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Reading Time: 6 min
Last Updated: March 5, 2026
Main Ideas: 5
Reading Time: 6 min
Last Updated: March 5, 2026
Main Ideas: 5

Topic 5.12 Notes – Exploring Behaviors of Implicit Relations

Verified for 2027 AP® Calculus AB Exam
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Topic 5.12 extends everything you know about derivatives to implicit relations. Even when an equation isn’t written as y=f(x)y=f(x), you can still find slopes, critical points, and concavity using implicit differentiation. The ideas are the same as before, but the algebra and interpretation require more care.

1. Implicit Relations and Their Derivatives

An implicit relation is an equation involving both xx and yy, like
F(x,y)=0(example: x2+xy+y2=7) F(x,y)=0 \quad \text{(example: } x^2+xy+y^2=7\text{)}

It might not pass the vertical line test, but locally you can still treat yy as a function of xx.

Implicit differentiation reminder

When you differentiate with respect to xx:

  • Differentiate every term.
  • Treat yy as a function of xx.
  • Every time you differentiate yy, multiply by dydx\frac{dy}{dx} (chain rule).
  • Solve for dydx\frac{dy}{dx}.

Quick example:

Given x2+y2=10x^2 + y^2 = 10

Differentiate:
2x+2ydydx=0 2x + 2y\frac{dy}{dx} = 0

Solve:
dydx=−xy \frac{dy}{dx} = -\frac{x}{y}

That expression gives the slope of the tangent line at any point on the curve.

Below is the circle x2+y2=10x^2 + y^2 = 10 with two sample tangent lines drawn using dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.

Circle x2+y2=10x^2 + y^2 = 10 with tangent lines

At (1,3)(1,3), the slope is negative. At (−1,3)(-1,3), the slope is positive. The formula tells you the slope at every point without ever solving for yy explicitly.

Even though this isn’t written as y=f(x)y=f(x), slopes still exist at most points.

Everything you already know about increasing, decreasing, extrema, and concavity now comes from this implicitly found derivative.

2. Critical Points of an Implicit Relation

A critical point occurs where:

  • dydx=0\frac{dy}{dx} = 0, or
  • dydx\frac{dy}{dx} does not exist

and the point satisfies the original equation.

How to find them

  1. Find dydx\frac{dy}{dx} implicitly.
  2. Set the numerator equal to 0 → horizontal tangents.
  3. Set the denominator equal to 0 → undefined slopes (often vertical tangents).
  4. Plug back into the original equation to confirm the point lies on the curve.

Important subtlety:

  • Undefined slope does not automatically mean max or min.
  • It may be a vertical tangent line.

On tests, students often forget to check where the derivative is undefined. That costs easy points.

To classify a critical point, use a sign chart around that xx-value. If the sign of dydx\frac{dy}{dx} changes:

  • +→−+\to- → local maximum
  • −→+-\to+ → local minimum
  • No change → neither

Because this is implicit, you may need to check nearby points that satisfy the original equation.

3. Using the First Derivative to Describe Behavior

Once you have dydx\frac{dy}{dx}:

  • dydx>0\frac{dy}{dx} > 0 → increasing
  • dydx<0\frac{dy}{dx} < 0 → decreasing

You analyze signs just like with explicit functions.

On FRQs, you must justify conclusions. That means explicitly stating the sign of dydx\frac{dy}{dx} and what it implies about behavior.

For example:
“If dydx>0\frac{dy}{dx} > 0 for x>2x>2, the function is increasing for x>2x>2.”

Keep your reasoning tied directly to the derivative.

4. Second Derivative and Concavity of Implicit Relations

Concavity still comes from the second derivative.

Steps

  1. Find dydx\frac{dy}{dx}.
  2. Differentiate again with respect to xx.
  3. Solve for d2ydx2\frac{d^2y}{dx^2}.

The second derivative may involve:

  • xx
  • yy
  • dydx\frac{dy}{dx}

That’s completely normal.

Interpreting it

  • d2ydx2>0\frac{d^2y}{dx^2} > 0 → concave up
  • d2ydx2<0\frac{d^2y}{dx^2} < 0 → concave down

A point of inflection occurs where concavity changes. Setting d2ydx2=0\frac{d^2y}{dx^2}=0 only finds candidates. You must confirm a sign change.

Here’s a visual reminder of concavity and an inflection point.

Study guide illustration

Concavity change at an inflection point

In the graph, the curve changes from concave down to concave up at the labeled point. That change in shape is what your sign chart for d2ydx2\frac{d^2y}{dx^2} must show.

On exams, students lose points by stopping after solving d2ydx2=0\frac{d^2y}{dx^2}=0. Always check the sign change.

5. Extending Derivative Applications to Implicit Functions

All derivative applications still work:

  • Increasing/decreasing
  • Relative extrema
  • Concavity
  • Inflection points
  • Related rates

Related rates with implicit equations

Often the equation naturally relates two variables, like geometry problems.

You differentiate with respect to time tt:

dydt=dydx⋅dxdt \frac{dy}{dt} = \frac{dy}{dx}\cdot \frac{dx}{dt}

This is just the chain rule.

Important habits that matter on quizzes and FRQs:

  • Differentiate before plugging in numbers.
  • Keep track of signs (increasing vs decreasing).
  • Clearly state units in your final answer.

Key Takeaways

A critical point of an implicit relation occurs where dydx=0\frac{dy}{dx}=0 or is undefined and the point lies on the original equation.
Undefined slope often means a vertical tangent, not automatically a max or min.
Classification requires a sign change in dydx\frac{dy}{dx}, not just solving dydx=0\frac{dy}{dx}=0.
The second derivative of an implicit function may include xx, yy, and dydx\frac{dy}{dx}, and that is expected.
For inflection points, a sign change in d2ydx2\frac{d^2y}{dx^2} must be shown.
In related rates, differentiate first and substitute values after.

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