Product Rule and Quotient Rule: When to Use Each, with Revenue and Average-Cost Examples

In 30 seconds

Use the product rule when a function is one expression multiplied by another: $ (uv)’ = u’v + uv’ $.
Use the quotient rule when one expression is divided by another: $ \left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2} $.
Revenue is price times quantity, so when both change over time, the rate of change of revenue needs the product rule.
In our case, revenue is still rising by $80 a month at month 3, and average cost is lowest at 40 units, where it equals marginal cost at $60.
Exam point: in the quotient rule the order on the top matters. Write $ u’v – uv’ $, never the other way round, and keep the square on the bottom.

The case: a course shop that cuts its price while sales grow

A small online shop sells a revision course. Each month it trims the price to attract more buyers. The owner tracks two numbers: the price and the number of courses sold. Both change over time, and she wants to know one thing. Is revenue still going up, and how fast?

Let $ t $ be the number of months since launch. The price, in dollars, is $ P(t) = 40 – 2t $. Monthly sales, in courses, are $ Q(t) = 100 + 10t $. Monthly revenue is price times quantity:

\[ R(t) = P(t)\,Q(t) = (40 – 2t)(100 + 10t) \]

At launch the price is $40 and she sells 100 courses, so revenue is $4,000. By month 3 the price is $34 and she sells 130, so revenue is $4,420. Revenue went up. But the price is falling every month, so at some point the cuts will cost more than the extra sales bring in. She needs the rate of change of revenue, $ R'(t) $, at month 3, and she needs to know when that rate hits zero.

The shop also has a cost question. Making and supporting $ q $ courses costs $ C(q) = 0.5q^2 + 20q + 800 $ dollars a month. The $800 is fixed (hosting, software, the website). The average cost per course is total cost divided by quantity:

\[ AC(q) = \frac{C(q)}{q} = \frac{0.5q^2 + 20q + 800}{q} \]

At what output is the average cost per course lowest? That needs the derivative of a fraction. (The figures are illustrative.)

So we have one product and one quotient. These two shapes are exactly what the product rule and the quotient rule are for.

The concept: differentiating products and quotients

You already know how to differentiate a sum: differentiate each part and add. It is tempting to think a product works the same way, so that the derivative of $ uv $ is $ u’v’ $. It does not. Try it with $ x \cdot x = x^2 $. The true derivative is $ 2x $. The “multiply the derivatives” idea gives $ 1 \times 1 = 1 $. That is wrong for every $ x $ except $ x = \tfrac12 $.

The reason is easy to see with a rectangle. Let one side be $ u $ and the other side be $ v $, so the area is $ uv $. Now let both sides grow a little. The area grows in two strips: one strip of size $ (\text{change in } u) \times v $, and another of size $ u \times (\text{change in } v) $. There is also a tiny corner piece, the product of the two small changes. When the changes shrink towards zero, that corner piece becomes negligible. So the rate of change of the area is “rate of change of $ u $ times $ v $” plus “$ u $ times rate of change of $ v $”. That is the product rule.

Revenue is exactly such a rectangle. Price is one side and quantity is the other. When the price falls, you lose a strip of revenue on all the units you were already selling. When quantity rises, you gain a strip of revenue at the current price. The product rule adds the two effects together, with their signs.

Key termProduct rule: if $ y = u(x)\,v(x) $, where $ u $ and $ v $ are both differentiable, then $ \frac{dy}{dx} = u’v + uv’ $. In words: derivative of the first times the second, plus the first times the derivative of the second. Quotient rule: if $ y = \frac{u(x)}{v(x)} $ with $ v(x) \neq 0 $, then $ \frac{dy}{dx} = \frac{u’v – uv’}{v^2} $.

\[ \frac{d}{dx}\big(uv\big) = \frac{du}{dx}\,v + u\,\frac{dv}{dx} \qquad\qquad \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{\dfrac{du}{dx}\,v – u\,\dfrac{dv}{dx}}{v^2} \]

Here $ u $ and $ v $ are the two functions of $ x $ that are multiplied or divided. In the quotient, $ u $ is the numerator (the top) and $ v $ is the denominator (the bottom). $ \frac{du}{dx} $ and $ \frac{dv}{dx} $ are their ordinary derivatives, often written $ u’ $ and $ v’ $. The minus sign in the quotient rule is what makes order matter: swapping the two terms changes the sign of the whole answer. The $ v^2 $ on the bottom is the denominator squared, not differentiated. A useful memory line is “low d-high minus high d-low, over the square of what’s below” ^[1].

Mathematics Made Simple explains the product and quotient rules with worked examples and graphs.Get the textbook · $12

The quotient rule is not really a new idea. You can write $ \frac{u}{v} = u \cdot v^{-1} $ and use the product rule together with the chain rule. The derivative of $ v^{-1} $ is $ -v^{-2}v’ $, so you get $ u’v^{-1} – uv’v^{-2} $. Put everything over $ v^2 $ and you have $ \frac{u’v – uv’}{v^2} $. Knowing this is useful in an exam if you forget the order of the top line: you can rebuild it in three lines.

Why this method fits the case

Revenue is a product of two things that both depend on time. Neither factor is a constant. If the price were fixed at $40, revenue would be $ 40Q(t) $ and you would just multiply the derivative of $ Q $ by 40. Because the price moves too, you need both strips of the rectangle. The product rule also tells a story that a single number cannot. It splits the change in revenue into a “price effect” ($ P’Q $) and a “quantity effect” ($ PQ’ $). Managers care about that split.

Average cost is a ratio, total cost over quantity. Both the top and the bottom change with $ q $. That is the quotient rule’s job. In economics, many useful measures are ratios: average cost, average product, revenue per customer, profit margin. Whenever you want the rate of change of a ratio, the quotient rule is the natural tool ^[4].

When is it not the right tool? There are three common cases.

When you can simplify first. $ (40 – 2t)(100 + 10t) $ expands to $ 4000 + 200t – 20t^2 $. Differentiating that is quicker than the product rule. In the same way, $ \frac{0.5q^2 + 20q + 800}{q} $ splits into $ 0.5q + 20 + 800q^{-1} $, and the power rule finishes the job. If the algebra is easy, do the algebra.
When one factor is a constant. $ 5x^3 $ does not need the product rule. A constant multiple just comes along for the ride.
When the function is a composition, not a product. $ (3x + 1)^5 $ is one function inside another. That is a chain rule job. Many students reach for the product rule here and get lost.

The product and quotient rules really earn their keep when expanding is messy or impossible, for example $ x^2 e^x $, $ x \sin x $, $ \frac{\ln x}{x} $ or $ \frac{e^x}{1 + e^x} $. We kept the case simple so you can check the rule against an expansion and see that both give the same answer.

Setting it up

Before differentiating, name every piece. This one habit removes most product and quotient rule errors.

Quantity	Symbol	Expression	Derivative
Price (dollars)	$ u = P(t) $	$ 40 – 2t $	$ u’ = -2 $
Courses sold per month	$ v = Q(t) $	$ 100 + 10t $	$ v’ = 10 $
Revenue (dollars per month)	$ R = uv $	$ (40 – 2t)(100 + 10t) $	$ u’v + uv’ $
Total cost (numerator)	$ u = C(q) $	$ 0.5q^2 + 20q + 800 $	$ u’ = q + 20 $
Quantity (denominator)	$ v = q $	$ q $	$ v’ = 1 $
Average cost (dollars per course)	$ AC = u/v $	$ \frac{0.5q^2 + 20q + 800}{q} $	$ \frac{u’v – uv’}{v^2} $

Notice that $ u’ = q + 20 $ is also the marginal cost, the extra cost of one more course. That will matter when we interpret the answer.

Solving it

Worked example

Find the rate of change of revenue at month 3, and the month when revenue stops rising.

Step 1Write $ u = 40 – 2t $, $ u’ = -2 $, $ v = 100 + 10t $, $ v’ = 10 $.

Step 2Apply the product rule: $ R'(t) = u’v + uv’ = (-2)(100 + 10t) + (40 – 2t)(10) $.

Step 3Simplify: $ -200 – 20t + 400 – 20t = 200 – 40t $.

Step 4At $ t = 3 $: price effect $ = (-2)(130) = -260 $; quantity effect $ = (34)(10) = 340 $. So $ R'(3) = -260 + 340 = 80 $. Check: $ 200 – 40(3) = 80 $. ✓

Step 5Set $ R'(t) = 0 $: $ 200 – 40t = 0 $, so $ t = 5 $. Then $ P(5) = 30 $, $ Q(5) = 150 $, $ R(5) = 4500 $.

Step 6Check by expanding: $ R = 4000 + 200t – 20t^2 $, so $ R’ = 200 – 40t $. Same answer. ✓

At month 3, revenue is rising at $80 per month; it peaks at $4,500 in month 5, after which further price cuts reduce revenue.

Worked example

Find $ \frac{d(AC)}{dq} $ using the quotient rule. Evaluate it at $ q = 20 $, then find the output that minimises average cost.

Step 1Top: $ u = 0.5q^2 + 20q + 800 $, $ u’ = q + 20 $. Bottom: $ v = q $, $ v’ = 1 $.

Step 2Quotient rule: $ AC'(q) = \frac{(q + 20)q – (0.5q^2 + 20q + 800)(1)}{q^2} $.

Step 3Simplify the top: $ q^2 + 20q – 0.5q^2 – 20q – 800 = 0.5q^2 – 800 $. So $ AC'(q) = \frac{0.5q^2 – 800}{q^2} $.

Step 4At $ q = 20 $: $ \frac{0.5(400) – 800}{400} = \frac{-600}{400} = -1.5 $. Average cost is falling by $1.50 per extra course. $ AC(20) = 10 + 20 + 40 = 70 $.

Step 5Set the top to zero: $ 0.5q^2 = 800 $, so $ q^2 = 1600 $ and $ q = 40 $ (quantity must be positive).

Step 6$ AC(40) = \frac{800 + 800 + 800}{40} = 60 $. Marginal cost at 40 is $ 40 + 20 = 60 $. They are equal.

Average cost is lowest at 40 courses a month, where it is $60 per course and exactly equals marginal cost.

Worked example

A pure-maths check of the quotient rule. Find the gradient of $ y = \frac{2x + 1}{x^2 + 3} $ at $ x = 1 $.

Step 1$ u = 2x + 1 $, $ u’ = 2 $; $ v = x^2 + 3 $, $ v’ = 2x $.

Step 2$ y’ = \frac{2(x^2 + 3) – (2x + 1)(2x)}{(x^2 + 3)^2} $.

Step 3Top: $ 2x^2 + 6 – 4x^2 – 2x = -2x^2 – 2x + 6 $.

Step 4At $ x = 1 $: top $ = -2 – 2 + 6 = 2 $; bottom $ = (1 + 3)^2 = 16 $. So $ y’ = \frac{2}{16} = \frac18 $.

The gradient at $ x = 1 $ is $ \frac18 = 0.125 $; here you could not simply expand, so the quotient rule is the efficient route.

Make the product and quotient rules automatic with the Practice Questions book: 1,565 questions, all fully worked.Get 1,565 questions · $9

What the answer tells us

The revenue result splits neatly. At month 3, the $2 price cut loses $260 a month on the 130 courses already selling. The extra 10 courses a month bring in $ 10 \times 34 = 340 $ dollars. The gain beats the loss by $80, so revenue still rises. But look at how the two effects move. The price effect, $ -2Q(t) $, gets bigger as sales grow, because each cut hits more buyers. The quantity effect, $ 10P(t) $, gets smaller, because each new buyer pays less. By month 5 they cancel exactly. After that, the owner should stop cutting the price, or find another way to grow sales.

Is $80 reasonable? Revenue at month 3 is $4,420 and at month 4 it is $ 32 \times 140 = 4480 $. That is a rise of $60, a little less than $80. That makes sense: the derivative is the rate at the instant $ t = 3 $, and the rate is falling, so the actual change over the next whole month is a bit smaller. The average of $ R'(3) = 80 $ and $ R'(4) = 40 $ is 60, which matches exactly, as it should for a quadratic.

The cost result is a classic piece of economics. Figure 1 plots average cost and marginal cost. Where marginal cost is below average cost, each extra course is cheaper than the current average, so it pulls the average down. Where marginal cost is above average cost, it pulls the average up. So average cost is lowest exactly where the two curves cross. The quotient rule proves this in general. The top of $ AC’ $ is $ qC’ – C $. Dividing by $ q $, it is zero exactly when $ C’ = C/q $, that is, when marginal cost equals average cost ^[4].

Average cost is lowest where marginal cost crosses it $ per course

Average cost AC = C/qMarginal cost MC = q + 20

Figure 1. Average cost $ AC(q) = 0.5q + 20 + 800/q $ and marginal cost $ MC = q + 20 $. The curves cross at $ q = 40 $, where average cost is at its minimum of $60. (Illustrative figures.)

How sensitive is the answer? The minimum is at $ q = \sqrt{1600} = 40 $, and 1600 is $ 800 / 0.5 $, the fixed cost divided by the coefficient of $ q^2 $. If fixed costs rose to $1,250, the best output would rise to $ \sqrt{2500} = 50 $. Higher fixed costs mean you need more units to spread them over. Also notice how flat the curve is near the bottom: average cost is $61 at 32 or 50 courses, only $1 above the minimum. In practice, any output from about 32 to 50 is close to the lowest average cost.

Common mistakeWriting the derivative of a product as the product of the derivatives. For $ R = (40 – 2t)(100 + 10t) $, the wrong method gives $ (-2)(10) = -20 $ for every month, suggesting revenue always falls. The true value at month 3 is $ +80 $. Always use two terms: $ u’v + uv’ $.

Common mistakeSwapping the order on top of the quotient rule, writing $ uv’ – u’v $. This gives the right size but the wrong sign, so a falling average cost looks like a rising one. Also watch the bottom: it is $ v^2 $, not $ v’ $ and not $ v $.

Exam tipWrite the four pieces $ u, u’, v, v’ $ in a small list before you apply either rule, and show the unsimplified line. Examiners in AP Calculus, Cambridge 9709 and IB give method marks for a correct application of the rule even if later algebra slips ^[2]^[3]. In AP free-response questions you may leave a derivative unsimplified unless the question asks you to simplify.

To build the full set of differentiation rules step by step, the Mathematics Made Simple textbook covers the power, product, quotient and chain rules with graphs and economic examples. For drilling, the Mathematics Made Simple Practice Questions book has hundreds of differentiation questions with full worked answers.

Master differentiation rules with both Mathematics Made Simple books for $16.Both books · $16

Practice questions

Try each question before opening the worked answer.

EasyQ1. Use the product rule to differentiate $ y = x^3(2x + 5) $. Check by expanding. Show worked answer

$ u = x^3 $, $ u’ = 3x^2 $; $ v = 2x + 5 $, $ v’ = 2 $.
$ y’ = 3x^2(2x + 5) + x^3(2) = 6x^3 + 15x^2 + 2x^3 = 8x^3 + 15x^2 $.
Check: $ y = 2x^4 + 5x^3 $, so $ y’ = 8x^3 + 15x^2 $. ✓

EasyQ2. Use the quotient rule to differentiate $ y = \frac{x}{x + 1} $. Show worked answer

$ u = x $, $ u’ = 1 $; $ v = x + 1 $, $ v’ = 1 $.
$ y’ = \frac{1(x + 1) – x(1)}{(x + 1)^2} = \frac{1}{(x + 1)^2} $.

MediumQ3. Find the gradient of $ y = x^2 \sin x $ at $ x = \frac{\pi}{2} $. Show worked answer

$ u = x^2 $, $ u’ = 2x $; $ v = \sin x $, $ v’ = \cos x $.
$ y’ = 2x\sin x + x^2\cos x $.
At $ x = \frac{\pi}{2} $: $ \sin = 1 $, $ \cos = 0 $, so $ y’ = 2 \cdot \frac{\pi}{2} \cdot 1 + 0 = \pi \approx 3.14 $.

MediumQ4. A firm’s revenue is $ R(q) = \frac{120q}{q + 2} $ (in thousands of dollars). Find marginal revenue $ R'(q) $ and evaluate it at $ q = 8 $. Show worked answer

$ u = 120q $, $ u’ = 120 $; $ v = q + 2 $, $ v’ = 1 $.
$ R'(q) = \frac{120(q + 2) – 120q}{(q + 2)^2} = \frac{240}{(q + 2)^2} $.
At $ q = 8 $: $ \frac{240}{100} = 2.4 $, i.e. about $2,400 of extra revenue per extra unit.

ExamQ5. A firm’s total cost is $ C(q) = 2q^2 + 50q + 1800 $. (a) Use the quotient rule to show that $ \frac{d}{dq}\left(\frac{C}{q}\right) = \frac{2q^2 – 1800}{q^2} $. (b) Find the output that minimises average cost and the minimum average cost. (c) Show that marginal cost equals average cost at this output. [6] Show worked answer

(a) $ u = 2q^2 + 50q + 1800 $, $ u’ = 4q + 50 $; $ v = q $, $ v’ = 1 $. [1]
$ \frac{(4q + 50)q – (2q^2 + 50q + 1800)}{q^2} $ [1] $ = \frac{2q^2 – 1800}{q^2} $. [1]
(b) $ 2q^2 = 1800 \Rightarrow q = 30 $. [1] $ AC(30) = \frac{1800 + 1500 + 1800}{30} = 170 $. [1]
(c) $ MC = 4q + 50 = 170 $ at $ q = 30 $, equal to $ AC $. [1]

Frequently asked questions

How do I know whether to use the product rule or the chain rule?

Ask how the pieces are joined. If two functions of $ x $ are multiplied side by side, like $ x^2 \cdot e^x $, use the product rule. If one function sits inside another, like $ e^{x^2} $ or $ (3x + 1)^5 $, use the chain rule. Some functions need both, such as $ x^2(3x + 1)^5 $.

Can I always avoid the quotient rule?

Often, yes. You can rewrite $ \frac{u}{v} $ as $ u \cdot v^{-1} $ and use the product and chain rules. If the denominator is a single power of $ x $, just divide each term first. But when the denominator is a sum, like $ x^2 + 3 $, the quotient rule is usually the cleanest route.

Does the order matter in the product rule?

No. $ u’v + uv’ $ and $ uv’ + u’v $ are the same because addition does not care about order. The order only matters in the quotient rule, because of the minus sign.

Is there a product rule for three functions?

Yes. For $ uvw $, the derivative is $ u’vw + uv’w + uvw’ $: differentiate one factor at a time and add the three results. You can get it by applying the two-factor rule twice.

References

Strang, G. and Herman, E. (2016) Calculus Volume 1. Houston, TX: OpenStax. Available at: https://openstax.org/details/books/calculus-volume-1.
College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description. New York: College Board.
Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment. Available at: cambridgeinternational.org.
Sydsæter, K., Hammond, P., Strøm, A. and Carvajal, A. (2016) Essential Mathematics for Economic Analysis. 5th edn. Harlow: Pearson.

Quantity	Symbol	Expression	Derivative
Price (dollars)	\( u = P(t) \)	\( 40 – 2t \)	\( u’ = -2 \)
Courses sold per month	\( v = Q(t) \)	\( 100 + 10t \)	\( v’ = 10 \)
Revenue (dollars per month)	\( R = uv \)	\( (40 – 2t)(100 + 10t) \)	\( u’v + uv’ \)
Total cost (numerator)	\( u = C(q) \)	\( 0.5q^2 + 20q + 800 \)	\( u’ = q + 20 \)
Quantity (denominator)	\( v = q \)	\( q \)	\( v’ = 1 \)
Average cost (dollars per course)	\( AC = u/v \)	\( \frac{0.5q^2 + 20q + 800}{q} \)	\( \frac{u’v – uv’}{v^2} \)

Product Rule and Quotient Rule: When to Use Each, with Revenue and Average-Cost Examples

The case: a course shop that cuts its price while sales grow

The concept: differentiating products and quotients

Why this method fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

How do I know whether to use the product rule or the chain rule?

Can I always avoid the quotient rule?

Does the order matter in the product rule?

Is there a product rule for three functions?

References

Keep reading

Contact Info

Quick Links

Success!