Implicit Differentiation: Step-by-Step Examples & AP Practice

Q: What is implicit differentiation used for?

It finds dy/dx when an equation mixes x and y and you can't (or don't want to) solve for y first, such as circles, ellipses and curves like x^3 + y^3 = 6xy. On the AP exam it is used for tangent lines, horizontal and vertical tangents, concavity, and related rates.

Q: Why do you multiply by dy/dx when differentiating y?

Because y is secretly a function of x. Differentiating y^2 with respect to x is the Chain Rule: the outside derivative 2y times the inside derivative dy/dx.

Q: How do you find horizontal and vertical tangent lines with implicit differentiation?

Write dy/dx as a fraction. Horizontal tangents: set the numerator equal to 0, solve together with the original equation, and check the denominator is not 0. Vertical tangents: set the denominator equal to 0 and check the numerator is not 0.

Q: How do you find the second derivative implicitly?

Differentiate the equation (or your dy/dx formula) again with respect to x, remembering that every y and every dy/dx needs the Chain Rule. Then substitute the known value of dy/dx. Example (e) above shows the full process.

Standard differentiation needs $y$ written as a function of $x$ (like $y = x^2$ ). Implicit differentiation finds $\frac{dy}{dx}$ even when $x$ and $y$ are mixed together, like $x^2 + xy + y^2 = 3$ . It is how you find tangent lines, horizontal and vertical tangents, and second derivatives of curves on the AP Calculus AB exam.

Core Theorem

Treat

y

as a function of

x

. Differentiate both sides with respect to

x

, and apply the Chain Rule to every term that contains

y

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

For example,

\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}

and

\frac{d}{dx}(xy) = y + x\frac{dy}{dx}

Step-by-Step SOP

1
Differentiate both sides
Take $\frac{d}{dx}$ of every term on both sides of the equation. A constant on either side becomes $0$ .
2
Chain Rule on every y
Each term with $y$ gets a $\frac{dy}{dx}$ : $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$ . Terms like $xy$ or $x^2y$ also need the Product Rule.
3
Collect the dy/dx terms
Move every term containing $\frac{dy}{dx}$ to one side and everything else to the other side.
4
Factor and solve
Factor out $\frac{dy}{dx}$ and divide. Your answer can (and usually will) contain both $x$ and $y$ .
5
Plug in the point (if asked)
To get a slope at a point, substitute both coordinates $(x, y)$ . On the AP exam you may plug in before solving for $\frac{dy}{dx}$ if it makes the algebra easier.

Practice Exercises

Example 01Easy

Find the equation of the tangent line to the circle

x^2 + y^2 = 25

at the point

(3, -4)

NEED A HINT?

Differentiate both sides with respect to

x

. The derivative of the constant

25

0

SHOW DETAILED EXPLANATION

Step 1: Differentiate both sides

\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)

gives

2x + 2y\frac{dy}{dx} = 0

Step 2: Solve for dy/dx

2y\frac{dy}{dx} = -2x

, so

\frac{dy}{dx} = -\frac{x}{y}

Step 3: Slope at the point

(3, -4)

\frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}

Step 4: Tangent line

Point-slope form:

y + 4 = \frac{3}{4}(x - 3)

Example 02Easy

Find

\frac{dy}{dx}

y^2 = \frac{1}{2x + 5}

NEED A HINT?

Rewrite the right side as

(2x+5)^{-1}

and use the Chain Rule on both sides.

SHOW DETAILED EXPLANATION

Step 1: Rewrite

y^2 = (2x + 5)^{-1}

Step 2: Differentiate both sides

Left:

2y\frac{dy}{dx}

. Right:

-(2x+5)^{-2} \cdot 2 = -\frac{2}{(2x+5)^2}

Step 3: Solve for dy/dx

2y\frac{dy}{dx} = -\frac{2}{(2x+5)^2}

, so

\frac{dy}{dx} = -\frac{1}{y(2x+5)^2}

Check with the explicit form

For the upper branch

y = (2x+5)^{-1/2}

, the formula gives

-\frac{1}{(2x+5)^{-1/2}(2x+5)^{2}} = -(2x+5)^{-3/2}

, which matches differentiating

y = (2x+5)^{-1/2}

directly.

If the equation is y² = ½x + 5 instead

Then

2y\frac{dy}{dx} = \frac{1}{2}

, so

\frac{dy}{dx} = \frac{1}{4y}

. Read the problem carefully:

\frac{1}{2x+5}

and

\frac{1}{2}x + 5

are different curves.

Example 03Medium

Find

\frac{dy}{dx}

for the curve

x^3 + y^3 = 6xy

(the Folium of Descartes).

NEED A HINT?

The term

6xy

is a product of

6x

and

y

, so it needs the Product Rule.

SHOW DETAILED EXPLANATION

Step 1: Differentiate the left side

\frac{d}{dx}(x^3 + y^3) = 3x^2 + 3y^2\frac{dy}{dx}

Step 2: Differentiate the right side (Product Rule)

\frac{d}{dx}(6xy) = 6y + 6x\frac{dy}{dx}

Step 3: Collect the dy/dx terms

3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2

Step 4: Factor and solve

\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2

, so

\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}

Example 04AB/BC Standard

AP FRQ style. Consider the curve

x^2 + xy + y^2 = 3

.
(a) Show that

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

.
(b) Write the equation of the tangent line at

(1, 1)

.
(c) Find all points on the curve where the tangent line is horizontal.
(d) Find all points where the tangent line is vertical.
(e) Find

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

(1, 1)

NEED A HINT?

Horizontal tangent: numerator of

\frac{dy}{dx}

0

(denominator not

0

). Vertical tangent: denominator is

0

(numerator not

0

). Each time, substitute back into the original equation to find the actual points.

SHOW DETAILED EXPLANATION

(a) Differentiate implicitly

2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0

. So

\frac{dy}{dx}(x + 2y) = -(2x + y)

and

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

(b) Tangent line at (1, 1)

Check the point:

1 + 1 + 1 = 3

. Slope:

-\frac{2+1}{1+2} = -1

. Tangent line:

y - 1 = -(x - 1)

(c) Horizontal tangents

Set the numerator to

0

2x + y = 0

, so

y = -2x

. Substitute:

x^2 + x(-2x) + 4x^2 = 3x^2 = 3

, so

x = \pm 1

. Points:

(1, -2)

and

(-1, 2)

. The denominator

x + 2y

-3

and

3

there, not

0

, so both are horizontal tangents.

(d) Vertical tangents

Set the denominator to

0

x = -2y

. Substitute:

4y^2 - 2y^2 + y^2 = 3y^2 = 3

, so

y = \pm 1

. Points:

(-2, 1)

and

(2, -1)

. The numerator

2x + y

-3

and

3

there, not

0

, so both are vertical tangents.

(e) Second derivative at (1, 1)

Differentiate

2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0

again:

2 + 2\frac{dy}{dx} + x\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} = 0

. At

(1, 1)

with

\frac{dy}{dx} = -1

2 - 2 + \frac{d^2y}{dx^2} + 2 + 2\frac{d^2y}{dx^2} = 0

, so

3\frac{d^2y}{dx^2} = -2

and

\frac{d^2y}{dx^2} = -\frac{2}{3}

. The curve is concave down there.

Example 05Medium

Practice: Find the slope of the tangent line to

x^2 y + y^3 = 10

at the point

(3, 1)

NEED A HINT?

x^2 y

needs the Product Rule. Check that

(3, 1)

is on the curve first:

9 + 1 = 10

SHOW DETAILED EXPLANATION

Step 1: Differentiate

2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0

Step 2: Plug in the point first

(3, 1)

6 + 9\frac{dy}{dx} + 3\frac{dy}{dx} = 0

, so

12\frac{dy}{dx} = -6

Answer

\frac{dy}{dx} = -\frac{1}{2}

(3, 1)

Common Pitfalls

⚠
Forgetting dy/dx on y terms $\frac{d}{dx}(y^2)$ is $2y\frac{dy}{dx}$ , not $2y$ . $y$ is a function of $x$ , so it never disappears without leaving a $\frac{dy}{dx}$ behind.
⚠
Skipping the Product Rule $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ . Writing just $\frac{dy}{dx}$ or $1 \cdot \frac{dy}{dx}$ is the most common lost point.
⚠
Not differentiating the constantIn $x^2 + y^2 = 25$ , the right side becomes $0$ , not $25$ .
⚠
Horizontal and vertical tangents without checkingA horizontal tangent needs numerator $= 0$ AND denominator $\neq 0$ (vertical is the reverse). If both are $0$ , the test is inconclusive. Also substitute back into the original equation: a point must be on the curve.
⚠
Second derivative with dy/dx left inWhen you find $\frac{d^2y}{dx^2}$ , substitute your expression or value for $\frac{dy}{dx}$ back in. AP graders expect the final answer in terms of $x$ and $y$ (or a number).

Frequently Asked Questions

What is implicit differentiation used for?

It finds

\frac{dy}{dx}

when an equation mixes

x

and

y

and you can't (or don't want to) solve for

y

first, such as circles, ellipses and curves like

x^3 + y^3 = 6xy

. On the AP exam it is used for tangent lines, horizontal and vertical tangents, concavity, and related rates.

Why do you multiply by dy/dx when differentiating y?

Because

y

is secretly a function of

x

. Differentiating

y^2

with respect to

x

is the Chain Rule: the outside derivative

2y

times the inside derivative

\frac{dy}{dx}

How do you find horizontal and vertical tangent lines with implicit differentiation?

Write

\frac{dy}{dx}

as a fraction. Horizontal tangents: set the numerator equal to

0

, solve together with the original equation, and check the denominator is not

0

. Vertical tangents: set the denominator equal to

0

and check the numerator is not

0

How do you find the second derivative implicitly?

Differentiate the equation (or your

\frac{dy}{dx}

formula) again with respect to

x

, remembering that every

y

and every

\frac{dy}{dx}

needs the Chain Rule. Then substitute the known value of

\frac{dy}{dx}

. Example (e) above shows the full process.

Is implicit differentiation on the AP Calculus AB exam?

Yes. It is part of Unit 3 (Differentiation: Composite, Implicit, and Inverse Functions) and shows up in multiple choice and in free-response questions about curves. See the real past AP questions linked below.

Ready to test it?

Practice Derivatives multiple-choice questions, timed and fully explained.

MCQ Practice

On the AP exam

See how this shows up on the free-response section: step-by-step method, scoring tips and practice.

FRQ Practice

Past AP Exams

Real AP exam questions on this topic

Official AP Calculus AB free-response questions, each with a full step-by-step solution.

In the 30-Day AP Calc AB Plan

Day 8: Implicit & Exponential/Log Derivatives

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Differentiate both sides

Chain Rule on every y

Collect the dy/dx terms

Factor and solve

Plug in the point (if asked)

Practice Exercises

Step 1: Differentiate both sides

Step 2: Solve for dy/dx

Step 3: Slope at the point

Step 4: Tangent line

Step 1: Rewrite

Step 2: Differentiate both sides

Step 3: Solve for dy/dx

Check with the explicit form

If the equation is y² = ½x + 5 instead

Step 1: Differentiate the left side

Step 2: Differentiate the right side (Product Rule)

Step 3: Collect the dy/dx terms

Step 4: Factor and solve

(a) Differentiate implicitly

(b) Tangent line at (1, 1)

(c) Horizontal tangents

(d) Vertical tangents

(e) Second derivative at (1, 1)

Step 1: Differentiate

Step 2: Plug in the point first

Answer

What is implicit differentiation used for?

Why do you multiply by dy/dx when differentiating y?

How do you find horizontal and vertical tangent lines with implicit differentiation?

How do you find the second derivative implicitly?

Is implicit differentiation on the AP Calculus AB exam?

Real AP exam questions on this topic

Gary Chang

Want a Structured Way to Score a 5 in ap calculus ab?