Why does d2ydx2 have dydx in it?

Because y is a function of x. Differentiating any term with y in it brings out a factor of dydx, and differentiating dydx itself gives d2ydx2.

Unit 5 · Topic 5.12 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.12: Exploring Behaviors of Implicit Relations

Q: Do I have to solve the equation for y before I can use the derivative tests?

No. Implicit differentiation gives dydx and d2ydx2 directly, and the tests use only their values and signs near the point. Solving for y is often impossible anyway.

Q: Is a point with a vertical tangent a maximum or a minimum?

It is a critical point of the relation, because dydx does not exist there. Near a vertical tangent the curve is usually not the graph of a function of x, so the maximum and minimum tests for a function of x do not apply. On x2 + xy + y2 = 3, the vertical tangents at (2, -1) and (-2, 1) are the curve's farthest points right and left.

Q: Which test should I use, the first derivative test or the second?

At a point where dydx = 0, the second derivative test is usually quicker, because the terms with dydx drop out. If d2ydx2 is 0 there, that test says nothing, and the first derivative test, the sign of dydx on each side, decides.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 5: 15 to 20% of the multiple-choice section

What you'll be able to do

Find the critical points of a curve given by an equation in $x$ and $y$ , the points where $\frac{d y}{d x}$ is zero or does not exist.
Justify conclusions about a function defined implicitly, such as where it increases, how it bends and where it has a relative maximum or minimum, from its first and second derivatives.

Key vocabulary

Term	What it means	The exam trap
Implicit relation	An equation in $x$ and $y$ , such as $x^{2} + x y + y^{2} = 3$ , that is not solved for $y$ .	One $x$ -value can go with two or more $y$ -values, so name each point by both coordinates.
Critical point of an implicit relation	A point on the relation where the first derivative $\frac{d y}{d x}$ equals zero or does not exist.	Put each candidate point back into the original equation before you call it a critical point.
Implicitly defined function	The function of $x$ whose graph is the part of the curve near a given point, used without solving the equation for $y$ .	Near a vertical tangent the curve is usually not the graph of a function of $x$ , so say which point you are working near.
Second derivative $\frac{d^{2} y}{d x^{2}}$ of an implicit relation	The derivative of $\frac{d y}{d x}$ , found by differentiating implicitly again; it may be a relation of $x$ , $y$ and $\frac{d y}{d x}$ .	It needs the point's $x$ , its $y$ and the value of $\frac{d y}{d x}$ there, not $x$ alone.

Graph of the curve x squared plus xy plus y squared equals 3, a tilted oval. Short horizontal tangent segments mark the points (-1, 2) and (1, -2), where dy/dx = 0, and short vertical tangent segments mark (2, -1) and (-2, 1), where dy/dx does not exist.

Part 1 Think first The north end of a tilted flower bed

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A gardener plans an oval flower bed. On her plan, $x$ is meters east of the center and $y$ is meters north, and the edge of the bed is $x^{2} - x y + y^{2} = 12$ . Where is the edge farthest north?

In short. A critical point of an implicit relation is a point on the curve where $\frac{d y}{d x}$ is 0 or does not exist, and the derivative tests then describe the function the curve defines near that point.

Find $\frac{d y}{d x}$ by implicit differentiation, set its numerator or its denominator equal to 0, and keep only the points that satisfy the original equation.
Whether $\frac{d y}{d x}$ changes sign at a critical point decides between a relative maximum, a relative minimum and neither. Where $\frac{d y}{d x} = 0$ , a positive or negative $\frac{d^{2} y}{d x^{2}}$ decides between the first two, and a zero one decides nothing.
A second derivative found implicitly is usually a relation of $x$ , $y$ and $\frac{d y}{d x}$ , so all three values go into it.

Part 2 Learn it Three short ideas, each checked as you go

5a. Critical points of a curve given implicitly

Earlier in this unit a critical point of $f$ was an input where $f^{'}$ is 0 or does not exist. A curve given by an equation in $x$ and $y$ , an implicit relation, works the same way with $\frac{d y}{d x}$ in place of $f^{'}$ : a point on the relation where $\frac{d y}{d x}$ equals zero or does not exist is a critical point.

Take the curve at the top, $x^{2} + x y + y^{2} = 3$ . Differentiating, $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ , so

\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y} .

$\frac{d y}{d x} = 0$ where the numerator is 0 and the denominator is not: $y = - 2 x$ . In the curve's equation that gives $x^{2} - 2 x^{2} + 4 x^{2} = 3$ , so $x = 1$ or $x = - 1$ , and the points are $(1, - 2)$ and $(- 1, 2)$ . The tangent is horizontal there.
$\frac{d y}{d x}$ does not exist where the denominator is 0: $x = - 2 y$ , which gives $(2, - 1)$ and $(- 2, 1)$ . The tangent is vertical there.

Two habits matter. A point counts only if it is on the curve, so put every candidate back into the original equation. And $\frac{d y}{d x}$ usually contains $y$ , so a critical point is named by both coordinates, not by $x$ alone.

Quick check · Sort it

No calculator. For the circle $(x - 1)^{2} + (y + 2)^{2} = 25$ , implicit differentiation gives $\frac{d y}{d x} = \frac{1 - x}{y + 2}$ . Tap a point, then tap where it belongs.

Critical point, $\frac{d y}{d x} = 0$

Critical point, $\frac{d y}{d x}$ does not exist

Not a critical point

5b. The derivative tests work on implicit functions too

Near most points, a curve given implicitly is the graph of a function of $x$ , even if you never solve for $y$ . Everything this unit taught about $f^{'}$ carries over to $\frac{d y}{d x}$ . Where $\frac{d y}{d x} > 0$ that function is increasing, where $\frac{d y}{d x} < 0$ it is decreasing, and the first derivative test classifies a critical point by whether $\frac{d y}{d x}$ changes sign there.

Go back to $x^{2} + x y + y^{2} = 3$ near $(- 1, 2)$ . The denominator $x + 2 y$ is close to 3 there, so it is positive. The numerator $2 x + y$ is negative just to the left of $x = - 1$ on the curve and positive just to the right. With the minus sign in front,

\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y}

changes from positive to negative at $x = - 1$ . The function the curve defines near $(- 1, 2)$ has a relative maximum there, and the picture agrees: $(- 1, 2)$ is the top of the curve.

Read the sign of the whole fraction, numerator and denominator together. On some curves the denominator is negative, and then $\frac{d y}{d x}$ has the opposite sign to its numerator.

Quick check · Decide and justify

No calculator. The relation $3 y^{2} = 2 (x - 2)^{3} + 27$ defines $y$ as a function of $x$ near the point $(2, 3)$ , with $\frac{d y}{d x} = \frac{(x - 2)^{2}}{y}$ . At $(2, 3)$ , $\frac{d y}{d x} = 0$ . Does this function have a relative minimum, a relative maximum or neither at $x = 2$ ?

Which reason supports your answer?

5c. The second derivative: x, y and dy/dx together

To find $\frac{d^{2} y}{d x^{2}}$ , differentiate again. Every $y$ is a function of $x$ , so differentiating a term with $y$ in it brings out another $\frac{d y}{d x}$ . That is why a second derivative found implicitly is usually a relation of $x$ , $y$ and $\frac{d y}{d x}$ .

Start from $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ and differentiate each term. The product $x \frac{d y}{d x}$ gives $\frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}}$ , and the product $2 y \frac{d y}{d x}$ gives $2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}}$ . Collected,

2 + 2 \frac{d y}{d x} + 2 {(\frac{d y}{d x})}^{2} + (x + 2 y) \frac{d^{2} y}{d x^{2}} = 0.

At $(- 1, 2)$ , $\frac{d y}{d x} = 0$ and $x + 2 y = 3$ , so $2 + 3 \frac{d^{2} y}{d x^{2}} = 0$ and $\frac{d^{2} y}{d x^{2}} = - \frac{2}{3}$ . It is negative, so by the second derivative test the function has a relative maximum at $x = - 1$ , which agrees with the sign change you found in the first derivative.

At a critical point with $\frac{d y}{d x} = 0$ , the terms with $\frac{d y}{d x}$ drop out, which makes the arithmetic short. At any other point they stay, so keep every one of them in your work until you substitute.

Quick check · Multiple choice

No calculator. For the curve $y^{2} = x^{3} + 4$ , implicit differentiation gives $2 y \frac{d y}{d x} = 3 x^{2}$ . Which expression is $\frac{d^{2} y}{d x^{2}}$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Setting the numerator of $\frac{d y}{d x}$ equal to 0 and reporting the solutions without checking them in the curve's equation.	A solution of the numerator's equation alone may not be on the curve at all. Only points of the relation can be its critical points.	Substitute each candidate into the original equation, and keep only the points that satisfy it.
Dropping a factor of $\frac{d y}{d x}$ when differentiating a $y$ term a second time, as in writing $6 y$ for the derivative of $3 y^{2}$ .	$3 y^{2}$ is a function of $x$ through $y$ , so the chain rule gives $6 y \frac{d y}{d x}$ .	Write $\frac{d y}{d x}$ after the derivative of every expression in $y$ , then substitute $x$ , $y$ and $\frac{d y}{d x}$ last.
Reading the sign of $\frac{d y}{d x}$ from its numerator alone.	A negative denominator flips the sign of the fraction.	Find the sign of the numerator and of the denominator near the point, then combine them.

Part 3 Watch it done Find and classify a critical point

Watch each step, then do the last one yourself.

The relation $3 x^{2} + y^{3} - 3 y = 18$ defines $y$ as a function of $x$ near the point $(0, 3)$ . Show that the function has a critical point at $x = 0$ , and use the second derivative to decide whether it is a relative maximum or a relative minimum.

Step 1 · Differentiate implicitly
$6 x + 3 y^{2} \frac{d y}{d x} - 3 \frac{d y}{d x} = 0$ , so $(3 y^{2} - 3) \frac{d y}{d x} = - 6 x$ and $\frac{d y}{d x} = \frac{- 6 x}{3 y^{2} - 3}$ .
Step 2 · Check the critical point
$(0, 3)$ is on the curve: $0 + 27 - 9 = 18$ . There $\frac{d y}{d x} = \frac{0}{27 - 3} = 0$ , and the denominator 24 is not 0, so $(0, 3)$ is a critical point with a horizontal tangent.
Step 3 · Differentiate again
Differentiate $6 x + 3 y^{2} \frac{d y}{d x} - 3 \frac{d y}{d x} = 0$ term by term: $6 + 6 y {(\frac{d y}{d x})}^{2} + 3 y^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d^{2} y}{d x^{2}} = 0$ , that is, $6 + 6 y {(\frac{d y}{d x})}^{2} + (3 y^{2} - 3) \frac{d^{2} y}{d x^{2}} = 0$ .
Why this step?
Why does $3 y^{2} \frac{d y}{d x}$ turn into two terms, one of them with ${(\frac{d y}{d x})}^{2}$ ?
It is a product of $3 y^{2}$ and $\frac{d y}{d x}$ , so the product rule gives two terms. The derivative of $3 y^{2}$ needs the chain rule, $6 y \frac{d y}{d x}$ , and multiplied by $\frac{d y}{d x}$ that is $6 y {(\frac{d y}{d x})}^{2}$ . The other term is $3 y^{2} \frac{d^{2} y}{d x^{2}}$ .
Your step · Calculate

No calculator. Finish it. At $(0, 3)$ , $\frac{d y}{d x} = 0$ and $6 + 6 y {(\frac{d y}{d x})}^{2} + (3 y^{2} - 3) \frac{d^{2} y}{d x^{2}} = 0$ . What is $\frac{d^{2} y}{d x^{2}}$ at $(0, 3)$ ?

Your answer

Part 4 Try it A new curve

Try it · Calculate

No calculator. For the curve $y^{2} + 2 y = x^{3} - 12 x$ , implicit differentiation gives $\frac{d y}{d x} = \frac{3 x^{2} - 12}{2 y + 2}$ . The curve has horizontal tangents at points with only one $x$ -coordinate. What is it?

Your answer

Try it · Decide and justify

No calculator. The same curve, $y^{2} + 2 y = x^{3} - 12 x$ , has a horizontal tangent at the point where $x = - 2$ and $y = - 1 + \sqrt{17}$ , and near that point $y$ is a function of $x$ . Does that function have a relative minimum, a relative maximum or neither at $x = - 2$ ?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A designer draws the edge of a metal bracket on a grid, with $x$ and $y$ measured in centimeters. The edge is the part of the curve $y^{2} + x y - x^{2} = 45$ with $y > 0$ , and along it $y$ is a function of $x$ .

(a) Verify that $\frac{d y}{d x} = \frac{2 x - y}{2 y + x}$ .

(b) Find the coordinates of the point on the edge where the tangent line is horizontal.

(c) Find the value of $\frac{d^{2} y}{d x^{2}}$ at the point from part (b). Does the edge have a relative minimum, a relative maximum or neither at that point? Justify your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Differentiate both sides with respect to $x$ , using the product rule on $x y$ : $2 y \frac{d y}{d x} + y + x \frac{d y}{d x} - 2 x = 0$ . So $(2 y + x) \frac{d y}{d x} = 2 x - y$ , and $\frac{d y}{d x} = \frac{2 x - y}{2 y + x}$ .

(b) $\frac{d y}{d x} = 0$ where $2 x - y = 0$ , so $y = 2 x$ . In the curve's equation, $4 x^{2} + 2 x^{2} - x^{2} = 45$ , so $x^{2} = 9$ . With $y > 0$ , $x = 3$ and $y = 6$ ; there $2 y + x = 15$ , not 0. The point is $(3, 6)$ , in centimeters.

(c) Differentiate $2 y \frac{d y}{d x} + y + x \frac{d y}{d x} - 2 x = 0$ again: $2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}} - 2 = 0$ . At $(3, 6)$ , $\frac{d y}{d x} = 0$ , so $15 \frac{d^{2} y}{d x^{2}} = 2$ and $\frac{d^{2} y}{d x^{2}} = \frac{2}{15}$ . The edge has a relative minimum at $(3, 6)$ , because $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} = \frac{2}{15} > 0$ there.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Differentiates implicitly with the product rule on $x y$ , and solves to $\frac{d y}{d x} = \frac{2 x - y}{2 y + x}$ .
Sets $2 x - y = 0$ , substitutes $y = 2 x$ into the curve, and gives $(3, 6)$ in centimeters, rejecting $x = - 3$ because there $y = - 6$ .
Differentiates a second time, keeping the ${(\frac{d y}{d x})}^{2}$ and $\frac{d^{2} y}{d x^{2}}$ terms, and gets $\frac{d^{2} y}{d x^{2}} = \frac{2}{15}$ at $(3, 6)$ .
Concludes a relative minimum, citing $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} > 0$ at $(3, 6)$ .

Part 6 Check yourself Six points on one curve

The two case questions use the same curve and table. Answer them from it.

A curve is given by $x^{2} - 2 x y + 4 y^{2} = 12$ . Implicit differentiation gives

\frac{d y}{d x} = \frac{y - x}{4 y - x} and \frac{d^{2} y}{d x^{2}} = \frac{4 \frac{d y}{d x} - 2 - 8 {(\frac{d y}{d x})}^{2}}{8 y - 2 x} .

The table lists six points on the curve, with the numerator and the denominator of $\frac{d y}{d x}$ at each.

Six points on the curve, with the numerator and denominator of dy/dx
Point $(x, y)$	Numerator $y - x$	Denominator $4 y - x$
$(- 4, - 1)$	3	0
$(- 2, - 2)$	0	$- 6$
$(- 2, 1)$	3	6
$(2, - 1)$	$- 3$	$- 6$
$(2, 2)$	0	6
$(4, 1)$	$- 3$	0

Case question 1 · Calculate

No calculator. How many of the six points in the table are critical points of the relation?

Your answer

Case question 2 · Multiple choice

No calculator. Near the point $(2, - 1)$ , $y$ is a function of $x$ . Which describes the graph of that function at $(2, - 1)$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $\frac{d^{2} y}{d x^{2}}$ at the point $(1, 2)$ on the curve $x^{2} + y^{3} = 9$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

A critical point of an implicit relation is a point where $\frac{d y}{d x}$ is 0 or . If $\frac{d y}{d x} = 0$ and $\frac{d^{2} y}{d x^{2}} < 0$ at such a point, the function the relation defines near it has a there.

Part 7 Keep it Bring back earlier ideas

This lesson stands on the second derivative test and on implicit differentiation. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. A function $g$ is twice differentiable, with $g^{'} (4) = 0$ and $g^{''} (4) = - 3$ . Which statement about $g$ must be true?

Bring it back · Calculate

No calculator. For the curve $x^{3} + y^{3} = 9$ , find $\frac{d y}{d x}$ at the point $(2, 1)$ .

Your answer

In three lines

A critical point of an implicit relation is a point on the curve where $\frac{d y}{d x}$ is 0 or does not exist.
The first and second derivative tests classify a critical point of a function defined implicitly, as they do for any function.
A second derivative found implicitly is a relation of $x$ , $y$ and $\frac{d y}{d x}$ , so substitute all three.

Unit 5 check: 12 questions from every lesson in Unit 5

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks whether a function defined implicitly has a relative maximum or minimum at a point, show $\frac{d y}{d x} = 0$ there, find $\frac{d^{2} y}{d x^{2}}$ with the point's $x$ , $y$ and $\frac{d y}{d x}$ substituted, and write the conclusion with its reason, naming $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ rather than saying "it".

Questions students ask

Do I have to solve the equation for

y

before I can use the derivative tests?

No. Implicit differentiation gives $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ directly, and the tests use only their values and signs near the point. Solving for $y$ is often impossible anyway.

Is a point with a vertical tangent a maximum or a minimum?

It is a critical point of the relation, because $\frac{d y}{d x}$ does not exist there. Near a vertical tangent the curve is usually not the graph of a function of $x$ , so the maximum and minimum tests for a function of $x$ do not apply. On $x^{2} + x y + y^{2} = 3$ , the vertical tangents at $(2, - 1)$ and $(- 2, 1)$ are the curve's farthest points right and left.

Why does

\frac{d^{2} y}{d x^{2}}

have

\frac{d y}{d x}

in it?

Because $y$ is a function of $x$ . Differentiating any term with $y$ in it brings out a factor of $\frac{d y}{d x}$ , and differentiating $\frac{d y}{d x}$ itself gives $\frac{d^{2} y}{d x^{2}}$ .

Which test should I use, the first derivative test or the second?

At a point where $\frac{d y}{d x} = 0$ , the second derivative test is usually quicker, because the terms with $\frac{d y}{d x}$ drop out. If $\frac{d^{2} y}{d x^{2}}$ is 0 there, that test says nothing, and the first derivative test, the sign of $\frac{d y}{d x}$ on each side, decides.

Course alignment, for teachers

AP Calculus AB topic 5.12, Unit 5: Analytical Applications of Differentiation.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.
Skill 3.E: Provide reasons or rationales for solutions and conclusions.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first The north end of a tilted flower bed

Part 2 Learn it Three short ideas, each checked as you go

5a. Critical points of a curve given implicitly

5b. The derivative tests work on implicit functions too

5c. The second derivative: x, y and dy/dx together

Part 3 Watch it done Find and classify a critical point

Part 4 Try it A new curve

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself Six points on one curve

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic