Implicitly defined functions

AP Precalculus· AP Precalculus CED — Functions Involving Parameters, Vectors, and Matrices· 14 min read

1. What Are Implicitly Defined Functions?★☆☆☆☆⏱ 3 min

An explicitly defined function writes the dependent output variable (usually $y$ ) directly as a function of the independent input variable (usually $x$ ), in the form $y = f (x)$ , the standard form you have used for most of the course. An implicitly defined function is a relation between $x$ and $y$ written as a single equation $F (x, y) = 0$ , where $y$ is treated as a function of $x$ even if we cannot solve for $y$ in terms of elementary functions.

📘 Definition

Locally Implicit Function

$F (x, y) = 0$

An implicit relation that defines $y$ as a function of $x$ in a small neighborhood around a given point on the curve, even if it cannot be solved for $y$ explicitly globally.

Example:

The full unit circle $x^{2} + y^{2} = 1$ defines $y = 1 - x^{2}$ for all points above the $x$ -axis.

2. Implicit Differentiation via the Chain Rule★★☆☆☆⏱ 4 min

The core skill for working with implicit functions is implicit differentiation, which allows you to find $\frac{d y}{d x}$ without solving for $y$ explicitly. The key idea is that $y$ is a function of $x$ , so any term involving $y$ is a composite function of $x$ , which requires the chain rule when differentiating with respect to $x$ .

\frac{d}{dx}\left[y^n\right] = n y^{n-1} \cdot \frac{dy}{dx}

Terms that only involve $x$ are differentiated normally, just like with explicit functions. After differentiating both sides of the implicit equation with respect to $x$ , you rearrange terms to solve for $\frac{d y}{d x}$ , which will usually be an expression in both $x$ and $y$ — this is expected and acceptable.

📐 Worked Example

Find $\frac{d y}{d x}$ for the circle defined by $x^{2} + y^{2} = 25$ .

1
Differentiate both sides of the equation with respect to $x$ , applying the chain rule to terms with $y$ :
2
$\frac{d}{dx}\left[x^2\right] + \frac{d}{dx}\left[y^2\right] = \frac{d}{dx}\left[25\right]$
3
Compute each derivative, adding the $\frac{d y}{d x}$ factor for the $y$ term:
4
$2x + 2y \cdot \frac{dy}{dx} = 0$
5
Isolate the term with $\frac{d y}{d x}$ :
6
$2y \cdot \frac{dy}{dx} = -2x$
7
Divide both sides by $2 y$ to solve for $\frac{d y}{d x}$ :
8
$\frac{dy}{dx} = -\frac{x}{y}$

Exam tip:

Always keep the $\frac{d y}{d x}$ factor when differentiating any term that includes $y$ — if you forget it, you will end up with an incorrect constant slope.

3. Evaluating $\frac{dy}{dx}$ at a Point on an Implicit Curve★★★☆☆⏱ 3 min

Most AP Precalculus problems do not ask for the general form of $\frac{d y}{d x}$ ; instead, they ask for the slope of the tangent line at a given point that lies on the implicit curve. After you find the general expression for $\frac{d y}{d x}$ in terms of $x$ and $y$ , you simply substitute the coordinates of the given point $(x_{0}, y_{0})$ directly into the expression. A critical preliminary step most students skip is confirming the given point actually lies on the curve.

📐 Worked Example

Find the slope of the tangent line to the curve $x^{3} + y^{3} - 9 x y = 0$ at the point $(2, 4)$ .

1
First confirm the point is on the curve: $2^{3} + 4^{3} - 9 (2) (4) = 8 + 64 - 72 = 0$ , which matches the right-hand side, so the point is valid.
2
Differentiate both sides with respect to $x$ , applying the chain rule to $y$ terms and the product rule to the $9 x y$ term:
3
$3x^2 + 3y^2 \frac{dy}{dx} - 9\left(x \frac{dy}{dx} + y\right) = 0$
4
Expand and rearrange terms to collect $\frac{d y}{d x}$ on one side:
5
$\frac{dy}{dx} \left(3y^2 - 9x\right) = 9y - 3x^2$
6
Simplify the general derivative:
7
$\frac{dy}{dx} = \frac{3y - x^2}{y^2 - 3x}$
8
Substitute $x = 2, y = 4$ :
9
$\frac{dy}{dx}\bigg|_{(2,4)} = \frac{3(4) - 2^2}{4^2 - 3(2)} = \frac{8}{10} = \frac{4}{5}$

Exam tip:

Always simplify the general derivative before substituting the point to reduce arithmetic errors — factoring out common constants early cuts down on miscalculations.

4. Finding the Equation of a Tangent Line to an Implicit Curve★★★☆☆⏱ 4 min

One of the most common FRQ questions on this topic asks for the full equation of the tangent line to an implicit curve at a given point. This combines the skill of finding the slope via implicit differentiation with the point-slope form of a line you learned earlier in the course. The AP exam will usually accept either point-slope or slope-intercept form unless specified otherwise.

📐 Worked Example

Find the equation of the tangent line to the ellipse defined by $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ at the point $(2, \frac{3 3}{2})$ .

1
Confirm the point is on the ellipse: $\frac{2 ^{2}}{16} + \frac{( \frac{3 3}{2} ) ^{2}}{9} = \frac{1}{4} + \frac{3}{4} = 1$ , which checks out.
2
Differentiate both sides with respect to $x$ :
3
$\frac{x}{8} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0$
4
Solve for $\frac{d y}{d x}$ :
5
$\frac{dy}{dx} = -\frac{9x}{16y}$
6
Substitute the point to find the slope:
7
$m = -\frac{9(2)}{16\left(\frac{3\sqrt{3}}{2}\right)} = -\frac{\sqrt{3}}{4}$
8
Use point-slope form and simplify to slope-intercept:
9
$y - \frac{3\sqrt{3}}{2} = -\frac{\sqrt{3}}{4}(x - 2) \implies y = -\frac{\sqrt{3}}{4}x + 2\sqrt{3}$

Exam tip:

If the question asks for the tangent line, double-check that you did not mix up $x$ and $y$ when substituting into the derivative — swapping coordinates will give you the wrong slope.

5. AP-Style Concept Check★★★★☆⏱ 3 min

✓ Quick check

Test your understanding of implicit function techniques with this AP-style multiple choice question:

What is the slope of the tangent line to the curve $y^{2} - 2 x^{2} y = 8$ at the point $(1, 4)$ ?
- $\frac{4}{3}$
- $2$
- $\frac{8}{3}$
- $4$
Reveal answer
$\frac{8}{3}$ —
Correct: Confirm the point is on the curve, apply the product rule to $- 2 x^{2} y$ , isolate $\frac{d y}{d x}$ , and substitute to get the result.

6. Common Pitfalls

Wrong move:

Forgetting the chain rule factor of $\frac{d y}{d x}$ when differentiating $y^{n}$ , e.g., writing $\frac{d}{d x} [y^{2}] = 2 y$ instead of $2 y \frac{d y}{d x}$ .

Why:

Students are used to differentiating only terms with $x$ , so they automatically treat $y$ as a constant instead of a function of $x$ .

Correct move:

Every time you differentiate a term that includes $y$ , add $\frac{d y}{d x}$ as a factor before moving to the next term.

Wrong move:

Forgetting the product rule when differentiating cross terms like $x y$ or $x^{2} y^{3}$ , e.g., writing $\frac{d}{d x} [9 x y] = 9 x \frac{d y}{d x}$ instead of $9 x \frac{d y}{d x} + 9 y$ .

Why:

Students remember to add the $\frac{d y}{d x}$ for the $y$ term but forget that the $x$ term also needs to be differentiated.

Correct move:

For any product of a function of $x$ and a function of $y$ , apply the product rule first before adding the $\frac{d y}{d x}$ factor to the $y$ derivative.

Wrong move:

Trying to solve for $y$ explicitly before differentiating, leading to complicated square roots and algebraic errors.

Why:

Students are uncomfortable working with expressions in $x$ and $y$ , so they force a solution for $y$ even when it's unnecessary.

Correct move:

Always differentiate implicitly first, then substitute the point — you never need to solve for $y$ explicitly for AP problems.

Wrong move:

Using the original equation to eliminate $y$ from $\frac{d y}{d x}$ when only the value at a point is needed, leading to unnecessary algebraic complexity.

Why:

Students expect derivatives to be only in terms of $x$ , so they unnecessarily substitute to eliminate $y$ .

Correct move:

Leave $\frac{d y}{d x}$ in terms of both $x$ and $y$ and substitute the $y$ -coordinate of the point directly.

Wrong move:

Failing to confirm the point is on the curve before computing the derivative, leading to a wrong slope for a point that is not on the curve.

Why:

Students assume the point given is on the curve and skip the check.

Correct move:

Always plug the given point back into the original implicit equation to confirm it satisfies the equation before proceeding.

7. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Explicit function	$y = f (x)$	Dependent variable $y$ solved explicitly in terms of $x$
Implicit relation	$F (x, y) = 0$	$y$ is treated as a function of $x$ , no need to solve for $y$
Chain rule for $y^{n}$	$\frac{d}{d x} [y^{n}] = n y^{n - 1} \frac{d y}{d x}$	Always keep the $\frac{d y}{d x}$ factor for any power of $y$
Product rule for $x^{m} y^{n}$	$\frac{d}{d x} [x^{m} y^{n}] = m x^{m - 1} y^{n} + n x^{m} y^{n - 1} \frac{d y}{d x}$	Apply product rule before adding the $\frac{d y}{d x}$ factor
Slope at $(x_{0}, y_{0})$	$\frac{d y}{d x}_{(x_{0}, y_{0})}$	Substitute $x = x_{0}, y = y_{0}$ directly into general derivative
Tangent line equation	$y - y_{0} = m (x - x_{0})$	Confirm $(x_{0}, y_{0})$ is on the curve first; $m = \frac{d y}{d x}$ at the point
Horizontal tangent	$\frac{d y}{d x} = 0 ⟹$ numerator $= 0$	Denominator must be non-zero (singular point otherwise)
Vertical tangent	$\frac{d y}{d x}$ undefined $⟹$ denominator $= 0$	Numerator must be non-zero for a vertical tangent

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.

2024 · AP Precalculus
MCQ: slope of tangent to implicit curve
2023 · AP Precalculus
FRQ: tangent line to implicit curve

What's Next

Implicitly defined functions are the foundation for working with parametric curves, the next major topic in Unit 4 of AP Precalculus. You will use implicit differentiation to find derivatives of parametric functions, which relies on the core skill of differentiating $y$ as a function of $x$ without an explicit expression. Without mastering the techniques in this module, finding slopes of parametric curves and tangent lines to parametric trajectories — common free-response problems — will be much more error-prone. Beyond this unit, implicit functions are a core prerequisite for AP Calculus, where you will extend implicit differentiation to second derivatives and related rates problems, building on the local function definition you learned here.