Mathematical reference
- Product rule
- (fg)′ = f′g+fg′ for differentiable f and g
- At x=2
- For x·x²: 1×4+2×4 = 12, whereas multiplying side derivatives gives only 4
- Two squared factors
- For x²·x² at x=2: 4×4+4×4 = 32
Lesson transcript
This is the text of the narrated lesson. References to turning, highlighting or changing an object describe the linked interactive lesson; its demonstrations are not embedded in this reading page.
Two lengths change together
Watch the added pieces grow around this rectangle. The first side is x and the second is x squared. At x equal to two, their lengths are two and four, so the original area is eight.
Increasing the input changes both lengths. The new area therefore includes a strip along each original side and a corner where the two changes meet. Predict whether multiplying just the two length changes could account for the whole added area.
That product accounts only for the corner. Most of the finite change can be in the two strips, especially when the input change is small. We need to distinguish three quantities: the original lengths, their changes, and the area change.
The symbol delta x describes the input change. Delta f and delta g describe the changes in the two functions; they are not generally equal to delta x. Keep the original rectangle visible as a reference while identifying all three added regions.
Account for every added region
The same finite increase develops again. Call the old lengths f and g, and their changes delta f and delta g. One strip has area g times delta f. The other has area f times delta g.
The corner has area delta f times delta g. The new rectangle is f plus delta f multiplied by g plus delta g. Subtracting the old product f g leaves precisely those three contributions.
Emphasize the first strip and follow its unchanged side. Its height is the old g, not the new g plus delta g. Using the new height there would already include the corner. If we then added the corner separately, we would count it twice.
This is why a careful partition matters. The finite-change identity is exact. No limit or approximation has been taken yet. Explain which dimensions belong to each region, and check that the union covers the changed area without double counting.
A worked change at x equals two
At x equal to two and delta x equal to three tenths, the first length rises from two to two point three. Its change is three tenths. The squared length rises from four to five point two nine, a change of one point two nine.
Predict the three added areas. The first strip is four times three tenths, or one point two. The second is two times one point two nine, or two point five eight.
The corner is three tenths times one point two nine, or zero point three eight seven. Adding all three gives four point one six seven. Check using the complete products: two point three cubed is twelve point one six seven; subtract the original eight and obtain the same four point one six seven.
The corner is not an error in the picture. Omitting it would make the finite change wrong. The approximation will arise only when we compare change per unit input and take a limit.
Why the corner disappears from the derivative
Hold x fixed while delta x shrinks. The old rectangle stays the reference and the added regions become smaller. To form a rate, divide the exact area change by delta x.
The two strip terms become g times delta f over delta x and f times delta g over delta x. The corner becomes delta f times delta g divided by delta x. We must explain its limit, not merely notice that it looks small.
If both functions are differentiable at this input, delta f divided by delta x approaches f prime, while delta g approaches zero. Their product, the corner rate, therefore approaches zero. The strip rates approach g f prime and f g prime. This gives the product rule.
At a nonzero input change, the corner remains part of the exact finite result. The slider stops at a small positive value; it does not reach the mathematical limit. Differentiability and the limiting argument justify the rule beyond what a finite screen sample can establish.
Why multiplying the two derivatives gives the wrong rate
Use x times x squared at input two. The first side has local rate one, while the second has local rate four. Multiplying those rates gives four, but the product’s local area rate is twelve. A product of side rates does not count the two long strips that dominate the small area change.
Emphasize the strip produced by the first side change. Its old height is four. A first-side rate of one therefore contributes four units of area rate. The unchanged dimension must multiply the changing dimension’s rate. A side rate alone does not yet have the full area-rate meaning of this strip.
Emphasize the other strip. Its old width is two, and the second side’s local rate is four. This contributes eight units of area rate. Adding four and eight gives twelve. Each term holds one factor at its old value while taking the local rate of the other.
Show the corner and shrink the positive step. The product of the two side changes is needed for the exact finite area, but after division by the step its contribution approaches zero. Multiplying the two derivatives would preserve the wrong dimensional contribution and omit the original side lengths that determine the strip rates.
The same product can have different factor descriptions
Choose x squared for both factors, starting at input two. Their product is x to the fourth power. Each old side is four and each side’s local rate is four. The two product-rule terms are sixteen and sixteen, giving a total local rate of thirty-two.
Grow a one-tenth input step. Each side changes by zero point four one. The two strips contribute one point six four each, and the corner contributes zero point one six eight one. The finite area increase is three point four four eight one, giving an average rate of thirty-four point four eight one.
Now describe the same fourth-power product using x cubed times x. The old lengths become eight and two, but their product remains sixteen. After the same step, the lengths are nine point two six one and two point one. Their product is still nineteen point four four eight one, so the finite change agrees.
The strip contributions differ in this partition: two point five two two and eight tenths, with corner zero point one two six one. They still total three point four four eight one. At the local-rate level, two times twelve plus eight times one gives thirty-two. Different factor descriptions lead to the same derivative of the same product.
Predict and check a different local rate
Return to x times x squared, but start at input two point five. The old factors are two point five and six point two five. Their local rates are one and five. The product rule predicts six point two five plus twelve point five, or eighteen point seven five, before we shrink any displayed step.
Grow a finite step of two tenths. The new product is nineteen point six eight three and the old product fifteen point six two five. Their difference four point zero five eight, divided by two tenths, gives twenty point two nine. This finite average is greater than the predicted local rate.
Reduce the step to one tenth. The exact cubic difference quotient is three times the starting input squared, plus three times the starting input times the step, plus the step squared. Here it gives eighteen point seven five plus zero point seven five plus zero point zero one, or nineteen point five one.
Shrink to the native minimum step. The finite quotient is eighteen point seven eight seven five two five, still slightly above eighteen point seven five. The remaining terms have not become exactly zero on screen. Their explicit dependence on the step explains the limit and keeps the mathematical claim stronger and more precise than a visual impression of thinness.
Check x times x squared
The finite change shrinks again as we return to x times x squared. The product is x cubed, whose derivative should be three x squared. Apply the two-term rule: the first derivative is one and the second is two x.
We get x squared times one, plus x times two x, giving three x squared. At x equal to two, the rate is twelve.
For our earlier delta x of three tenths, the finite area change divided by delta x was thirteen point eight nine. It was not yet twelve. Expanding the cubic difference gives the exact quotient twelve plus six times delta x plus delta x squared at x equal to two. As delta x approaches zero, the extra terms approach zero.
At the displayed minimum of zero point zero zero five, the quotient is twelve point zero three zero zero two five. Try another starting input and predict its limiting rate. Explain both the finite corner and the limiting two-term rule; they answer related but different questions.