Implicit Differentiation

AP Calculus AB· AP Calculus AB CED — Differentiation: Composite, Implicit, and Inverse Functions· 14 min read

1. What Is Implicit Differentiation?★★☆☆☆⏱ 3 min

An explicit function is written in the form $y = f (x)$ , where $y$ is explicitly isolated on one side of the equation. However, many mathematical relations (such as circles, ellipses, and more complex curves) cannot be easily or fully solved for $y$ in terms of $x$ .

Implicit differentiation is a technique to find $r a c d y d x$ directly from the original implicit relation, without rearranging to isolate $y$ . It is not a new differentiation rule—it is simply a systematic application of the chain rule to implicit functions of $x$ . Because $y$ is treated as a function of $x$ even when not written explicitly, every time we differentiate a term containing $y$ , we must multiply by $r a c d y d x$ by the chain rule.

📘 Definition

Implicit Differentiation

A technique to find the derivative $r a c d y d x$ of an implicit relation between $x$ and $y$ , without needing to isolate $y$ explicitly as a function of $x$ .

Example:

Used to find derivatives of circles, ellipses, and other non-explicit curves.

2. The Core Implicit Differentiation Process★★☆☆☆⏱ 4 min

The entire technique relies on one key chain rule result for any differentiable function $f (y)$ , where $y$ is a function of $x$ :

\frac{d}{dx}\left[f(y)\right] = f'(y) \cdot \frac{dy}{dx}

Differentiate every term on both sides of the relation with respect to $x$
Move all terms that include $r a c d y d x$ to the left side of the equation, and all other terms to the right
Factor out $r a c d y d x$ from the left side
Divide both sides by the remaining factor to isolate $r a c d y d x$ as a function of $x$ and $y$

When you have products or quotients of $x$ and $y$ terms (like $x y$ , $x^{2} y$ , or $r a c y x$ ), you still apply the product rule or quotient rule as normal, before adding the $r a c d y d x$ factor for $y$ terms.

📐 Worked Example

Find $r a c d y d x$ for the relation $x^{2} + 3 x y + y^{2} = 6$ .

1
Differentiate every term on both sides with respect to $x$ :
2
$\frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(6)$
3
Differentiate each term, applying product rule to $3 x y$ and chain rule to $y^{2}$ :
4
$2x + 3\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0$
5
Collect all $r a c d y d x$ terms on the left, and constants/terms with only $x$ on the right:
6
$3x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - 3y$
7
Factor out $r a c d y d x$ and isolate:
8
$\frac{dy}{dx}(3x + 2y) = -(2x + 3y) \implies \frac{dy}{dx} = -\frac{2x + 3y}{3x + 2y}$

Exam tip:

Always apply product/quotient rule first for mixed $x$ - $y$ terms, then add the $r a c d y d x$ factor. Skipping the product rule (e.g. omitting the $y$ term for $r a c d d x (x y)$ ) is the most common first mistake on AP exams.

3. Tangent and Normal Lines to Implicit Curves★★★☆☆⏱ 3 min

One of the most common AP exam applications of implicit differentiation is finding the equation of a tangent or normal line to a point on an implicit curve. This works exactly the same way as finding tangent lines for explicit functions, once you have the slope $r a c d y d x$ from implicit differentiation.

Confirm the given point $(x_{0}, y_{0})$ lies on the original curve (AP sometimes tests this by giving a point not on the curve)
Substitute $x_{0}$ and $y_{0}$ into your expression for $r a c d y d x$ to get the tangent slope $m_{tan}$
The slope of the normal line (perpendicular to the tangent) is the negative reciprocal: $m_{norm} = - \frac{1}{m _{tan}}$
Use point-slope form $y - y_{0} = m (x - x_{0})$ to write the final equation

📐 Worked Example

Find the equation of the tangent line to the curve $x^{2} + 2 x y + y^{3} = 4$ at the point $(1, 1)$ .

1
Confirm the point is on the curve: $1^{2} + 2 (1) (1) + 1^{3} = 4$ , which matches the right-hand side.
2
Differentiate implicitly to find $r a c d y d x$ :
3
$2x + 2\left(x \frac{dy}{dx} + y\right) + 3y^2 \frac{dy}{dx} = 0$
4
Substitute $x = 1$ , $y = 1$ to solve for the tangent slope $m$ :
5
$2(1) + 2\left(1 \cdot m + 1\right) + 3(1)^2 m = 0 \implies 2 + 2m + 2 + 3m = 0 \implies 5m = -4 \implies m = -\frac{4}{5}$
6
Write the tangent line in point-slope form and simplify:
7
$y - 1 = -\frac{4}{5}(x - 1) \implies y = -\frac{4}{5}x + \frac{9}{5}$

Exam tip:

Always read the question carefully: if it asks for a normal line, not a tangent, you must use the negative reciprocal slope. AP exam writers regularly test this to catch students who skim the question.

4. Second Derivatives of Implicit Functions★★★☆☆⏱ 3 min

AP Calculus AB regularly asks for the second derivative $r a c d^{2} y d x^{2}$ of an implicit function, in terms of $x$ and $y$ . The process is straightforward, but requires an extra step that many students forget.

After finding the first derivative $r a c d y d x$ , you differentiate $r a c d y d x$ with respect to $x$ exactly as you differentiated the original equation: all terms containing $y$ or $r a c d y d x$ still require the chain rule, so you will get a $r a c d y d x$ factor when differentiating those terms. After differentiating, you must substitute the expression you already found for $r a c d y d x$ into the second derivative, so that the final result is only in terms of $x$ and $y$ , not $r a c d y d x$ . You can also use the original curve equation to simplify the final result by canceling constant terms.

📐 Worked Example

Find $r a c d^{2} y d x^{2}$ for the circle $x^{2} + y^{2} = 4$ , in terms of $x$ and $y$ .

1
Find the first derivative $r a c d y d x$ :
2
$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$
3
Differentiate $r a c d y d x$ with respect to $x$ using the quotient rule:
4
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{(1 \cdot y - x \cdot \frac{dy}{dx})}{y^2}$
5
Substitute $rac{dy}{dx} = - rac{x}{y}$ into the expression:
6
$\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{\frac{y^2 + x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}$
7
Simplify using the original equation $x^{2} + y^{2} = 4$ :
8
$\frac{d^2y}{dx^2} = -\frac{4}{y^3}$

Exam tip:

Never leave $r a c d y d x$ in your final answer for the second derivative. AP graders will deduct points for unsubstituted $r a c d y d x$ terms on FRQ. Always substitute immediately after differentiating the first derivative.

5. AP Style Practice Problems★★★★☆⏱ 3 min

✓ Quick check

Test your understanding with this multiple-choice problem:

Given $ sin (x y) = x + y$ , which of the following is the correct expression for $ \frac{d y}{d x}$ ?
- $\frac{1}{c o s ( x y ) - 1}$
- $\frac{1 - y c o s ( x y )}{x c o s ( x y ) - 1}$
- $\frac{y c o s ( x y ) - 1}{1 - x c o s ( x y )}$
- $1 - y cos (x y)$
Reveal answer
1 —
Correct! The full derivation differentiates both sides, applies product and chain rule, collects $ \frac{d y}{d x}$ terms, and isolates to get this result.

📐 Worked Example

Consider the curve defined by $y^{3} - x y + 2 x^{2} = 4$ . (a) Find $ \frac{d y}{d x}$ in terms of $x$ and $y$ . (b) Find the slope of the tangent line to the curve at the point $(1, 1)$ . (c) Find all points on the curve where the tangent line is horizontal.

1
Part (a): Differentiate both sides with respect to $x$ :
2
$3y^2 \frac{dy}{dx} - \left(x \frac{dy}{dx} + y\right) + 4x = 0$
3
Collect and factor $ \frac{d y}{d x}$ :
4
$\frac{dy}{dx} (3y^2 - x) = y - 4x \implies \frac{dy}{dx} = \frac{y - 4x}{3y^2 - x}$
5
Part (b): Substitute $x = 1$ , $y = 1$ into $ \frac{d y}{d x}$ :
6
$m = \frac{1 - 4(1)}{3(1)^2 - 1} = \frac{-3}{2}$
7
The slope of the tangent at $(1, 1)$ is $ - \frac{3}{2}$ .
8
Part (c): A horizontal tangent has slope $0$ , so set the numerator equal to $0$ and confirm denominator is non-zero: $y - 4 x = 0 ⟹ y = 4 x$ . Substitute into the original equation:
9
$(4x)^3 - x(4x) + 2x^2 = 4 \implies 64x^3 - 2x^2 - 4 = 0 \implies 32x^3 - x^2 - 2 = 0$
10
The only real solution is $x \approx 0.42$ , so $y = 4 x \approx 1.68$ . The only point with a horizontal tangent is approximately $(0.42, 1.68)$ .

6. Common Pitfalls

Wrong move:

When differentiating $x^{2} y$ , writing $2 x \frac{d y}{d x}$ and omitting the product rule term

Why:

Students focus so much on remembering the chain rule for $y$ -terms that they forget mixed $x$ - $y$ products require the product rule first

Correct move:

Always apply product/quotient rule to mixed terms first, then add the $ \frac{d y}{d x}$ factor for $y$ -terms: $ \frac{d}{d x} (x^{2} y) = 2 x y + x^{2} \frac{d y}{d x}$

Wrong move:

When differentiating $ cos (y)$ , writing $ - sin (y)$ and omitting the $ \frac{d y}{d x}$ factor

Why:

Students get accustomed to differentiating $x$ -terms and forget that every function of $y$ needs the chain rule factor

Correct move:

After differentiating any function of $y$ , immediately write the $ \frac{d y}{d x}$ factor as a routine step before moving to the next term

Wrong move:

When finding slope at a point, plugging in $x_{0}, y_{0}$ before isolating $ \frac{d y}{d x}$

Why:

Students think plugging in early simplifies the problem, but it leads to lost terms and messy algebra errors

Correct move:

Isolate $ \frac{d y}{d x}$ as a function of $x$ and $y$ first, then substitute the point values to get the numerical slope

Wrong move:

When finding horizontal tangents, setting the denominator of $ \frac{d y}{d x}$ equal to zero instead of the numerator

Why:

Students confuse horizontal and vertical tangent conditions

Correct move:

Slope zero means $ \frac{d y}{d x} = 0$ , which requires the numerator to be zero (and the denominator non-zero) for a rational $ \frac{d y}{d x}$

Wrong move:

When given only an $x$ -coordinate for a tangent point, picking any $y$ -value that solves the original equation

Why:

Implicit relations have multiple $y$ -values for one $x$ , and the wrong $y$ gives the wrong slope

Correct move:

Always use the given context (quadrant, position on the curve) to select the correct $y$ -value before calculating slope

7. Quick Reference Cheatsheet

Category	Formula / Process	Notes
Chain rule for $y$ -terms	$\frac{d}{d x} [f (y)] = f^{'} (y) \frac{d y}{d x}$	Applies to any function of $y$ , since $y$ is an implicit function of $x$
Core implicit differentiation	Differentiate both sides w.r.t $x$ ; 2. Collect $\frac{d y}{d x}$ terms left; 3. Factor $\frac{d y}{d x}$ ; 4. Isolate	No need to solve for $y$ first
Tangent line slope at $(x_{0}, y_{0})$	$m_{tan} = \frac{d y}{d x}_{(x_{0}, y_{0})}$	Always confirm $(x_{0}, y_{0})$ is on the original curve first
Normal line slope	$m_{norm} = - \frac{1}{m _{tan}}$	Undefined if tangent is horizontal, horizontal if tangent is vertical
Implicit second derivative	$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$	Substitute $\frac{d y}{d x}$ to get final answer in terms of $x$ and $y$ only
Horizontal tangent	$\frac{d y}{d x} = 0 ⟹ numerator = 0, denominator \neq = 0$	Only valid if denominator is non-zero at the point
Vertical tangent	$\frac{d y}{d x}$ undefined $⟹ denominator = 0, numerator \neq = 0$	Common AP follow-up question

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 dy/dx for implicit relation
2022 · FRQ
Tangent line to implicit curve

What's Next

Implicit differentiation is the foundational prerequisite for the remaining core topics in AP Calculus AB Unit 3: derivatives of inverse functions and related rates. Without mastering the chain rule application to implicit $y$ -terms, you will not be able to correctly derive derivative formulas for inverse trigonometric functions or solve related rate problems, which make up a large share of AP Calculus AB FRQ points. This topic also builds the critical conceptual shift from derivatives of functions to derivatives of relations, a key foundation for advanced calculus topics.