Blog · Calculus guides · 20 October 2026 · 4 min read

Implicit Differentiation: Steps, Examples & Practice Problems

Q: What is the difference between implicit and explicit differentiation?

Explicit means y is already on its own, y = x^2 + 1, and you differentiate directly. Implicit means x and y are tangled together and you differentiate the whole equation, solving for dy/dx afterwards.

Q: Why does dy/dx have y in it?

Because one x can sit under several points of the curve. On the circle x^2+y^2=25, at x = 3 there are two points, (3, 4) and (3, -4), with opposite slopes. The y in the answer tells them apart.

Q: How do you find the tangent line using implicit differentiation?

Find dy/dx, substitute the point to get the slope m, then use y - y_1 = m(x - x_1). For the folium at (3, 3) the slope is -1, so the tangent is y = -x + 6.

Q: How do you find horizontal and vertical tangents on an implicit curve?

Horizontal tangents are where the top of dy/dx is zero and vertical ones are where the bottom is zero, as long as the point is on the curve. For x^2+y^2=25, -x/y is zero at x = 0, the points (0, 5) and (0, -5), and undefined at (5, 0) and (-5, 0).

Q: How do you find the second derivative implicitly?

Differentiate dy/dx again with the quotient rule, again attaching dy/dx to each y, then substitute your first answer in for dy/dx. For the circle this gives (d^2y)/(dx^2) = -25/(y^3).

An illustration of Implicit differentiation

Implicit differentiation finds $\frac{d y}{d x}$ from an equation where $y$ is mixed in with $x$ , such as $x^{2} + y^{2} = 25$ , without solving for $y$ first. Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ so that every $y$ term picks up a factor of $\frac{d y}{d x}$ , then solve for $\frac{d y}{d x}$ . For the circle, $2 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{x}{y}$ .

The formula

\frac{d}{d x} [g (y)] = g^{'} (y) \frac{d y}{d x} for example \frac{d}{d x} [y^{3}] = 3 y^{2} \frac{d y}{d x}

This is just the chain rule with $y$ as the inside function. The $\frac{d y}{d x}$ is the inside derivative, turning up for work like it always does.

When to use it

Use implicit differentiation when the equation does not come as $y = \dots$ and getting it into that form is painful or impossible: circles and ellipses, curves like $x^{3} + y^{3} = 6 x y$ , anything with $sin y$ or $e^{x y}$ in it. It is also the engine behind related rates, where you differentiate with respect to time instead of $x$ . If you can solve for $y$ easily, you can, but implicit differentiation is often faster even then, and it gives you both halves of a circle at once.

How to do it, step by step

Differentiate every term on both sides with respect to $x$ . Terms in $x$ alone are done as usual. Constants become 0.
Attach $\frac{d y}{d x}$ to every $y$ term. Differentiate $y$ as if it were $x$ , then multiply by $\frac{d y}{d x}$ : $y^{2}$ becomes $2 y \frac{d y}{d x}$ . A term like $x y$ is a product, so it needs the product rule: $y + x \frac{d y}{d x}$ .
Collect the $\frac{d y}{d x}$ terms on one side. Everything with a $\frac{d y}{d x}$ goes left, everything else goes right.
Factor out $\frac{d y}{d x}$ and divide. The answer will usually contain both $x$ and $y$ . That is normal. To get a number, substitute a point on the curve.

Worked examples

Example 1: A circle

Easy

Find $\frac{d y}{d x}$ for the circle:

x^{2} + y^{2} = 25

Differentiate both sides. The $x^{2}$ gives $2 x$ , the $y^{2}$ gives $2 y \frac{d y}{d x}$ , and 25 gives 0.

2 x + 2 y \frac{d y}{d x} = 0

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

Divide by $2 y$ . At the point $(3, 4)$ the slope is $- \frac{3}{4}$ , which is exactly perpendicular to the radius, as a circle's tangent should be.

Answer

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

Example 2: With an xy term

Medium

Find $\frac{d y}{d x}$ for:

x^{2} + x y + y^{2} = 7

The middle term $x y$ is a product of $x$ and $y$ , so it differentiates to $1 \cdot y + x \cdot \frac{d y}{d x}$ .

2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0

Collect the $\frac{d y}{d x}$ terms and factor:

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

Divide by $x + 2 y$ . At the point $(1, 2)$ , which is on the curve because $1 + 2 + 4 = 7$ , the slope is $- \frac{2 + 2}{1 + 4} = - \frac{4}{5}$ .

Answer

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

Example 3: The folium of Descartes

Harder

Find $\frac{d y}{d x}$ , and the slope of the tangent at $(3, 3)$ :

x^{3} + y^{3} = 6 x y

Left side: $3 x^{2} + 3 y^{2} \frac{d y}{d x}$ . Right side: $6 x y$ is a product, so $6 y + 6 x \frac{d y}{d x}$ .

3 x^{2} + 3 y^{2} \frac{d y}{d x} = 6 y + 6 x \frac{d y}{d x}

Move the $\frac{d y}{d x}$ terms left and everything else right, then factor:

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

Divide through, and cancel the common 3. Then check $(3, 3)$ is on the curve ( $27 + 27 = 54 = 6 \cdot 9$ , it is) and substitute:

\frac{d y}{d x}_{(3, 3)} = \frac{6 - 9}{9 - 6} = - 1

Solving $x^{3} + y^{3} = 6 x y$ for $y$ would need the cubic formula. Nobody wants that, which is the whole case for doing it implicitly.

Answer

\frac{d y}{d x} = \frac{2 y - x ^{2}}{y ^{2} - 2 x}, slope - 1 at (3, 3)

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find $\frac{d y}{d x}$ for the ellipse:

4 x^{2} + 9 y^{2} = 36

Problem 2

Medium

Find $\frac{d y}{d x}$ :

3 x^{2} + 2 x y = 8

Problem 3

Harder

Find $\frac{d y}{d x}$ :

x^{2} y + y^{3} = 10

Common mistakes

Forgetting the $\frac{d y}{d x}$ . Writing $\frac{d}{d x} [y^{2}] = 2 y$ is the number one slip. Every time you differentiate something with $y$ in it, a $\frac{d y}{d x}$ comes out.
Treating $x y$ as one thing. It is a product. $\frac{d}{d x} [x y] = y + x \frac{d y}{d x}$ , two terms, not $\frac{d y}{d x}$ alone and not $1$ .
Not differentiating the right-hand side. A number on the right becomes 0, but an expression like $6 x y$ needs differentiating too.
Dividing before collecting. Get every $\frac{d y}{d x}$ term on one side and factor first. Dividing early leaves $\frac{d y}{d x}$ on both sides and a lot of crossing out.

Practice it now

Five minutes, no account

Implicit differentiation appears in a Sprint from difficulty 5 to 7: ellipses, an $x y$ term, sums of cubes and full quadratics. Type $\frac{d y}{d x}$ in terms of $x$ and $y$ ; any algebraically equal form is marked right.

Play a Sprint See it on the formula sheet

FAQ

What is the difference between implicit and explicit differentiation?: Explicit means $y$ is already on its own, $y = x^{2} + 1$ , and you differentiate directly. Implicit means $x$ and $y$ are tangled together and you differentiate the whole equation, solving for $\frac{d y}{d x}$ afterwards.
Why does dy/dx have y in it?: Because one $x$ can sit under several points of the curve. On the circle $x^{2} + y^{2} = 25$ , at $x = 3$ there are two points, $(3, 4)$ and $(3, - 4)$ , with opposite slopes. The $y$ in the answer tells them apart.
How do you find the tangent line using implicit differentiation?: Find $\frac{d y}{d x}$ , substitute the point to get the slope $m$ , then use $y - y_{1} = m (x - x_{1})$ . For the folium at $(3, 3)$ the slope is $- 1$ , so the tangent is $y = - x + 6$ .
How do you find horizontal and vertical tangents on an implicit curve?: Horizontal tangents are where the top of $\frac{d y}{d x}$ is zero and vertical ones are where the bottom is zero, as long as the point is on the curve. For $x^{2} + y^{2} = 25$ , $- \frac{x}{y}$ is zero at $x = 0$ , the points $(0, 5)$ and $(0, - 5)$ , and undefined at $(5, 0)$ and $(- 5, 0)$ .
How do you find the second derivative implicitly?: Differentiate $\frac{d y}{d x}$ again with the quotient rule, again attaching $\frac{d y}{d x}$ to each $y$ , then substitute your first answer in for $\frac{d y}{d x}$ . For the circle this gives $\frac{d ^{2} y}{d x ^{2}} = - \frac{25}{y ^{3}}$ .
Where can I find implicit differentiation practice problems?: Right on this page: "Your turn" has three implicit differentiation practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Implicit Differentiation: Steps, Examples & Practice Problems

When to use it

Use implicit differentiation when the equation does not come as

y = \dots

and getting it into that form is painful or impossible: circles and ellipses, curves like

x^{3} + y^{3} = 6 x y

, anything with

sin y

e^{x y}

in it. It is also the engine behind related rates, where you differentiate with respect to time instead of

x

. If you can solve for

y

easily, you can, but implicit differentiation is often faster even then, and it gives you both halves of a circle at once.

How to do it, step by step

Differentiate every term on both sides with respect to $x$ . Terms in

x

alone are done as usual. Constants become 0.

Attach $\frac{d y}{d x}$ to every $y$ term. Differentiate

y

as if it were

x

, then multiply by

\frac{d y}{d x}

y^{2}

becomes

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

. A term like

x y

is a product, so it needs the product rule:

y + x \frac{d y}{d x}

Collect the $\frac{d y}{d x}$ terms on one side. Everything with a

\frac{d y}{d x}

goes left, everything else goes right.

Factor out $\frac{d y}{d x}$ and divide. The answer will usually contain both

x

and

y

. That is normal. To get a number, substitute a point on the curve.

Worked examples

Example 1: A circle

Easy

Find $\frac{d y}{d x}$ for the circle:

x^{2} + y^{2} = 25

Differentiate both sides. The $x^{2}$ gives $2 x$ , the $y^{2}$ gives $2 y \frac{d y}{d x}$ , and 25 gives 0.

2 x + 2 y \frac{d y}{d x} = 0

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

Divide by $2 y$ . At the point $(3, 4)$ the slope is $- \frac{3}{4}$ , which is exactly perpendicular to the radius, as a circle's tangent should be.

Answer

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

Example 2: With an xy term

Medium

Find $\frac{d y}{d x}$ for:

x^{2} + x y + y^{2} = 7

The middle term $x y$ is a product of $x$ and $y$ , so it differentiates to $1 \cdot y + x \cdot \frac{d y}{d x}$ .

2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0

Collect the $\frac{d y}{d x}$ terms and factor:

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

Divide by $x + 2 y$ . At the point $(1, 2)$ , which is on the curve because $1 + 2 + 4 = 7$ , the slope is $- \frac{2 + 2}{1 + 4} = - \frac{4}{5}$ .

Answer

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

Example 3: The folium of Descartes

Harder

Find $\frac{d y}{d x}$ , and the slope of the tangent at $(3, 3)$ :

x^{3} + y^{3} = 6 x y

Left side: $3 x^{2} + 3 y^{2} \frac{d y}{d x}$ . Right side: $6 x y$ is a product, so $6 y + 6 x \frac{d y}{d x}$ .

3 x^{2} + 3 y^{2} \frac{d y}{d x} = 6 y + 6 x \frac{d y}{d x}

Move the $\frac{d y}{d x}$ terms left and everything else right, then factor:

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

Divide through, and cancel the common 3. Then check $(3, 3)$ is on the curve ( $27 + 27 = 54 = 6 \cdot 9$ , it is) and substitute:

\frac{d y}{d x}_{(3, 3)} = \frac{6 - 9}{9 - 6} = - 1

Solving $x^{3} + y^{3} = 6 x y$ for $y$ would need the cubic formula. Nobody wants that, which is the whole case for doing it implicitly.

Answer

\frac{d y}{d x} = \frac{2 y - x ^{2}}{y ^{2} - 2 x}, slope - 1 at (3, 3)

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find $\frac{d y}{d x}$ for the ellipse:

4 x^{2} + 9 y^{2} = 36

Problem 2

Medium

Find $\frac{d y}{d x}$ :

3 x^{2} + 2 x y = 8

Problem 3

Harder

Find $\frac{d y}{d x}$ :

x^{2} y + y^{3} = 10

Common mistakes

Forgetting the $\frac{d y}{d x}$ . Writing

\frac{d}{d x} [y^{2}] = 2 y

is the number one slip. Every time you differentiate something with

y

in it, a

\frac{d y}{d x}

comes out.

Treating $x y$ as one thing. It is a product.

\frac{d}{d x} [x y] = y + x \frac{d y}{d x}

, two terms, not

\frac{d y}{d x}

alone and not

1

Not differentiating the right-hand side. A number on the right becomes 0, but an expression like

6 x y

needs differentiating too.

Dividing before collecting. Get every

\frac{d y}{d x}

term on one side and factor first. Dividing early leaves

\frac{d y}{d x}

on both sides and a lot of crossing out.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

What is the difference between implicit and explicit differentiation?

Explicit means

y

is already on its own,

y = x^{2} + 1

, and you differentiate directly. Implicit means

x

and

y

are tangled together and you differentiate the whole equation, solving for

\frac{d y}{d x}

afterwards.

Why does dy/dx have y in it?

Because one

x

can sit under several points of the curve. On the circle

x^{2} + y^{2} = 25

, at

x = 3

there are two points,

(3, 4)

and

(3, - 4)

, with opposite slopes. The

y

in the answer tells them apart.

How do you find the tangent line using implicit differentiation?

Find

\frac{d y}{d x}

, substitute the point to get the slope

m

, then use

y - y_{1} = m (x - x_{1})

. For the folium at

(3, 3)

the slope is

- 1

, so the tangent is

y = - x + 6

How do you find horizontal and vertical tangents on an implicit curve?

Horizontal tangents are where the top of

\frac{d y}{d x}

is zero and vertical ones are where the bottom is zero, as long as the point is on the curve. For

x^{2} + y^{2} = 25

- \frac{x}{y}

is zero at

x = 0

, the points

(0, 5)

and

(0, - 5)

, and undefined at

(5, 0)

and

(- 5, 0)

How do you find the second derivative implicitly?

Differentiate

\frac{d y}{d x}

again with the quotient rule, again attaching

\frac{d y}{d x}

to each

y

, then substitute your first answer in for

\frac{d y}{d x}

. For the circle this gives

\frac{d ^{2} y}{d x ^{2}} = - \frac{25}{y ^{3}}

Where can I find implicit differentiation practice problems?

Right on this page: "Your turn" has three implicit differentiation practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Implicit Differentiation: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A circle

Example 2: With an xy term

Example 3: The folium of Descartes

Your turn

Common mistakes

FAQ

Keep going

Implicit Differentiation: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A circle

Example 2: With an xy term

Example 3: The folium of Descartes

Your turn

Common mistakes

FAQ

Keep going