How do you differentiate an implicit slope again?
You will be able to: Keep y dependent on x when finding a second derivative of a relation.
How do you differentiate an implicit slope again?
On the upper circle x²+y²=25, the tangent at (0,5) is level. Moving to the right makes that tangent slope decrease, so a zero first derivative can coexist with a nonzero second derivative.
A useful starting point: Why does a second derivative often need a new product rule? →
Words and symbols before equations
- Implicit second derivative
- The derivative with respect to x of an implicitly determined y′.
- y″
- The rate of change of y′ along the selected local branch.
- Substitution after differentiation
- Inserting point coordinates only after the changing expression has been differentiated.
- Branch condition
- A restriction ensuring the local formula is defined, here y≠0.
What this picture assumes
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. Circle x²+y²=25, equal coordinate scales. y′=−x/y and y″=−25/y³ only for y≠0. Second derivative is a change in slope, not its square.
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.
- Point (0, 5): y′=−x/y=0; y″=−25/y³=-0.2. Slope and change in slope are different quantities.
- 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 first derivative is y′=−x/y. When differentiating again, y still depends on x. The quotient rule gives y″=[−y+xy′]/y².
Substitute y′=−x/y to obtain y″=−(x²+y²)/y³=−25/y³. The original relation simplifies the numerator.
At (3,4), y″=−25/64. On the lower branch at (3,−4), it is +25/64. The sign describes how the slope changes as x increases along that branch.
Equivalently, differentiating 2x+2yy′=0 gives 2+2(y′)²+2yy″=0. The product yy′ produces both terms. These formulas require y≠0; the vertical endpoints do not have ordinary finite y′ or y″.
A worked example, step by step
Find y″ at (0,5) on x²+y²=25.
- First find y′=−x/y, so y′(0,5)=0.
- Differentiate the relation’s first derivative: 2+2(y′)²+2yy″=0.
- Substitute x=0, y=5 and y′=0: 2+10y″=0.
- Thus y″=−1/5: the initially horizontal slope decreases as x increases.
Do not substitute a point before differentiating or treat y in −x/y as a constant.
Why is d(yy′)/dx not just yy″?
Compare with an explanation
Both factors depend on x; the first factor contributes y′·y′ and the second contributes y·y″.
Predict. Change one thing. Explain.
Switch upper and lower branches and move x. Compare y′ and y″ separately. Explain why y″ changes sign between the two branches.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Point (0, 5): y′=−x/y=0; y″=−25/y³=-0.2. Slope and change in slope are different quantities.
Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. Circle x²+y²=25, equal coordinate scales. y′=−x/y and y″=−25/y³ only for y≠0. Second derivative is a change in slope, not its square.
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 questionFind y′ and y″ at (4,3) on x²+y²=25, and explain why the two values differ.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The point lies on the circle because 16+9=25.
- 1 point: y′=−4/3.
- 1 point: y″=−25/27.
- 1 point: The first is the tangent slope; the second is the local change of that slope per unit x, so they answer different questions.
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 1What remains variable in implicit second differentiation?
Both y and y′ depend on x.
RECALL 2When should coordinates be inserted?
After deriving the necessary changing expressions.
RECALL 3Why exclude y=0 in the circle formulas?
The ordinary finite derivative y(x) fails at the vertical endpoints.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you differentiate an implicit slope again?
- For x²+y²=R²: y′=−x/y and y″=−R²/y³, y≠0.
- d(yy′)/dx=(y′)²+yy″.
Remember: Do not substitute a point before differentiating or treat y in −x/y as a constant.
Conditions: Original mathematical model. Readouts are rounded; algebra supplies exact conclusions. Circle x²+y²=25, equal coordinate scales. y′=−x/y and y″=−25/y³ only for y≠0. Second derivative is a change in slope, not its square.
Refresh Kid · AP Calculus AB Unit 3 · Objectives FUN-3.F · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.6, FUN-3.F. 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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