Why does a changing product need two terms?
You will be able to: Apply and explain the product rule for two differentiable factors.
BC foundation: Unit 2 shares its derivative foundations with AB. Connect each rule to rates, graphs and its domain conditions. Use the written challenges to explain your reasoning. Chain-rule compositions, implicit and inverse differentiation, and advanced applications come later.
Why does a changing product need two terms?
A rectangle changes in both length and width. Its area changes through a strip along each dimension, not just through the tiny corner where the changes overlap.
A useful starting point: How can a known derivative evaluate a limit? →
Words and symbols before equations
- Product
- Two quantities multiplied, such as u(x)v(x).
- Factor
- One quantity in a product.
- Product rule
- The derivative u′v+uv′.
- First-order change
- A contribution proportional to a small input change, governing the local rate.
What this picture assumes
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. u=x² and v=x+1 at x=2; positive h=10⁻ᵖ. Rectangle dimensions u and v are abstract lengths in model units; area changes decompose exactly. The full added area is divided by h to estimate rate.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- h=0.1; [vΔu]/h=12.3; [uΔv]/h=4; overlap/h=0.41. Total area-change rate=16.71 tends to 12+4+0=16. All rectangles share a fixed length scale.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Expanding (u+Δu)(v+Δv)−uv gives vΔu+uΔv+ΔuΔv. The first two strips produce the two terms in the product rule; after division by the input increment, the overlap tends to zero when both factors are differentiable.
For u=x² and v=x+1, u′=2x and v′=1. Thus (uv)′=2x(x+1)+x².
Expanding first gives x³+x² and derivative 3x²+2x, confirming the same result. This check works for polynomials but does not replace the general rule.
Differentiating both factors and multiplying would keep neither full strip. In general (uv)′ is not u′v′.
A worked example, step by step
Differentiate f(x)=x²(x+1) and evaluate at 2.
- Name u=x² and v=x+1.
- Find u′=2x and v′=1.
- Combine u′v+uv′=2x(x+1)+x²=3x²+2x.
- At x=2 the derivative is 12+4=16.
Keep one factor unchanged in each term, and add the two contributions.
Why does multiplying u′ and v′ fail?
Compare with an explanation
It does not measure either main strip of area change; the rate is a sum of two differently weighted contributions.
Predict. Change one thing. Explain.
At x=2, shrink a positive input increment. Compare the two area strips and the overlap; explain why only the strip contributions survive in the limiting rate.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
h=0.1; [vΔu]/h=12.3; [uΔv]/h=4; overlap/h=0.41. Total area-change rate=16.71 tends to 12+4+0=16. All rectangles share a fixed length scale.
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. u=x² and v=x+1 at x=2; positive h=10⁻ᵖ. Rectangle dimensions u and v are abstract lengths in model units; area changes decompose exactly. The full added area is divided by h to estimate rate.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using a difference quotient, tangent slope, or derivative rule with its domain conditions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionDifferentiate p(x)=(x²+1)(x²−1) using the product rule and check by expansion.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Let u=x²+1 and v=x²−1, with u′=v′=2x.
- 1 point: Product rule gives 2x(x²−1)+(x²+1)2x.
- 1 point: Simplify to 4x³.
- 1 point: Expansion gives x⁴−1, whose derivative is also 4x³.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Product rule?
u′v+uv′.
RECALL 2What stays in each term?
One original unchanged factor.
RECALL 3Why two terms?
Either factor can contribute to the product’s first-order change.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does a changing product need two terms?
- (uv)′=u′v+uv′.
- Both factors must be differentiable at the point.
Remember: Keep one factor unchanged in each term, and add the two contributions.
Conditions: Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. u=x² and v=x+1 at x=2; positive h=10⁻ᵖ. Rectangle dimensions u and v are abstract lengths in model units; area changes decompose exactly. The full added area is divided by h to estimate rate.
Refresh Kid · AP Calculus BC Unit 2 · Objectives FUN-3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.8, FUN-3.B. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 2 has 10 official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Derivative definitions require finite two-sided limits at interior inputs. Graphs and finite tables supply evidence and estimates, not universal proofs. Continuity is necessary but insufficient for differentiability. Rules retain their original domain restrictions, and trigonometric inputs use radians. Chain-rule compositions, implicit differentiation, inverse-function differentiation and L’Hôpital’s rule are deferred to later units. Piecewise joins are checked for continuity before their side slopes are compared.
The Organic Chemistry Tutor video titles, creators and descriptions were checked; full videos were not reviewed. Khan Academy’s unit destination was checked; its JavaScript lesson content was not fully readable by the research tool. OpenStax Sections 3.1, 3.3 and 3.5 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. Graph geometry is generated from the stated original equations; self-hosted Three.js retains its MIT license. The optional 3D model uses separate planes for f and f′. Depth distinguishes panels, not an additional variable; camera rotation does not change their values. Use the complete labeled 2D graphs and text to read precise coordinates and slopes.
Independent teacher review and observation of students remain pending. Checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned to this lesson.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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