Unit 3 · Topic 3.2 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 3.2: Implicit Differentiation

Q: Why is the derivative of y2 not just 2y?

Because you are differentiating with respect to x, and y changes with x. y2 is a composition with y inside, so the chain rule multiplies 2y by dydx. The derivative is 2y only with respect to y.

Q: Is it wrong for my answer to have y in it?

No. An implicit derivative usually depends on both x and y. To get a number, put in both coordinates of a point on the curve.

Q: Is y' the same as dydx?

Yes. Both name the derivative of y with respect to x. Use one of them consistently in a solution.

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

Unit 3: 5 to 10% of the multiple-choice section

What you'll be able to do

Find $\frac{d y}{d x}$ for a function defined implicitly by an equation in $x$ and $y$ , using the chain rule on every term that contains $y$ , and evaluate it at a point of the curve.

Key vocabulary

Term	What it means	The exam trap
Implicitly defined function	A function $y$ of $x$ given by an equation in $x$ and $y$ , such as $x^{2} + y^{2} = 100$ , instead of by a formula $y = \dots$ .	You do not need to solve the equation for $y$ before you differentiate it, and often you could not.
Implicit differentiation	Differentiating both sides of an equation in $x$ and $y$ with respect to $x$ , treating $y$ as a differentiable function of $x$ , and then solving for $\frac{d y}{d x}$ .	Differentiate both sides, including a side that is a constant, whose derivative is 0.
Chain rule	The rule $\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)$ : the derivative of the outer function at the inside, times the derivative of the inside.	Before you differentiate a term, name its inside function, then multiply by that inside's derivative.
Derivative $\frac{d y}{d x}$	The derivative of $y$ with respect to $x$ , also written $y^{'}$ .	When you solve for it, treat it as one unknown: get its terms together, factor it out, then divide.

The curve x squared plus xy plus y squared equals 7, a tilted oval. A dashed vertical line at x = 1 meets it at (1, 2) and (1, -3), and the tangent lines at those two points have slopes -0.8 and -0.2.

Part 1 Think first The slope of a circle at one point

Make a guess first. Committing to an answer, even a wrong one, helps the lesson stick.

Predict first

A round pond is drawn on a map with its center at the origin, and its edge is the circle $x^{2} + y^{2} = 100$ , in meters. A bench sits on the edge at $(6, 8)$ . What is the slope of the line tangent to the edge at the bench?

In short. Implicit differentiation finds $\frac{d y}{d x}$ from an equation in $x$ and $y$ by differentiating both sides with respect to $x$ , treating $y$ as a function of $x$ , so the chain rule applies to every term that contains $y$ .

By the chain rule, $\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$ , and a product such as $x y$ also needs the product rule.
After differentiating, collect the terms with $\frac{d y}{d x}$ on one side, factor it out and divide.
The result can contain both $x$ and $y$ , so its value at a point uses both coordinates.

Part 2 Learn it Three steps to an implicit derivative

3a. Every y term gets the chain rule

An equation such as $x^{2} + y^{2} = 100$ ties $y$ to $x$ without ever saying $y = \dots$ . Near most points of its graph, though, $y$ behaves as a function of $x$ : on the top half of that circle $y = \sqrt{100 - x^{2}}$ . A function given this way, inside an equation instead of by a formula for $y$ , is implicitly defined.

To differentiate such an equation, treat $y$ as a differentiable function of $x$ that you never write out. Then a term like $y^{3}$ is a composition: cubing is the outer function and $y$ is the inside. The chain rule gives

\frac{d}{d x} (y^{3}) = 3 y^{2} \cdot \frac{d y}{d x} .

The factor $\frac{d y}{d x}$ is the derivative of the inside. Any function of $y$ works the same way: $\frac{d}{d x} (\sin y) = \cos y \frac{d y}{d x}$ and $\frac{d}{d x} (e^{y}) = e^{y} \frac{d y}{d x}$ . Even $y$ on its own has derivative $\frac{d y}{d x}$ .

A term with only $x$ in it differentiates as it always has, $\frac{d}{d x} (x^{3}) = 3 x^{2}$ , with no extra factor. The chain rule is still there in principle, but the inside is $x$ , and the derivative of $x$ with respect to $x$ is 1.

Quick check · Sort it

No calculator. Here $y$ is a differentiable function of $x$ . Sort each expression by its derivative with respect to $x$ . Tap an expression, then a bin.

Its derivative has a dy/dx factor

Its derivative has no dy/dx factor

3b. Differentiate both sides, then solve

To find $\frac{d y}{d x}$ from an equation, differentiate both sides with respect to $x$ , then solve the result for $\frac{d y}{d x}$ .

Take $y^{3} + 2 x y = 8$ . The term $2 x y$ is a constant times a product of two functions of $x$ , namely $x$ and $y$ . So it needs the product rule, and the chain rule turns the derivative of $y$ into $\frac{d y}{d x}$ :

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

The right side, 8, is a constant, so its derivative is 0. Altogether,

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

Now solve, as you would for any unknown. Move the term without $\frac{d y}{d x}$ to the other side, factor $\frac{d y}{d x}$ out of what is left, and divide:

\frac{d y}{d x} (3 y^{2} + 2 x) = - 2 y, \frac{d y}{d x} = \frac{- 2 y}{3 y^{2} + 2 x} .

The answer has $y$ in it as well as $x$ . That is normal for implicit differentiation, and it is why you never needed a formula for $y$ .

Quick check · Multiple choice

No calculator. The equation $y^{2} - x y = 4$ defines $y$ as a differentiable function of $x$ . Which expression is $\frac{d y}{d x}$ ?

3c. A slope at a point needs both coordinates

Because $\frac{d y}{d x}$ usually contains both $x$ and $y$ , its value at a point needs both coordinates, and a value of $x$ alone may not be enough. On the graph at the top, the line $x = 1$ meets the curve $x^{2} + x y + y^{2} = 7$ twice, at $(1, 2)$ and $(1, - 3)$ . Implicit differentiation gives $\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y}$ , which is $- \frac{4}{5}$ at the first point and $- \frac{1}{5}$ at the second. Same $x$ , different slopes.

So to find the slope of a curve at a point:

Check that the point is on the curve, by putting both coordinates into the equation.
Put both coordinates into $\frac{d y}{d x}$ .

For $y^{3} + 2 x y = 8$ , the point $(0, 2)$ is on the curve, since $2^{3} + 2 (0) (2) = 8$ . There $\frac{d y}{d x} = \frac{- 2 (2)}{3 (2)^{2} + 2 (0)} = - \frac{1}{3}$ , the slope of the line tangent to the curve at $(0, 2)$ .

The formula gives a slope only where its denominator is not 0, and only where $y$ is a differentiable function of $x$ near the point. Every point in this lesson's questions is one where both hold.

Quick check · Calculate

No calculator. The point $(2, 1)$ lies on the curve $x^{3} + y^{3} = 9$ . What is the slope of the curve at $(2, 1)$ ?

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing the derivative of $y^{2}$ as $2 y$ .	$y$ is a function of $x$ , so $y^{2}$ is a composition, and the chain rule multiplies by the derivative of the inside, $\frac{d y}{d x}$ .	Each time you differentiate a term with $y$ in it, write $\frac{d y}{d x}$ beside the result before you move on.
Differentiating $x y$ as $x \frac{d y}{d x}$ and losing the $y$ .	$x y$ is a product of two functions of $x$ , and the product rule gives two terms.	Write $\frac{d}{d x} (x y) = x \frac{d y}{d x} + y \cdot 1$ in full, then simplify.
Evaluating $\frac{d y}{d x}$ from the $x$ -coordinate alone.	The derivative usually contains $y$ as well, and one value of $x$ can belong to more than one point of the curve.	Check that the point is on the curve, then use both of its coordinates.

Part 3 Watch it done The slope of a curve, start to finish

Watch each step, then do the last one yourself.

The curve $x^{2} y + y^{3} = 10$ passes through the point $(3, 1)$ . Find $\frac{d y}{d x}$ , and then the slope of the curve at $(3, 1)$ .

Step 1 · Check the point
$3^{2} (1) + 1^{3} = 9 + 1 = 10$ , so $(3, 1)$ is on the curve.
Step 2 · Differentiate both sides
$\frac{d}{d x} (x^{2} y) + \frac{d}{d x} (y^{3}) = \frac{d}{d x} (10)$ , which gives $(2 x y + x^{2} \frac{d y}{d x}) + 3 y^{2} \frac{d y}{d x} = 0$ .
Why this step?
Why does $x^{2} y$ need the product rule, and why does each $y$ term pick up a factor of $\frac{d y}{d x}$ ?
$x^{2} y$ is a product of two functions of $x$ , $x^{2}$ and $y$ , so it takes the product rule. Each derivative of a $y$ term is a chain rule derivative with $y$ as the inside function, and the derivative of that inside is $\frac{d y}{d x}$ .
Step 3 · Collect and factor
Move $2 x y$ to the right side and factor: $\frac{d y}{d x} (x^{2} + 3 y^{2}) = - 2 x y$ .
Step 4 · Solve
$\frac{d y}{d x} = \frac{- 2 x y}{x^{2} + 3 y^{2}}$ .
Your step · Calculate

No calculator. Finish it. For the curve $x^{2} y + y^{3} = 10$ , $\frac{d y}{d x} = \frac{- 2 x y}{x^{2} + 3 y^{2}}$ . What is the slope of the curve at $(3, 1)$ ? Give an exact decimal.

Your answer

Part 4 Try it Two new curves

Try it · Calculate

No calculator. The point $(2, 0)$ lies on the curve $x e^{y} - y = 2$ . What is the value of $\frac{d y}{d x}$ at $(2, 0)$ ?

Your answer

Try it · Decide and justify

No calculator. For the curve $y^{2} = x^{3} + 1$ , implicit differentiation gives $\frac{d y}{d x} = \frac{3 x^{2}}{2 y}$ . A classmate says the curve has slope 2 wherever $x = 2$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, 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 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. On a map of a state park, a hiking trail follows the curve $y^{3} + x y = 12$ , where $x$ and $y$ are measured in kilometers east and north of the ranger station. Along the trail, $y$ is a differentiable function of $x$ .

(a) Show that $\frac{d y}{d x} = \frac{- y}{3 y^{2} + x}$ .

(b) The point $(2, 2)$ is on the trail. Write an equation for the line tangent to the trail at $(2, 2)$ .

(c) The trail passes through exactly one point with $x = 11$ . Find the $y$ -coordinate of that point and the value of $\frac{d y}{d x}$ there. Using correct units, interpret the meaning of that value in the context of the problem.

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

Model answer

(a) Differentiate both sides with respect to $x$ : $3 y^{2} \frac{d y}{d x} + (x \frac{d y}{d x} + y) = 0$ . Then $\frac{d y}{d x} (3 y^{2} + x) = - y$ , so $\frac{d y}{d x} = \frac{- y}{3 y^{2} + x}$ .

(b) At $(2, 2)$ , $\frac{d y}{d x} = \frac{- 2}{3 (2)^{2} + 2} = - \frac{2}{14} = - \frac{1}{7}$ . The tangent line is $y - 2 = - \frac{1}{7} (x - 2)$ .

(c) With $x = 11$ the equation is $y^{3} + 11 y = 12$ , and $y = 1$ satisfies it, since $1 + 11 = 12$ . There $\frac{d y}{d x} = \frac{- 1}{3 (1)^{2} + 11} = - \frac{1}{14}$ . At the point $(11, 1)$ , the trail's distance north of the ranger station is decreasing at a rate of $\frac{1}{14}$ kilometer per kilometer the trail moves east.

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

In (a), differentiates $y^{3}$ as $3 y^{2} \frac{d y}{d x}$ and $x y$ as $x \frac{d y}{d x} + y$ , with 0 on the right side.
In (a), factors out $\frac{d y}{d x}$ and solves to $\frac{- y}{3 y^{2} + x}$ .
In (b), gets the slope $- \frac{1}{7}$ at $(2, 2)$ and writes $y - 2 = - \frac{1}{7} (x - 2)$ .
In (c), finds $y = 1$ and $\frac{d y}{d x} = - \frac{1}{14}$ at $(11, 1)$ .
In (c), says the distance north is decreasing at $\frac{1}{14}$ kilometer per kilometer east, with those units.

Part 6 Check yourself One curve, several points

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

A curve is given by $y^{3} - x y = 6$ . Near each point in the table, $y$ is a differentiable function of $x$ . The last column shows that each point satisfies the equation.

Points on the curve $y^{3} - x y = 6$
$x$	$y$	$y^{3} - x y$
7	3	6
7	-1	6
7	-2	6
-5	1	6
1	2	6

Case question 1 · Calculate

No calculator. What is the slope of the curve $y^{3} - x y = 6$ at the point $(7, - 1)$ ? Give an exact decimal.

Your answer

Case question 2 · Match

No calculator. Use the curve $y^{3} - x y = 6$ and the table. Tap a point, then the slope of the curve there.

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

Check yourself · Spot the error

No calculator. A student finds $\frac{d y}{d x}$ for the curve $x^{2} + y^{4} = 3 x y$ . Tap the first line with a mistake.

Check yourself · Fill the statement

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

In $x^{2} + x y = 5$ , the term $x y$ is a product of two functions of $x$ , so differentiating it takes the rule. After both sides are differentiated, the terms with $\frac{d y}{d x}$ go on one side, and $\frac{d y}{d x}$ is out before dividing.

Check yourself · Multiple choice

No calculator. Here $y$ is a differentiable function of $x$ . Which expression is $\frac{d}{d x} (\sin (y^{2}))$ ?

Part 7 Keep it Bring back earlier ideas

Implicit differentiation runs on the chain rule and the product rule. These two questions bring them back.

Bring it back · Multiple choice

No calculator. Let $h (x) = {(3 x^{2} + 1)}^{4}$ . Which expression is $h^{'} (x)$ ?

From 2.8 The Product Rule · Calculate

No calculator. Let $h (x) = (x^{2} + 1) (3 x - 2)$ . What is $h^{'} (2)$ ?

Your answer

In three lines

Treat $y$ as a function of $x$ , so every term with $y$ in it gets a factor of $\frac{d y}{d x}$ from the chain rule.
Differentiate both sides of the equation, collect the $\frac{d y}{d x}$ terms, factor and divide.
Evaluate $\frac{d y}{d x}$ at a point with both of its coordinates, after checking the point is on the curve.

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 for $\frac{d y}{d x}$ from an equation in $x$ and $y$ , write the line where you differentiate both sides before you solve, so each chain rule factor and each product rule term can be seen. For a slope at a point, show the substitution of both coordinates before the number.

Questions students ask

Do I have to solve the equation for

y

first?

No. Implicit differentiation works on the equation as it stands. Solving for $y$ is often impractical, as with the cubic $y^{3} + 2 x y = 8$ , and sometimes impossible with the functions you know, as with $x e^{y} - y = 2$ . Where it is possible, as with a circle, it splits the curve into pieces and gives the same slopes.

Why is the derivative of

y^{2}

not just

2 y

Because you are differentiating with respect to $x$ , and $y$ changes with $x$ . $y^{2}$ is a composition with $y$ inside, so the chain rule multiplies $2 y$ by $\frac{d y}{d x}$ . The derivative is $2 y$ only with respect to $y$ .

Is it wrong for my answer to have

y

in it?

No. An implicit derivative usually depends on both $x$ and $y$ . To get a number, put in both coordinates of a point on the curve.

y^{'}

the same as

\frac{d y}{d x}

Yes. Both name the derivative of $y$ with respect to $x$ . Use one of them consistently in a solution.

What if the denominator of

\frac{d y}{d x}

is 0 at a point?

Then the formula gives no value there, so you cannot read a slope from it at that point. Every slope you report should come from a point where the denominator is not 0.

Course alignment, for teachers

AP Calculus AB topic 3.2, Unit 3: Differentiation: Composite, Implicit, and Inverse Functions.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is 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 slope of a circle at one point

Part 2 Learn it Three steps to an implicit derivative

3a. Every y term gets the chain rule

3b. Differentiate both sides, then solve

3c. A slope at a point needs both coordinates

Part 3 Watch it done The slope of a curve, start to finish

Part 4 Try it Two new curves

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself One curve, several points

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic