Unit 3Second Derivatives of Implicit Functions

Concept 4.98Built for May 2027 AP Exam

Second Derivatives of Implicit Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The second derivative of an implicit curve is built by differentiating the first derivative again. Take $\frac{dy}{dx}$ , differentiate it with respect to $x$ , and treat $y$ as a function one more time, which means the quotient rule and the chain rule both fire on the same line. The result still contains $\frac{dy}{dx}$ , and that expression has to be substituted back before the answer counts as finished. What remains is allowed to contain both $x$ and $y$ . A second derivative left in terms of $\frac{dy}{dx}$ is incomplete, and the rubric says so explicitly.

On the AP Exam

This is a two-stage computation, and the exam is careful about which stage it pays for. In nine released exams no free-response part asks for $\frac{d^2y}{dx^2}$ from an implicit equation, so the skill lives in multiple choice and then hides inside one justification: deciding whether a horizontal tangent is a maximum or a minimum without leaving the curve.

What you’ll see: An implicit equation followed by a request for $\frac{d^2y}{dx^2}$ , either as a general expression or as a value at a named point.
What you’ll do: Find $\frac{dy}{dx}$ , differentiate it again with the quotient rule, then substitute $\frac{dy}{dx}$ and simplify to $x$ and $y$ .
Memory hook: Differentiate, substitute, simplify. The middle word is the one students skip.
Use it on the exam: For MCQs: Any option that still contains $\frac{dy}{dx}$ is wrong by construction, so that single check eliminates most distractors. For FRQs: No scored part in the released sets asks for $\frac{d^2y}{dx^2}$ outright. What the skill buys you is a one-line justification: evaluated at the point where $\frac{dy}{dx}=0$ , the sign of $\frac{d^2y}{dx^2}$ names the extremum without any table of nearby values.

Concept Review

The Second Pass

Figure 1 — The second derivative comes from differentiating the first one and then replacing it. The substitution is not cosmetic; it is what turns a two-variable expression into an answer.

DIFFERENTIATE THE DERIVATIVE

Two Curves, Worked Once

Equation	$\frac{dy}{dx}$	$\frac{d^2y}{dx^2}$
$x^2 + y^2 = 25$	$-\dfrac{x}{y}$	$-\dfrac{25}{y^3}$
$xy = 6$	$-\dfrac{y}{x}$	$\dfrac{12}{x^3}$
$y^2 = x$	$\dfrac{1}{2y}$	$-\dfrac{1}{4y^3}$

For the circle, the quotient rule gives $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$ , and substituting $\frac{dy}{dx} = -\frac{x}{y}$ produces $-\frac{y^2 + x^2}{y^3}$ , which collapses to $-\frac{25}{y^3}$ because the original equation says $x^2 + y^2 = 25$ .

Why the Substitution Is Not Optional

A second derivative answers a question about concavity at a point on the curve. An expression containing $\frac{dy}{dx}$ cannot be evaluated from the coordinates alone, because the slope would also have to be supplied. Substituting removes that dependence and leaves something the point alone can evaluate.

Common Misconception Check

Myth 1: "The second derivative is the derivative of $\frac{dy}{dx}$ with $y$ held constant." Truth: $y$ is still a function of $x$ , so every $y$ -term picks up another tag during the second pass.
Myth 2: "Substituting $\frac{dy}{dx}$ early is fine as long as the answer works out." Truth: Early substitution usually turns a differentiable expression into a number and makes the remaining differentiation meaningless.
Myth 3: "A second derivative in terms of $x$ and $y$ is not simplified enough." Truth: That form is the expected one, and the original equation may be used to collapse it further when the algebra allows.

Topic-Specific Study Method

Differentiate Twice, Substitute Once

Borrowed idea: What happens when the derivative you just found is differentiated a second time? The second derivative of an implicit curve is built exactly that way. Take $\frac{dy}{dx}$ , differentiate it with respect to $x$ , and treat $y$ as a function once more. The result still contains $\frac{dy}{dx}$ , which has to be replaced before the answer is finished. Why does that substitution matter so much? Because a second derivative written in terms of the first is not accepted as an answer.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general request	"Find $\frac{d^2y}{dx^2}$ "	Two passes, then substitution	Differentiate, substitute, simplify
A value at a point	"the concavity at $(3, 4)$ "	Substitute the point only at the end	Keep the expression symbolic until then
A circle	$x^2 + y^2 = r^2$	The original equation collapses the answer	Use $x^2 + y^2 = r^2$ in the final step
A product curve	$xy = k$	The second pass is pure quotient rule	Expect $x$ only in the answer

① Take the circle and compute $\frac{d^2y}{dx^2}$ twice: once substituting $\frac{dy}{dx}$ and once leaving it in place. Compare the two results and notice that only the first one can be evaluated from a point.

② Write out the quotient rule on $\frac{dy}{dx} = -\frac{x}{y}$ slowly, tagging the $y$ in the denominator. The denominator is where the second tag is usually forgotten.

③ On any concavity question, write the three words "differentiate, substitute, simplify" above your work before starting. The middle step is the one that gets dropped under time pressure.

Training & Error Diagnosis

Core Archetype (Simplest Form): Given $x^2 + y^2 = 25$ , find $\frac{d^2y}{dx^2}$ .

Standard Execution:

First pass. $\frac{dy}{dx} = -\frac{x}{y}$ .

Set up the second pass. Treat $-\frac{x}{y}$ as a quotient and apply the quotient rule.

Tag the denominator. $\frac{d}{dx}[y] = \frac{dy}{dx}$ , which is where the second tag belongs.

Substitute. Replace $\frac{dy}{dx}$ with $-\frac{x}{y}$ and simplify using $x^2 + y^2 = 25$ .

Finish. The result is $-\frac{25}{y^3}$ .

Exam Variation (Context / Parameter Shift): The curve $xy = 6$ passes through $(2, 3)$ . Find $\frac{d^2y}{dx^2}$ at that point.

Solution: The first pass gives $\frac{dy}{dx} = -\frac{y}{x}$ . Differentiating again gives $\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2}$ , and substituting $\frac{dy}{dx} = -\frac{y}{x}$ produces $\frac{2y}{x^2}$ . At $(2, 3)$ this is $\frac{6}{4} = \frac{3}{2}$ , which is positive, so the curve is concave up there.

Common Traps & Failure Signals:

The surviving $\frac{dy}{dx}$ (Signal: an answer that still names the first derivative): the substitution step never happened.
The constant $y$ (Signal: a second derivative that looks like $-\frac{1}{y}$ ): $y$ was treated as a constant during the second pass.
The early number (Signal: a plain value where an expression belongs): the point went in before the second differentiation was complete.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Skipped Substitution $\rightarrow$ Check the answer for $\frac{dy}{dx}$ before moving on.
Constant $y$ Again: The old habit of treating $y$ as a number returns during the second pass, precisely because the first pass already felt like the hard part and attention drops once the result looks familiar.
Premature Point: You substituted the coordinates before the expression was finished.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part asks for a second derivative of an implicit relation; the one extremum question that comes close is scored on the sign of the first derivative
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Circle

If $x^2 + y^2 = 25$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 2

A Hyperbola

Given $xy = 6$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 3

Reading the Answer

A student obtains $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$ for a curve. Why is this not a finished answer?

FRQ 4

Concavity on an Implicit Curve

Consider the curve $x^2 + xy + y^2 = 7$ , which passes through $(1, 2)$ .

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ .

(b) Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ .

(c) Evaluate $\frac{d^2y}{dx^2}$ at $(1, 2)$ and state whether the curve is concave up or concave down there.

(d) A classmate substitutes $x = 1$ and $y = 2$ into $\frac{dy}{dx}$ before differentiating again. Explain why that shortcut cannot produce $\frac{d^2y}{dx^2}$ .

Cheatsheet

Second Implicit Derivatives

	What to write
First derivative	Tag every $y$ , isolate $\frac{dy}{dx}$
Second pass	Quotient rule on $\frac{dy}{dx}$ , tag every $y$ again
Substitution	Replace $\frac{dy}{dx}$ with its expression
Simplification	Use the original equation to collapse terms
Final form	$x$ and $y$ only, or a number at a point

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Unit 3Second Derivatives of Implicit Functions

Concept 4.98Built for May 2027 AP Exam

Second Derivatives of Implicit Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: An implicit equation followed by a request for $\frac{d^2y}{dx^2}$ , either as a general expression or as a value at a named point.
What you’ll do: Find $\frac{dy}{dx}$ , differentiate it again with the quotient rule, then substitute $\frac{dy}{dx}$ and simplify to $x$ and $y$ .
Memory hook: Differentiate, substitute, simplify. The middle word is the one students skip.
Use it on the exam: For MCQs: Any option that still contains $\frac{dy}{dx}$ is wrong by construction, so that single check eliminates most distractors. For FRQs: No scored part in the released sets asks for $\frac{d^2y}{dx^2}$ outright. What the skill buys you is a one-line justification: evaluated at the point where $\frac{dy}{dx}=0$ , the sign of $\frac{d^2y}{dx^2}$ names the extremum without any table of nearby values.

Concept Review

The Second Pass

Figure 1 — The second derivative comes from differentiating the first one and then replacing it. The substitution is not cosmetic; it is what turns a two-variable expression into an answer.

DIFFERENTIATE THE DERIVATIVE

Two Curves, Worked Once

Equation	$\frac{dy}{dx}$	$\frac{d^2y}{dx^2}$
$x^2 + y^2 = 25$	$-\dfrac{x}{y}$	$-\dfrac{25}{y^3}$
$xy = 6$	$-\dfrac{y}{x}$	$\dfrac{12}{x^3}$
$y^2 = x$	$\dfrac{1}{2y}$	$-\dfrac{1}{4y^3}$

Why the Substitution Is Not Optional

Common Misconception Check

Myth 1: "The second derivative is the derivative of $\frac{dy}{dx}$ with $y$ held constant." Truth: $y$ is still a function of $x$ , so every $y$ -term picks up another tag during the second pass.
Myth 2: "Substituting $\frac{dy}{dx}$ early is fine as long as the answer works out." Truth: Early substitution usually turns a differentiable expression into a number and makes the remaining differentiation meaningless.
Myth 3: "A second derivative in terms of $x$ and $y$ is not simplified enough." Truth: That form is the expected one, and the original equation may be used to collapse it further when the algebra allows.

Topic-Specific Study Method

Differentiate Twice, Substitute Once

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general request	"Find $\frac{d^2y}{dx^2}$ "	Two passes, then substitution	Differentiate, substitute, simplify
A value at a point	"the concavity at $(3, 4)$ "	Substitute the point only at the end	Keep the expression symbolic until then
A circle	$x^2 + y^2 = r^2$	The original equation collapses the answer	Use $x^2 + y^2 = r^2$ in the final step
A product curve	$xy = k$	The second pass is pure quotient rule	Expect $x$ only in the answer

② Write out the quotient rule on $\frac{dy}{dx} = -\frac{x}{y}$ slowly, tagging the $y$ in the denominator. The denominator is where the second tag is usually forgotten.

③ On any concavity question, write the three words "differentiate, substitute, simplify" above your work before starting. The middle step is the one that gets dropped under time pressure.

Training & Error Diagnosis

Core Archetype (Simplest Form): Given $x^2 + y^2 = 25$ , find $\frac{d^2y}{dx^2}$ .

Standard Execution:

First pass. $\frac{dy}{dx} = -\frac{x}{y}$ .

Set up the second pass. Treat $-\frac{x}{y}$ as a quotient and apply the quotient rule.

Tag the denominator. $\frac{d}{dx}[y] = \frac{dy}{dx}$ , which is where the second tag belongs.

Substitute. Replace $\frac{dy}{dx}$ with $-\frac{x}{y}$ and simplify using $x^2 + y^2 = 25$ .

Finish. The result is $-\frac{25}{y^3}$ .

Exam Variation (Context / Parameter Shift): The curve $xy = 6$ passes through $(2, 3)$ . Find $\frac{d^2y}{dx^2}$ at that point.

Solution: The first pass gives $\frac{dy}{dx} = -\frac{y}{x}$ . Differentiating again gives $\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2}$ , and substituting $\frac{dy}{dx} = -\frac{y}{x}$ produces $\frac{2y}{x^2}$ . At $(2, 3)$ this is $\frac{6}{4} = \frac{3}{2}$ , which is positive, so the curve is concave up there.

Common Traps & Failure Signals:

The surviving $\frac{dy}{dx}$ (Signal: an answer that still names the first derivative): the substitution step never happened.
The constant $y$ (Signal: a second derivative that looks like $-\frac{1}{y}$ ): $y$ was treated as a constant during the second pass.
The early number (Signal: a plain value where an expression belongs): the point went in before the second differentiation was complete.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Skipped Substitution $\rightarrow$ Check the answer for $\frac{dy}{dx}$ before moving on.
Constant $y$ Again: The old habit of treating $y$ as a number returns during the second pass, precisely because the first pass already felt like the hard part and attention drops once the result looks familiar.
Premature Point: You substituted the coordinates before the expression was finished.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part asks for a second derivative of an implicit relation; the one extremum question that comes close is scored on the sign of the first derivative
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Circle

If $x^2 + y^2 = 25$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 2

A Hyperbola

Given $xy = 6$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 3

Reading the Answer

A student obtains $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$ for a curve. Why is this not a finished answer?

FRQ 4

Concavity on an Implicit Curve

Consider the curve $x^2 + xy + y^2 = 7$ , which passes through $(1, 2)$ .

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ .

(b) Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ .

(c) Evaluate $\frac{d^2y}{dx^2}$ at $(1, 2)$ and state whether the curve is concave up or concave down there.

(d) A classmate substitutes $x = 1$ and $y = 2$ into $\frac{dy}{dx}$ before differentiating again. Explain why that shortcut cannot produce $\frac{d^2y}{dx^2}$ .

Cheatsheet

Second Implicit Derivatives

	What to write
First derivative	Tag every $y$ , isolate $\frac{dy}{dx}$
Second pass	Quotient rule on $\frac{dy}{dx}$ , tag every $y$ again
Substitution	Replace $\frac{dy}{dx}$ with its expression
Simplification	Use the original equation to collapse terms
Final form	$x$ and $y$ only, or a number at a point

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Unit 3Second Derivatives of Implicit Functions

Concept 4.98Built for May 2027 AP Exam

Second Derivatives of Implicit Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: An implicit equation followed by a request for $\frac{d^2y}{dx^2}$ , either as a general expression or as a value at a named point.
What you’ll do: Find $\frac{dy}{dx}$ , differentiate it again with the quotient rule, then substitute $\frac{dy}{dx}$ and simplify to $x$ and $y$ .
Memory hook: Differentiate, substitute, simplify. The middle word is the one students skip.
Use it on the exam: For MCQs: Any option that still contains $\frac{dy}{dx}$ is wrong by construction, so that single check eliminates most distractors. For FRQs: No scored part in the released sets asks for $\frac{d^2y}{dx^2}$ outright. What the skill buys you is a one-line justification: evaluated at the point where $\frac{dy}{dx}=0$ , the sign of $\frac{d^2y}{dx^2}$ names the extremum without any table of nearby values.

Concept Review

The Second Pass

Figure 1 — The second derivative comes from differentiating the first one and then replacing it. The substitution is not cosmetic; it is what turns a two-variable expression into an answer.

DIFFERENTIATE THE DERIVATIVE

Two Curves, Worked Once

Equation	$\frac{dy}{dx}$	$\frac{d^2y}{dx^2}$
$x^2 + y^2 = 25$	$-\dfrac{x}{y}$	$-\dfrac{25}{y^3}$
$xy = 6$	$-\dfrac{y}{x}$	$\dfrac{12}{x^3}$
$y^2 = x$	$\dfrac{1}{2y}$	$-\dfrac{1}{4y^3}$

Why the Substitution Is Not Optional

Common Misconception Check

Myth 1: "The second derivative is the derivative of $\frac{dy}{dx}$ with $y$ held constant." Truth: $y$ is still a function of $x$ , so every $y$ -term picks up another tag during the second pass.
Myth 2: "Substituting $\frac{dy}{dx}$ early is fine as long as the answer works out." Truth: Early substitution usually turns a differentiable expression into a number and makes the remaining differentiation meaningless.
Myth 3: "A second derivative in terms of $x$ and $y$ is not simplified enough." Truth: That form is the expected one, and the original equation may be used to collapse it further when the algebra allows.

Topic-Specific Study Method

Differentiate Twice, Substitute Once

Exam disguise	What you see on the page	What your brain must translate	Required execution
A general request	"Find $\frac{d^2y}{dx^2}$ "	Two passes, then substitution	Differentiate, substitute, simplify
A value at a point	"the concavity at $(3, 4)$ "	Substitute the point only at the end	Keep the expression symbolic until then
A circle	$x^2 + y^2 = r^2$	The original equation collapses the answer	Use $x^2 + y^2 = r^2$ in the final step
A product curve	$xy = k$	The second pass is pure quotient rule	Expect $x$ only in the answer

② Write out the quotient rule on $\frac{dy}{dx} = -\frac{x}{y}$ slowly, tagging the $y$ in the denominator. The denominator is where the second tag is usually forgotten.

③ On any concavity question, write the three words "differentiate, substitute, simplify" above your work before starting. The middle step is the one that gets dropped under time pressure.

Training & Error Diagnosis

Core Archetype (Simplest Form): Given $x^2 + y^2 = 25$ , find $\frac{d^2y}{dx^2}$ .

Standard Execution:

First pass. $\frac{dy}{dx} = -\frac{x}{y}$ .

Set up the second pass. Treat $-\frac{x}{y}$ as a quotient and apply the quotient rule.

Tag the denominator. $\frac{d}{dx}[y] = \frac{dy}{dx}$ , which is where the second tag belongs.

Substitute. Replace $\frac{dy}{dx}$ with $-\frac{x}{y}$ and simplify using $x^2 + y^2 = 25$ .

Finish. The result is $-\frac{25}{y^3}$ .

Exam Variation (Context / Parameter Shift): The curve $xy = 6$ passes through $(2, 3)$ . Find $\frac{d^2y}{dx^2}$ at that point.

Solution: The first pass gives $\frac{dy}{dx} = -\frac{y}{x}$ . Differentiating again gives $\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2}$ , and substituting $\frac{dy}{dx} = -\frac{y}{x}$ produces $\frac{2y}{x^2}$ . At $(2, 3)$ this is $\frac{6}{4} = \frac{3}{2}$ , which is positive, so the curve is concave up there.

Common Traps & Failure Signals:

The surviving $\frac{dy}{dx}$ (Signal: an answer that still names the first derivative): the substitution step never happened.
The constant $y$ (Signal: a second derivative that looks like $-\frac{1}{y}$ ): $y$ was treated as a constant during the second pass.
The early number (Signal: a plain value where an expression belongs): the point went in before the second differentiation was complete.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Skipped Substitution $\rightarrow$ Check the answer for $\frac{dy}{dx}$ before moving on.
Constant $y$ Again: The old habit of treating $y$ as a number returns during the second pass, precisely because the first pass already felt like the hard part and attention drops once the result looks familiar.
Premature Point: You substituted the coordinates before the expression was finished.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part asks for a second derivative of an implicit relation; the one extremum question that comes close is scored on the sign of the first derivative
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; All 54 free-response questions from the nine released sets were read part by part
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Circle

If $x^2 + y^2 = 25$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 2

A Hyperbola

Given $xy = 6$ , what is $\frac{d^2y}{dx^2}$ ?

MCQ 3

Reading the Answer

A student obtains $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$ for a curve. Why is this not a finished answer?

FRQ 4

Concavity on an Implicit Curve

Consider the curve $x^2 + xy + y^2 = 7$ , which passes through $(1, 2)$ .

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ .

(b) Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ .

(c) Evaluate $\frac{d^2y}{dx^2}$ at $(1, 2)$ and state whether the curve is concave up or concave down there.

(d) A classmate substitutes $x = 1$ and $y = 2$ into $\frac{dy}{dx}$ before differentiating again. Explain why that shortcut cannot produce $\frac{d^2y}{dx^2}$ .

Cheatsheet

Second Implicit Derivatives

	What to write
First derivative	Tag every $y$ , isolate $\frac{dy}{dx}$
Second pass	Quotient rule on $\frac{dy}{dx}$ , tag every $y$ again
Substitution	Replace $\frac{dy}{dx}$ with its expression
Simplification	Use the original equation to collapse terms
Final form	$x$ and $y$ only, or a number at a point

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