How do you combine several derivative rules without losing a term?
You will be able to: Organize product, quotient and chain rules in a nested calculation.
How do you combine several derivative rules without losing a term?
An expression x²e^(3x) changes because its x² factor changes and because its exponential factor changes. The exponential also has its own inner rate.
A useful starting point: How do you choose the first derivative rule? →
Words and symbols before equations
- Product contribution
- The term obtained by differentiating one factor and retaining the other.
- Nested rule
- A rule applied inside a larger differentiation step.
- Quotient denominator
- The squared original denominator in the quotient rule.
- Factored derivative
- An equivalent derivative written with shared factors extracted.
What this picture assumes
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. y=x²e^(3x), all real x. Product-rule contributions may have opposite signs; their sum is the derivative. Finite graph window may clip large heights.
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.
- First contribution 2x e^(3x)=4.48169; second contribution 3x²e^(3x)=3.36127; sum y′=7.84296.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For y=x²e^(3x), start with the outer product. Its derivative is 2x·e^(3x)+x²·[3e^(3x)]. The chain factor 3 belongs inside the second contribution.
Factoring gives e^(3x)(2x+3x²). Factoring is optional; both forms describe the same derivative.
For a quotient u/v, keep parentheses around the numerator u′v−uv′ and divide by v². If u or v is itself composed, apply its chain rule inside that expression.
A useful check is to compare methods at a simple input or differentiate an expanded equivalent form. Such checks can reveal mistakes but do not prove every algebraic formula correct.
A worked example, step by step
Differentiate y=sin(2x)/(x+1), stating its domain.
- Take u=sin(2x), u′=2cos(2x), v=x+1 and v′=1.
- Apply the quotient rule: (u′v−uv′)/v².
- Obtain [2(x+1)cos(2x)−sin(2x)]/(x+1)².
- Keep x≠−1 and use radians for the trigonometric input.
Do not differentiate a product by multiplying derivatives, or a quotient by dividing derivatives.
At x=0, are both contributions zero in this example?
Compare with an explanation
Yes: 2x e^(3x) and 3x²e^(3x) both vanish. That is specific to this expression and input.
Predict. Change one thing. Explain.
Move x for x²e^(3x). Compare both product contributions and their sum. Explain why one contribution vanishing does not force the other to vanish.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
First contribution 2x e^(3x)=4.48169; second contribution 3x²e^(3x)=3.36127; sum y′=7.84296.
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. y=x²e^(3x), all real x. Product-rule contributions may have opposite signs; their sum is the derivative. Finite graph window may clip large heights.
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 y=x sin(x²), then find y′(0).
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Compare with the answer and four-point rubric
- 1 point: Use the product rule on x and sin(x²).
- 1 point: Differentiate the second factor with the chain rule: 2x cos(x²).
- 1 point: y′=sin(x²)+2x²cos(x²).
- 1 point: At x=0, y′=0+0=0.
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 1Where does a chain factor go in a product rule?
Inside the derivative of the composed factor.
RECALL 2Must you factor the final answer?
No. Equivalent correct expressions are acceptable.
RECALL 3Why keep numerator parentheses?
They ensure the entire difference is divided by the squared denominator.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you combine several derivative rules without losing a term?
- (uv)′=u′v+uv′.
- (u/v)′=(u′v−uv′)/v² for v≠0.
- Apply chain factors inside each needed derivative.
Remember: Do not differentiate a product by multiplying derivatives, or a quotient by dividing derivatives.
Conditions: Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. y=x²e^(3x), all real x. Product-rule contributions may have opposite signs; their sum is the derivative. Finite graph window may clip large heights.
Refresh Kid · AP Calculus AB Unit 3 · Objectives FUN-3.A–E; Mathematical Practice 1.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.5, FUN-3.A–E; Mathematical Practice 1.C. 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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