5x^{2}+2x+200\right)}{x^{2}} \]
Simplify the numerator
Why: The two x terms cancel and half of the square survives, leaving the fixed cost as the negative part.
\[ A'(x) = \frac{0.5x^{2}-200}{x^{2}} \]
Set the numerator to zero and solve
Why: The denominator is a positive square, so the derivative is zero only where the numerator is. Negative quantities are not real production levels.
\[ 0.5x^{2} = 200 \;\Longrightarrow\; x^{2} = 400 \;\Longrightarrow\; x = 20 \ \text{units} \]
Verify it is a minimum by comparing neighbours
Why: Average cost is 22 dollars at 20 units and slightly higher on both sides, so 20 units really is the cheapest per unit. The numerator also changes from negative to positive there, confirming the turn.
| units | total cost | average cost per unit |
|---|---|---|
| 19 | 418.50 | about 22.026 |
| 20 | 440 | 22 |
| 21 | 462.50 | about 22.024 |
Picture it
Animation
Shows: Each line of the worked example "Minimizing average cost", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Average cost is 22 dollars at 20 units and slightly higher on both sides, so 20 units really is the cheapest per unit. The numerator also changes from negative to positive there, confirming the turn.
Concept
Look at what the quotient rule produced. Setting it to zero says the bottom times the derivative of the top equals the top.
\[ A'(x) = 0 \iff x\,C'(x) - C(x) = 0 \iff C'(x) = \frac{C(x)}{x} \]
In our numbers, the marginal cost at 20 units is 20 plus 2, which is 22, and the average cost there was also 22. They meet exactly at the minimum.
The idea is the same as a grade average: a new score above your average pulls it up, below it pulls it down, and it stops moving exactly when the new score equals the average. The quotient rule proved that in one line.
Counterexample
Discussion prompt
Look at what the quotient rule produced. Setting it to zero says the bottom times the derivative of the top equals the top.
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:
In our numbers, the marginal cost at 20 units is 20 plus 2, which is 22, and the average cost there was also 22. They meet exactly at the minimum.
Prediction
Predict first
What is the marginal revenue when 20 units are sold?
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: 40 dollars per unit
Why: The product rule gives R prime of x equal to (1)(60 - 0.5x) plus x times (-0.5), which simplifies to 60 - x, so at 20 units the marginal revenue is 40 dollars per unit. Real revenue rises by 39.50 dollars going from 20 to 21 units, confirming it.
Check
Same shop, same demand equation. Do the product rule before you look at the choices.
\[ p(x) = 60-0.5x, \qquad R(x) = x\,p(x) \]
Check your understanding
What is the marginal revenue when 20 units are sold?
Answer: A
Why: The product rule gives R prime of x equal to (1)(60 - 0.5x) plus x times (-0.5), which simplifies to 60 - x, so at 20 units the marginal revenue is 40 dollars per unit. Real revenue rises by 39.50 dollars going from 20 to 21 units, confirming it.
Section
Section 6
Ranking
Put in order
Put the moves of Where is the tangent horizontal? 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. Bottom times derivative of top is two x times the quantity x minus two; top times derivative of bottom is the numerator itself.
Worked example
A classic follow-up question. Find every x where this curve has a horizontal tangent.
\[ f(x) = \frac{x^{2}+1}{x-2} \]
Differentiate with the quotient rule
Why: Bottom times derivative of top is two x times the quantity x minus two; top times derivative of bottom is the numerator itself.
\[ f'(x) = \frac{2x(x-2) - \left(x^{2}+1\right)(1)}{(x-2)^{2}} \]
Simplify the numerator
Why: Two x squared minus four x, minus x squared minus one, leaves a quadratic on top.
\[ f'(x) = \frac{x^{2}-4x-1}{(x-2)^{2}} \]
A fraction is zero only where its numerator is zero
Why: The denominator being a square means it can never make the fraction zero. It only tells us where the derivative fails to exist, at the asymptote x equal to 2.
\[ x^{2}-4x-1 = 0 \]
Solve with the quadratic formula
Why: The discriminant is sixteen plus four, which is twenty, and the root of twenty is two roots of five, so the twos divide out.
\[ x = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5} \]
Verify both roots satisfy the numerator exactly
Why: Squaring two plus root five gives nine plus four root five; subtracting four times two plus root five gives eight plus four root five; the root-five parts cancel and nine minus eight minus one is zero. The same cancellation works for the other sign, so both are genuine flat spots, at about 4.236 and about -0.236.
\[ \left(2+\sqrt5\right)^{2} - 4\left(2+\sqrt5\right) - 1 = \left(9+4\sqrt5\right) - \left(8+4\sqrt5\right) - 1 = 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Where is the tangent horizontal?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Squaring two plus root five gives nine plus four root five; subtracting four times two plus root five gives eight plus four root five; the root-five parts cancel and nine minus eight minus one is zero. The same cancellation works for the other sign, so both are genuine flat spots, at about 4.236 and about -0.236.
Ranking
Put in order
These are the steps of Self-audit before you hand it in, 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
Four questions, ten seconds, most of the lost points recovered.
Then one sign check: pick a convenient value, decide whether the function should be rising or falling there, and see whether your derivative agrees.
Real world
Discussion prompt
Outside this lesson: where does The Product and Quotient Rules actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Self-audit before you hand it in is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck explains why the derivative of a product is not the product of the derivatives, then gives the product rule with its expanding-rectangle picture and the quotient rule, paying close attention to the order of the terms in its numerator. It covers when to simplify instead of grinding, products of three factors, and combining the rules, and finishes with rate-of-change applications such as marginal revenue and average cost. It targets the product-of-derivatives error, the reversed numerator, the unsquared denominator, and reaching for the quotient rule when plain division is easier.
Concept
| product rule | quotient rule | |
|---|---|---|
| shape | f prime g plus f g prime | (f prime g minus f g prime) over g squared |
| how many terms | one per factor | always two, in the numerator |
| order matters | no, addition commutes | yes, subtraction does not |
| denominator | none | squared, always |
One rule forgives sloppiness and the other does not. That asymmetry is worth memorizing on its own.
Both rules will be inside almost everything that follows: the chain rule, implicit differentiation, related rates, optimization. Getting them automatic now pays every week for the rest of the course.
Comparison
Comparison matrix
From The two rules side by side: refill the quotient rule column from what you know. The rest of the table is as it appeared.
| product rule | quotient rule | |
|---|---|---|
| shape | f prime g plus f g prime | (f prime g minus f g prime) over g squared |
| how many terms | one per factor | always two, in the numerator |
| order matters | no, addition commutes | yes, subtraction does not |
| denominator | none | squared, always |
Picture it
Animation
Shows: The two rules side by side — a rendered Manim animation.
Rendered with Manim.
Takeaway: Plus for products, minus for quotients — and order only matters below.
Elimination
Eliminate the wrong options
At which x values does this curve have a horizontal tangent?
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: The product rule gives 2x times e to the x plus x squared times e to the x, which factors as x times e to the x times the quantity x plus 2. The exponential is never zero, so the derivative vanishes exactly when x is 0 or x is -2.
Check
Differentiate, factor, and then think about which factors can actually be zero.
\[ f(x) = x^{2}e^{x} \]
Check your understanding
At which x values does this curve have a horizontal tangent?
Answer: A
Why: The product rule gives 2x times e to the x plus x squared times e to the x, which factors as x times e to the x times the quantity x plus 2. The exponential is never zero, so the derivative vanishes exactly when x is 0 or x is -2.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Why a Product Needs Its Own Rule · Products You Cannot Multiply Out · The Quotient Rule · Choosing the Efficient Path · Combining the Rules · Locking It In. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Two rules, four errors, one habit of checking. That is the whole deck.
| you see | you do |
|---|---|
| two short polynomials multiplied | expand, or product rule |
| a polynomial times an exponential or a root | product rule |
| a fraction with one term underneath | divide through, or negative exponents |
| a fraction with a sum underneath | quotient rule, denominator squared first |
| a product on top of a fraction | quotient rule outside, product rule inside |
Next up is the chain rule, which handles a function tucked inside another function. It is the last of the three big rules, and it shows up inside product and quotient problems constantly.
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