How do product and chain rules work together implicitly?
You will be able to: Differentiate mixed x–y products and collect all y′ terms.
How do product and chain rules work together implicitly?
Suppose two measurements satisfy x²+xy+y²=7. Near (1,2), both measurements change together; the mixed term xy changes through both factors.
A useful starting point: Why does differentiating y² produce 2y y′? →
Words and symbols before equations
- Mixed product
- A product such as x·y(x).
- Collect terms
- Move all terms containing y′ to the same side.
- Factor
- Extract a shared multiplier to solve an equation.
- Local solvability
- The ability to express the relation as a differentiable y(x) near a point.
What this picture assumes
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. xy=6, so y=6/x. Only the positive branch is shown. Dimensionless coordinates; full relation excludes x=0.
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.
- x=2, y=3, y′=-1.5. Product-rule contributions: y=3 and xy′=-3 sum to zero.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The product rule gives d(xy)/dx=1·y+x·y′. Differentiating only y, or treating xy as y times a constant, loses one contribution.
Differentiating the relation yields 2x+y+xy′+2yy′=0. Collecting gives (x+2y)y′=−(2x+y).
Thus y′=−(2x+y)/(x+2y) where the denominator is nonzero. Substitute the point only after differentiating; early substitution turns a changing equation into constants.
A zero denominator does not by itself classify every possible singular point. Check the relation and its local geometry before claiming a vertical tangent.
A worked example, step by step
Find the slope of x²+xy+y²=7 at (1,2).
- Check 1+2+4=7.
- Differentiate to 2x+y+xy′+2yy′=0.
- Factor y′: (x+2y)y′=−2x−y.
- At (1,2), y′=−4/5.
d(xy)/dx is y+xy′, not just xy′ and not 1·y′.
Why do the contributions cancel for xy=6?
Compare with an explanation
The product is constant, so its total derivative is zero even though each factor changes.
Predict. Change one thing. Explain.
Inspect points on xy=6. Compare the two product-rule contributions y and xy′; explain why they sum to zero.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=2, y=3, y′=-1.5. Product-rule contributions: y=3 and xy′=-3 sum to zero.
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. xy=6, so y=6/x. Only the positive branch is shown. Dimensionless coordinates; full relation excludes x=0.
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 x²y+y²=12 and find the slope at (2,2).
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The point satisfies 4·2+4=12.
- 1 point: Differentiate: 2xy+x²y′+2yy′=0.
- 1 point: Collect: (x²+2y)y′=−2xy.
- 1 point: At (2,2), y′=−8/8=−1.
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 1How many terms does d(xy)/dx have?
Two: y and xy′.
RECALL 2When should you substitute the point?
After differentiating the variable relation.
RECALL 3Why factor y′?
It isolates the unknown derivative for solving.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do product and chain rules work together implicitly?
- Product rule: d(xy)/dx=y+xy′.
- Collect and factor y′ before dividing; keep the denominator condition.
Remember: d(xy)/dx is y+xy′, not just xy′ and not 1·y′.
Conditions: Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. xy=6, so y=6/x. Only the positive branch is shown. Dimensionless coordinates; full relation excludes x=0.
Refresh Kid · AP Calculus AB Unit 3 · Objectives FUN-3.D · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.2, FUN-3.D. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 3 has six official topics. Topic 3.5 integrates existing derivative objectives and Mathematical Practice 1.C; it does not introduce a separate new objective. Focused lesson titles and questions are original Refresh Kid teaching material.
Chain rule factors are evaluated at their correct nested inputs. Implicit derivatives refer to local branches and retain denominator conditions. Inverse derivatives require the corresponding preimage and a nonzero original derivative under the local inverse conditions. Trigonometric inputs and returned angles use radians; inverse branch conventions and domains are stated. Related rates and L’Hôpital’s rule remain in later units. Higher derivatives are repeated differentiation, not powers.
The Organic Chemistry Tutor video creator and relevant descriptions were checked (the implicit video link is labeled descriptively); 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 3.6, 3.7 and 3.8 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 folds a graph 180 degrees around y=x to construct its inverse reflection. Intermediate depth is a geometric construction, not an extra function variable. Camera rotation only changes the view. Equal-scale labeled 2D graphs and text provide the complete explanation.
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 AB 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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