Behaviors of implicit relations

AP Calculus BC· AP Calculus BC CED — Analytical Applications of Differentiation· 14 min read

1. Implicit Differentiation for First and Second Derivatives★★☆☆☆⏱ 4 min

Implicit relations are equations relating $x$ and $y$ that cannot be rearranged to write $y$ explicitly as a function of $x$ . Even though they are often multi-valued (one $x$ maps to multiple $y$ values), we can still analyze their geometric behavior using derivatives via the technique of implicit differentiation.

📘 Definition

Implicit Differentiation

A technique to find $\frac{d y}{d x}$ for implicit relations that relies entirely on the chain rule: any term containing $y$ is differentiated with respect to $y$ first, then multiplied by $\frac{d y}{d x}$ to account for $y$ being a function of $x$ .

For any implicit equation $F (x, y) = C$ (constant), differentiate every term on both sides with respect to $x$ . For terms containing both $x$ and $y$ , apply product/quotient rule first, then chain rule to any derivative of $y$ . Collect all terms with $\frac{d y}{d x}$ on one side, factor out $\frac{d y}{d x}$ , and solve to get $\frac{d y}{d x}$ as an expression in both $x$ and $y$ . For the second derivative, differentiate $\frac{d y}{d x}$ again, then substitute the known expression for $\frac{d y}{d x}$ back into the result.

📐 Worked Example

Find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ for the relation $x^{2} + 3 x y + y^{2} = 11$ at the point $(2, 1)$ .

1
Differentiate both sides with respect to $x$ , applying product rule to $3 x y$ and chain rule to y-terms:
2
$2x + 3\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0$
3
Collect terms with $\frac{d y}{d x}$ and solve:
4
$\frac{dy}{dx}(3x + 2y) = -2x - 3y \implies \frac{dy}{dx} = \frac{-2x - 3y}{3x + 2y}$
5
Evaluate at $(2, 1)$ :
6
$\frac{dy}{dx} = \frac{-4 - 3}{6 + 2} = -\frac{7}{8}$
7
Differentiate $\frac{d y}{d x}$ with the quotient rule to get the second derivative:
8
$\frac{d^2y}{dx^2} = \frac{(3x + 2y)(-2 - 3\frac{dy}{dx}) - (-2x - 3y)(3 + 2\frac{dy}{dx})}{(3x + 2y)^2}$
9
Substitute $x = 2, y = 1, \frac{d y}{d x} = - \frac{7}{8}$ : numerator simplifies to $\frac{55}{4}$ , denominator is $64$ , so final result:
10
$\frac{d^2y}{dx^2} = \frac{55}{256} \approx 0.215$

Exam tip:

Always evaluate $\frac{d y}{d x}$ at the given point before finding the second derivative, rather than substituting the point at the end. This drastically simplifies arithmetic and reduces algebra errors on the AP exam.

2. Tangent and Normal Lines, Horizontal and Vertical Tangents★★☆☆☆⏱ 4 min

Once you have $\frac{d y}{d x}$ at a point on an implicit curve, $\frac{d y}{d x}$ is exactly the slope of the tangent line at that point, just like for explicit functions. The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope (when tangent slope is non-zero and defined). We use point-slope form $y - y_{0} = m (x - x_{0})$ to write the equation of either line.

AP exam questions very frequently ask for all points on an implicit curve that have horizontal or vertical tangents. A horizontal tangent has slope zero, so this occurs when $\frac{d y}{d x} = 0$ (numerator of $\frac{d y}{d x}$ is zero, denominator is non-zero). A vertical tangent has undefined slope, so this occurs when the denominator of $\frac{d y}{d x}$ is zero, and the numerator is non-zero.

📐 Worked Example

Find the equations of the tangent line and normal line to the ellipse $2 x^{2} + y^{2} = 18$ at the point $(2, 10)$ .

1
Differentiate implicitly with respect to $x$ :
2
$4x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{y}$
3
Evaluate the tangent slope at $(2, 10)$ :
4
$m_{\text{tan}} = -\frac{2(2)}{\sqrt{10}} = -\frac{2\sqrt{10}}{5}$
5
Write the tangent line using point-slope form and simplify:
6
$y - \sqrt{10} = -\frac{2\sqrt{10}}{5}(x - 2) \implies 2\sqrt{10}x + 5y = 9\sqrt{10}$
7
Calculate the normal line slope as the negative reciprocal:
8
$m_{\text{norm}} = \frac{\sqrt{10}}{4}$
9
Write and simplify the normal line equation:
10
$y - \sqrt{10} = \frac{\sqrt{10}}{4}(x - 2) \implies \sqrt{10}x - 4y + 2\sqrt{10} = 0$

Exam tip:

If asked to find all points with horizontal/vertical tangents, always check that your candidate points actually lie on the original implicit curve. Solving the slope condition is not enough for a valid answer.

3. Analyzing Critical Points, Extrema, and Concavity★★★☆☆⏱ 4 min

The same rules we use to analyze extrema and concavity for explicit functions apply directly to implicit relations. A critical point on an implicit curve is a point where $\frac{d y}{d x} = 0$ (horizontal tangent, candidate for local maximum or minimum of $y$ with respect to $x$ ) or $\frac{d y}{d x}$ is undefined (vertical tangent, candidate for extremum of $x$ with respect to $y$ ).

To classify critical points, we use the second derivative test just like for explicit functions: if $\frac{d y}{d x} = 0$ at a critical point, then $\frac{d ^{2} y}{d x ^{2}} > 0$ means the point is a local minimum, and $\frac{d ^{2} y}{d x ^{2}} < 0$ means the point is a local maximum. For concavity, $\frac{d ^{2} y}{d x ^{2}} > 0$ means concave up, $\frac{d ^{2} y}{d x ^{2}} < 0$ means concave down, and inflection points occur where concavity changes.

📐 Worked Example

Find all local maxima and minima of the relation $x^{2} - x y + y^{2} = 3$ .

1
Differentiate implicitly and solve for $\frac{d y}{d x}$ :
2
$2x - \left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y - 2x}{2y - x}$
3
Critical points for extrema of $y$ occur when $\frac{d y}{d x} = 0$ , so numerator $y - 2 x = 0 ⟹ y = 2 x$ . We confirm denominator $2 y - x \neq = 0$ for these points.
4
Substitute $y = 2 x$ into the original equation to find valid points on the curve:
5
$x^2 - x(2x) + (2x)^2 = 3 \implies 3x^2 = 3 \implies x = \pm 1$
6
This gives critical points $(1, 2)$ and $(- 1, - 2)$ . We use the second derivative test, substituting $\frac{d y}{d x} = 0$ to simplify calculation:
7
At $(1, 2)$ :
8
$\frac{d^2y}{dx^2} = -\frac{2}{3} < 0$
9
So $(1, 2)$ is a local maximum. At $(- 1, - 2)$ :
10
$\frac{d^2y}{dx^2} = \frac{2}{3} > 0$
11
So $(- 1, - 2)$ is a local minimum.

Exam tip:

When evaluating the second derivative at a critical point, remember $\frac{d y}{d x} = 0$ , so this term will drop out of most expressions, drastically simplifying calculation. Always substitute $\frac{d y}{d x} = 0$ first before doing any other arithmetic.

4. AP Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

The curve defined by $x^{3} + 2 x y + y^{2} = 5$ contains the point $(1, 1)$ . What is the slope of the tangent line to the curve at this point?
- $- \frac{5}{4}$
- $- \frac{5}{2}$
- $\frac{5}{4}$
- $- 1$

5. Common Pitfalls

Wrong move:

After differentiating a product term like $2 x y$ , you write $2 x \frac{d y}{d x}$ and forget the product rule term $2 y$ .

Why:

Students remember to apply chain rule to y but forget that terms with both x and y require the product rule before applying the chain rule.

Correct move:

Always use the product/quotient rule first for any term that has both x and y as factors, then apply the chain rule to any derivative of y.

Wrong move:

When finding $\frac{d ^{2} y}{d x ^{2}}$ , you differentiate $\frac{d y}{d x}$ and leave out the chain rule factor of $\frac{d y}{d x}$ when differentiating terms that include y.

Why:

Students forget that y is still a function of x when taking the second derivative, so all y terms still require a $\frac{d y}{d x}$ factor.

Correct move:

Whenever you differentiate a term containing y at any derivative order, multiply by the appropriate derivative of y per the chain rule.

Wrong move:

When asked for points with horizontal tangents, you set the denominator of $\frac{d y}{d x}$ equal to zero instead of the numerator.

Why:

Confusion between the conditions for horizontal vs vertical tangents, common when memorizing conditions without context.

Correct move:

Remember $\frac{d y}{d x} =$ slope: horizontal slope $= 0$ , so numerator $= 0$ (denominator $\neq = 0$ ); vertical slope is undefined, so denominator $= 0$ (numerator $\neq = 0$ ).

Wrong move:

When finding extrema, you keep $\frac{d y}{d x}$ as a variable when evaluating $\frac{d ^{2} y}{d x ^{2}}$ at a critical point, leading to complicated unsimplified expressions.

Why:

Students follow a generic formula instead of using the fact that $\frac{d y}{d x} = 0$ at all critical points from the first derivative condition.

Correct move:

Substitute $\frac{d y}{d x} = 0$ into the second derivative expression immediately before evaluating, which eliminates most terms and simplifies arithmetic.

Wrong move:

After finding that all horizontal tangents occur along a line like $y = 2 x$ , you report all points on that line instead of finding specific points on the original curve.

Why:

Students think satisfying the derivative condition is enough, and don't realize only specific points on the line lie on the original implicit curve.

Correct move:

Always substitute the derivative condition back into the original curve equation to solve for the specific x and y coordinates of valid points.

6. Quick Reference Cheatsheet

Category	Rule/Condition	Notes
Implicit differentiation of $f (y)$	$\frac{d}{d x} [f (y)] = f^{'} (y) \cdot \frac{d y}{d x}$	Apply chain rule to all y-containing terms; use product/quotient rule first for terms with both x and y
First derivative of implicit relation	Solve for $\frac{d y}{d x}$ after differentiating all terms	Result is an expression in both x and y, not just x
Second derivative of implicit relation	Differentiate $\frac{d y}{d x}$ again, substitute $\frac{d y}{d x}$ back	Evaluate $\frac{d y}{d x}$ at the point first to simplify calculation
Tangent slope at $(x_{0}, y_{0})$	$m = \frac{d y}{d x}_{(x_{0}, y_{0})}$	Only valid if $(x_{0}, y_{0})$ lies on the original curve
Normal slope at $(x_{0}, y_{0})$	$m_{normal} = - \frac{1}{m _{tangent}}$	Undefined if tangent is horizontal; zero if tangent is vertical
Horizontal tangent condition	$\frac{d y}{d x} = 0 ⟹ num (\frac{d y}{d x}) = 0, den (\frac{d y}{d x}) \neq = 0$	Candidates for local maxima/minima of y with respect to x
Vertical tangent condition	$\frac{d y}{d x}$ undefined $⟹ den (\frac{d y}{d x}) = 0, num (\frac{d y}{d x}) \neq = 0$	Candidates for local maxima/minima of x with respect to y
Second derivative test for extrema	If $\frac{d y}{d x} = 0$ : $\frac{d ^{2} y}{d x ^{2}} > 0 \to$ local min; $\frac{d ^{2} y}{d x ^{2}} < 0 \to$ local max	Same rule as for explicit functions
Concavity for implicit curves	$\frac{d ^{2} y}{d x ^{2}} > 0 \to$ concave up; $\frac{d ^{2} y}{d x ^{2}} < 0 \to$ concave down	Inflection points occur where concavity changes

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Find tangent slope at given point
2022 · FRQ
Find all horizontal tangent points

What's Next

Mastering the analysis of implicit relations is a foundational skill for many subsequent topics in AP Calculus BC. The core technique of implicit differentiation you learned here is directly applied to related rates problems, where you relate the rates of change of multiple quantities that are implicitly connected to each other. It also underpins the study of parametric and polar curves, which are unique to AP Calculus BC, where derivatives are calculated implicitly for non-cartesian curves. This topic also reinforces your understanding of derivative applications for curve sketching, which is a recurring theme across multiple-choice and free-response questions. Building a strong grasp of implicit behaviors will make more advanced topics in differentiation much easier to master as you progress through your exam preparation.