Implicitly defined functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of implicit vs explicit functions, distinguishing dependent and independent variables, implicit differentiation via the chain rule, finding tangent lines to implicit curves, and evaluating derivatives at points on implicit relations.

You should already know: Chain rule for composite functions, explicit function notation, point-slope form for linear equations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Implicitly defined functions?

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 $y=f(x)$, which is the form you have worked with 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.

Unlike explicit functions, an implicit relation can describe entire curves that are not functions (like the full unit circle, which fails the vertical line test) while still defining one or more functions locally around any point on the curve that is not a vertical tangent. This local definition is key to AP Precalculus problems, which always ask about behavior at a specific point on the curve, so you do not need to solve for $y$ explicitly to find its rate of change.

According to the AP Precalculus CED, this topic makes up approximately 8-10% of Unit 4’s exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam.

2. Implicit Differentiation via the Chain Rule

The core skill for working with implicit functions is implicit differentiation, which allows you to find $\frac{dy}{dx}$ 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 $x$.

For example, if you have a term $y^n$, differentiating with respect to $x$ gives: $$\frac{d}{dx}\left[y^n\right]=n{y}^{n-1}\cdot \frac{dy}{dx}$$ This follows directly from the chain rule: the derivative of the outer function (the power of $y$) is $n{y}^{n-1}$, multiplied by the derivative of the inner function $y$ with respect to $x$, which is $\frac{dy}{dx}$. 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 $\frac{dy}{dx}$, which will usually be an expression in both $x$ and $y$ — this is expected and acceptable for implicit derivatives.

Worked Example

Find $\frac{dy}{dx}$ 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 $y$: $$\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[y^2\right]=\frac{d}{dx}\left[25\right]$$
2. Compute each derivative, adding the $\frac{dy}{dx}$ factor for the $y$ term: $$2x+2y\cdot \frac{dy}{dx}=0$$
3. Isolate the term with $\frac{dy}{dx}$: $$2y\cdot \frac{dy}{dx}=-2x$$
4. Divide both sides by $2y$ to solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=-\frac{x}{y}$$

Exam tip: Always keep the $\frac{dy}{dx}$ factor when differentiating any term that includes $y$ — if you forget it, you will end up with an incorrect constant slope, so check your work by confirming you have a $\frac{dy}{dx}$ term for every $y$-containing term.

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

Most AP Precalculus problems do not ask for the general form of $\frac{dy}{dx}$; instead, they ask for the slope of the tangent line at a given point that lies on the implicit curve. The process is straightforward: after you find the general expression for $\frac{dy}{dx}$ in terms of $x$ and $y$, you simply substitute the coordinates of the given point $(x_0,y_0)$ directly into the expression. You do not need to do any extra algebra to eliminate $y$ or simplify further after substitution.

A critical preliminary step that most students skip is confirming the given point actually lies on the curve. If the point does not satisfy the original implicit equation, it is not on the curve, and the question is asking for a different calculation (like the tangent line from an external point). Always do this check first to avoid wasted work.

Worked Example

Find the slope of the tangent line to the curve $x^3+y^3-9xy=0$ 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 $9xy$ term: $$3x^2+3y^2\frac{dy}{dx}-9\left(x\frac{dy}{dx}+y\right)=0$$
3. Expand and rearrange terms to collect $\frac{dy}{dx}$ on one side: $$\frac{dy}{dx}\left(3y^2-9x\right)=9y-3x^2$$
4. Simplify the general derivative: $$\frac{dy}{dx}=\frac{3y-x^2}{y^2-3x}$$
5. Substitute $x=2,y=4$: $$\frac{dy}{dx}{|}_{(2,4)}=\frac{3(4)-2^2}{4^2-3(2)}=\frac{8}{10}=\frac{4}{5}$$

Exam tip: Always simplify the general derivative before substituting the point to reduce arithmetic errors — factoring out common constants early cuts down on the chance of miscalculating large numbers.

4. Finding the Equation of a Tangent Line to an Implicit Curve

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 process follows three core steps after confirming the point is on the curve: find the slope $m=\frac{dy}{dx}$ at the point, plug into point-slope form, and simplify to the requested form. The AP exam will usually accept either point-slope or slope-intercept form unless specified otherwise, but it is good practice to simplify to slope-intercept form when possible.

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 $\left(2,\frac{3\sqrt{3}}{2}\right)$.

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

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 $x_0$ and $y_0$ will give you the wrong slope every time.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the chain rule factor of $\frac{dy}{dx}$ when differentiating $y^n$, e.g., writing $\frac{d}{dx}[y^2]=2y$ instead of $2y\frac{dy}{dx}$. 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 $\frac{dy}{dx}$ as a factor before moving to the next term.
• Wrong move: Forgetting the product rule when differentiating cross terms like $xy$ or $x^2{y}^3$, e.g., writing $\frac{d}{dx}[9xy]=9x\frac{dy}{dx}$ instead of $9x\frac{dy}{dx}+9y$. Why: Students remember to add the $\frac{dy}{dx}$ 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 $\frac{dy}{dx}$ 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 $\frac{dy}{dx}$ 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 $\frac{dy}{dx}$ 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.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

What is the slope of the tangent line to the curve $y^2-2x^{2}y=8$ at the point $(1,4)$? A) $\frac{4}{3}$ B) $2$ C) $\frac{8}{3}$ D) $4$

Worked Solution: First, confirm $(1,4)$ is on the curve: $4^2-2(1{)}^2(4)=16-8=8$, which matches the right-hand side. Differentiate both sides with respect to $x$, applying the product rule to the $-2x^{2}y$ term: $2y\frac{dy}{dx}-2\left(2xy+x^2\frac{dy}{dx}\right)=0$. Simplify and factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y-2x^2)=4xy$. Substitute $x=1,y=4$: $\frac{dy}{dx}(8-2)=16⟹6\frac{dy}{dx}=16⟹\frac{dy}{dx}=\frac{8}{3}$. The correct answer is C.

Question 2 (Free Response)

Consider the folium curve defined by $x^3+y^3=4xy$. (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$. (b) Find the slope of the tangent line at the point $(2,2)$. (c) Find the $x$-coordinate of the point on the curve where $y>0$ and the tangent line is horizontal.

Worked Solution: (a) Differentiate both sides with respect to $x$: $$3x^2+3y^2\frac{dy}{dx}=4\left(y+x\frac{dy}{dx}\right)$$ Rearrange and factor to isolate $\frac{dy}{dx}$: $$\frac{dy}{dx}(3y^2-4x)=4y-3x^2⟹\frac{dy}{dx}=\frac{4y-3x^2}{3y^2-4x}$$

(b) Substitute $x=2,y=2$ into the derivative: $$\frac{dy}{dx}=\frac{4(2)-3(2{)}^2}{3(2{)}^2-4(2)}=\frac{8-12}{12-8}=-1$$ The slope at $(2,2)$ is $-1$.

(c) A horizontal tangent has slope $0$, which occurs when the numerator of $\frac{dy}{dx}$ is zero (denominator non-zero). Set $4y-3x^2=0⟹y=\frac{3x^2}{4}$. Substitute into the original curve equation: $$x^3+{\left(\frac{3x^2}{4}\right)}^3=4x\left(\frac{3x^2}{4}\right)⟹x^3\left(\frac{27x^3}{64}-2\right)=0$$ Solutions are $x=0$ (which gives $y=0$, discarded because $y>0$) and $x^3=\frac{128}{27}⟹x=\frac{4\sqrt[3]{2}}{3}$. This is the required $x$-coordinate.

Question 3 (Application / Real-World Style)

The trajectory of a comet passing through the solar system can be modeled by the hyperbolic relation $16y^2-9x^2=144$, where $x$ and $y$ are both measured in astronomical units (AU), and the sun is at the origin $(0,0)$. At the point $(4,3\sqrt{2})$ on the trajectory, what is the instantaneous rate of change of $y$ with respect to $x$? Interpret your answer in context.

Worked Solution: First confirm the point is on the trajectory: $16(3\sqrt{2}{)}^2-9(4{)}^2=16(18)-9(16)=288-144=144$, which matches the equation. Differentiate both sides with respect to $x$: $$32y\frac{dy}{dx}-18x=0⟹\frac{dy}{dx}=\frac{9x}{16y}$$ Substitute $x=4,y=3\sqrt{2}$: $$\frac{dy}{dx}=\frac{9(4)}{16(3\sqrt{2})}=\frac{36}{48\sqrt{2}}=\frac{3\sqrt{2}}{8}\approx 0.53$$ Interpretation: At the position 4 AU horizontally and $3\sqrt{2}$ AU vertically from the sun, the comet's vertical position increases by approximately 0.53 AU for every 1 AU increase in its horizontal position.

7. Quick Reference Cheatsheet

8. 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 chapter, finding slopes of parametric curves and tangent lines to parametric trajectories — common FRQ 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. Concretely, next you will connect implicit definitions to parametric representations of curves, building on the local function definition you learned here.

Parametric functions, Derivatives of parametric curves, Vectors in the plane, Matrix transformations