Exploring Behaviors of Implicit Relations
Analyze local implicit branches, tangent candidates, and second derivatives with point and domain checks.
College Board Topic 5.12 · CED p. 105 (PDF p. 110) · CED effective Fall 2020; project course outline 2026–27.
An implicit equation can describe more than one branch, so first decide which local function y(x) is being analyzed. Implicit differentiation produces a slope in terms of x and y. A horizontal-tangent candidate has y′ = 0, usually a numerator zero with a nonzero denominator; a zero denominator calls for checking whether the derivative fails to exist, whether the point is even on the relation, and whether there is a function branch nearby. A 0/0 expression needs more work.
For x² + y² = 25, implicit differentiation gives y′ = −x/y when y ≠ 0. On the upper branch, the point (0,5) has y′ = 0. At (±5,0), the slope formula is undefined and the circle has vertical tangents. Those are boundary candidates for the upper-branch closed interval; the interior stationary point is x = 0. The full circle is not a single-valued function y(x), so do not describe every point using one hidden branch.
Differentiate again carefully: 2 + 2(y′)² + 2yy″ = 0, giving y″ = −[1 + (y′)²]/y. Second derivatives of implicit relations can contain x, y, and y′. Substitution or the original relation may simplify the expression. On the upper circle branch, y > 0 makes y″ negative, so (0,5) is a local maximum and the branch is concave down. On the lower branch, the sign reverses.
A calculator plot may draw both branches or hide a vertical tangent. A numerical derivative near that tangent can become enormous without being a finite derivative at the point. Use the relation, domain, slope formula, and sign evidence to make the conclusion. Report an ordered pair when the relation contains more than one point at the same x.
Let's work through it
For x² + y² = 25 at (3,4), differentiate to get 2x + 2yy′ = 0 and y′ = −3/4.
Why: Treat y as a function of x and use the chain rule; the point lies on the upper branch.
Differentiate again: 2 + 2(y′)² + 2yy″ = 0.
Why: The product and chain rules retain both y′ and y″ terms.
At (3,4), y″ = −[1 + 9/16]/4 = −25/64.
Why: Substitute the point and the already-found first derivative.
The branch is decreasing and concave down at this point, under the local branch conditions.
Why: The first derivative is negative and the second derivative is negative; these are separate behaviors.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
In y′ = −x/y for a circle, what must be checked before calling x = 0 a horizontal tangent?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
In y′ = −x/y for a circle, what must be checked before calling x = 0 a horizontal tangent?
Worked answer and reason
The point must lie on the relation and y must be nonzero.
A zero numerator alone is not enough when the denominator also vanishes.
Retrieval 2
Why can y″ still include y′?
Worked answer and reason
Differentiating implicit expressions uses product and chain rules on y(x).
You can substitute a known y′ or simplify afterward.
Retrieval 3
Why state the branch?
Worked answer and reason
The same x can correspond to different y-values and derivative signs.
A relation need not define one global function.
Catch the tempting mistake
- A zero denominator always means a horizontal tangent.
- It signals a possible undefined slope; horizontal tangents require slope zero.
- Differentiate y² twice as 2yy″ only.
- The second derivative is 2(y′)² + 2yy″.
Calculator workspace and limits
For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.
Scientific calculator
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Put it to work
- Unit 5 problem 26 · No calculator · analytic
- Unit 5 problem 27 · No calculator · analytic
- Unit 5 problem 28 · No calculator · analytic
For the teacher
- Require the point-on-curve check before substituting into a derivative formula.
- Compare upper and lower circle branches to connect the same algebra with opposite concavity signs.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Function
A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.
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Implicit differentiation
Implicit differentiation finds a derivative from an equation relating variables without first solving explicitly for one variable. When y depends on x, differentiating a term containing y requires the chain rule and a factor of dy/dx. For example, x²+y²=1 gives 2x+2y(dy/dx)=0. Solving that relation for dy/dx gives −x/y only where y≠0 and the relation supports the local function being considered.
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Vertical tangent
A vertical tangent is a tangent line parallel to the vertical axis, with no finite slope. A graph can have a vertical tangent even though its derivative does not exist as a finite real number at that input. For example, y=∛x has the vertical tangent x=0 at the origin as its nearby slopes tend to positive infinity. This case is separate from the finite-slope tangent approximation available at a differentiable point.
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Interval
An interval is a set of real numbers containing every number between any two of its members. Parentheses exclude an endpoint, while square brackets include it. For example, (1,4] includes 4 and every number between 1 and 4, but excludes 1. An infinite endpoint always uses a parenthesis because infinity is not a real endpoint.
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Second derivative
A second derivative is the derivative of the first derivative. It measures how the first rate of change changes with the input; further repeated differentiation gives higher-order derivatives. For example, f(x)=x³ has f′(x)=3x² and f″(x)=6x. For position measured against time, the second derivative is acceleration.
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Local extremum
A local extremum is a value at least as large as, or at least as small as, all values at nearby allowed inputs. A local maximum compares nearby heights above, while a local minimum compares them below. For example, x² has a local minimum at 0. A local extremum need not be the greatest or least value over the entire domain.
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Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.