3.2 - Implicit Differentiation

Explicit vs. Implicit Equations

Explicit equations give you one variable directly in terms of another:

$f(t) = t^2 - 3t$

$y = 2x + 5$

$g(x) = \sin(x^2)$

Implicit equations have multiple variables mixed together:

$x^2 + y^2 = 9$ (a circle)

$x^3 + y^3 = 6xy$ (folium of Descartes)

$\sin(xy) = x + y$

Not every graph is a function, so we need implicit equations to describe curves like circles and ellipses. Implicit differentiation is how we get derivatives for those.

Understanding the Notation

Leibniz notation makes this click. Technically

\frac{dy}{dx}

isn’t a fraction, but it behaves like one.

With respect to $x$ :

\frac{d}{dx}[x] = \frac{dx}{dx} = 1 \qquad \frac{d}{dx}[y] = \frac{dy}{dx}

Power rule with chain rule:

\frac{d}{dx}[x^2] = 2x \cdot \frac{dx}{dx} = 2x \qquad \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}

Exponentials:

\frac{d}{dx}[e^{5x}] = 5e^{5x} \qquad \frac{d}{dx}[e^{5y}] = 5e^{5y} \cdot \frac{dy}{dx}

Key principle: whenever you differentiate a term containing

y

with respect to

x

, you have to multiply by

\frac{dy}{dx}

. That’s the chain rule doing its thing.

Why the chain rule? In an implicit equation,

y

is still a function of

x

, even when you can’t solve for it. So

y^2

is really

[y(x)]^2

, a composite function: the outer function is

u^2

and the inner function is

y(x)

. The chain rule says take the derivative of the outside,

2y

, and multiply by the derivative of the inside,

\frac{dy}{dx}

. That’s where every

\frac{dy}{dx}

comes from. An

x

term gets the same treatment, but the derivative of the inside is

\frac{dx}{dx} = 1

, so nothing extra shows up.

The Four-Step Process

Differentiate both sides with respect to $x$ . Tack on $\frac{dy}{dx}$ every time you differentiate a $y$ term.

Collect all $\frac{dy}{dx}$ terms on one side. Move everything else to the other side.

Factor out $\frac{dy}{dx}$ .

Solve for $\frac{dy}{dx}$ by dividing by whatever’s left.

Example 1: Basic Implicit Differentiation

Find

\frac{dy}{dx}

for

y^2 - 5x^3 = 3y

Step 1: Differentiate both sides.

2y\frac{dy}{dx} - 15x^2 = 3\frac{dy}{dx}

Step 2: Collect the

\frac{dy}{dx}

terms on the left.

2y\frac{dy}{dx} - 3\frac{dy}{dx} = 15x^2

Step 3: Factor.

\frac{dy}{dx}(2y - 3) = 15x^2

Step 4: Solve.

\frac{dy}{dx} = \frac{15x^2}{2y - 3}

Example 2: Implicit Differentiation with Trig and Product Rule

Find

\frac{dy}{dx}

for

\sin(xy) = 10x

Step 1: Differentiate both sides. The left side needs the chain rule and the product rule, since

xy

is a product.

\cos(xy) \cdot \frac{d}{dx}[xy] = 10

For

\frac{d}{dx}[xy]

, use the product rule:

\frac{d}{dx}[xy] = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}

So:

\cos(xy)\left(y + x\frac{dy}{dx}\right) = 10

Step 2: Divide both sides by

\cos(xy)

y + x\frac{dy}{dx} = \frac{10}{\cos(xy)} = 10\sec(xy)

Step 3: Isolate the

\frac{dy}{dx}

term.

x\frac{dy}{dx} = 10\sec(xy) - y

Step 4: Solve.

\frac{dy}{dx} = \frac{10\sec(xy) - y}{x}

You can write

\frac{dy}{dx}

y'

to save time, just don’t confuse

y

and

y'

Finding Equations of Tangent Lines

\frac{dy}{dx}

gives you the slope at any point

(x, y)

on the curve. Unlike explicit functions where an

x

-value is enough, implicit derivatives need both the

x

and

y

coordinates.

Example: Tangent Lines to a Circle

Find the equations of the tangent lines to

x^2 + y^2 = 4

x = 1

Find the $y$ -coordinates. Substitute

x = 1

1 + y^2 = 4 \quad \Rightarrow \quad y^2 = 3 \quad \Rightarrow \quad y = \pm\sqrt{3}

So the points are

(1, \sqrt{3})

and

(1, -\sqrt{3})

Find $\frac{dy}{dx}$ :

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

Find the slopes:

(1, \sqrt{3})

m_1 = -\frac{1}{\sqrt{3}}

(1, -\sqrt{3})

m_2 = \frac{1}{\sqrt{3}}

Write the tangent lines using point-slope form:

(1, \sqrt{3})

y - \sqrt{3} = -\frac{1}{\sqrt{3}}(x - 1)

(1, -\sqrt{3})

y + \sqrt{3} = \frac{1}{\sqrt{3}}(x - 1)

Horizontal and Vertical Tangent Lines

Horizontal tangent lines happen where the slope is zero:

\frac{dy}{dx} = 0

Vertical tangent lines happen where the slope is undefined.

So for a fraction

\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}

Horizontal tangents: set the numerator equal to 0

Vertical tangents: set the denominator equal to 0

Then check each point. It has to actually be on the curve (plug it into the original equation), and the other part of the fraction can’t also be 0 there. If the numerator and denominator are both 0 at the same point,

\frac{dy}{dx}

comes out as

\frac{0}{0}

, which tells you nothing yet. You can’t call that a horizontal or vertical tangent without more work.

Example: Finding Horizontal and Vertical Tangents

Given

3x^2 + 2y^2 = 16

Find $\frac{dy}{dx}$ :

6x + 4y\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{6x}{4y} = -\frac{3x}{2y}

Horizontal tangents. Set the numerator to 0:

-3x = 0 \quad \Rightarrow \quad x = 0

Find

y

when

x = 0

2y^2 = 16 \quad \Rightarrow \quad y^2 = 8 \quad \Rightarrow \quad y = \pm 2\sqrt{2}

Horizontal tangent lines:

y = 2\sqrt{2}

and

y = -2\sqrt{2}

Vertical tangents. Set the denominator to 0:

2y = 0 \quad \Rightarrow \quad y = 0

Find

x

when

y = 0

3x^2 = 16 \quad \Rightarrow \quad x^2 = \frac{16}{3} \quad \Rightarrow \quad x = \pm\frac{4\sqrt{3}}{3}

Vertical tangent lines:

x = \frac{4\sqrt{3}}{3}

and

x = -\frac{4\sqrt{3}}{3}

Important note: for now, whenever the derivative is undefined we’re treating it as a vertical tangent. Later in Unit 5 you’ll learn how to tell vertical tangents, cusps, and corners apart.

Practice Problem

Find

\frac{dy}{dx}

for

x^3 + y^3 = 4xy

Step 1: Differentiate both sides. The right side needs the product rule.

3x^2 + 3y^2\frac{dy}{dx} = 4\left(y + x\frac{dy}{dx}\right) = 4y + 4x\frac{dy}{dx}

Step 2: Collect the

\frac{dy}{dx}

terms.

3y^2\frac{dy}{dx} - 4x\frac{dy}{dx} = 4y - 3x^2

Step 3: Factor.

\frac{dy}{dx}(3y^2 - 4x) = 4y - 3x^2

Step 4: Solve.

\frac{dy}{dx} = \frac{4y - 3x^2}{3y^2 - 4x}

Key Takeaways

Explicit equations isolate one variable ( $y = f(x)$ ); implicit equations mix them ( $x^2 + y^2 = 9$ )

Every time you differentiate a $y$ term, multiply by $\frac{dy}{dx}$

Four-step process: differentiate both sides, collect $\frac{dy}{dx}$ terms, factor out $\frac{dy}{dx}$ , solve

To find a slope at a point, you need both the $x$ and $y$ values

Horizontal tangents: set the numerator of $\frac{dy}{dx}$ to 0, answer is $y = k$

Vertical tangents: set the denominator of $\frac{dy}{dx}$ to 0, answer is $x = k$

The product rule still applies: $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$

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