Implicit Differentiation

You have been finding derivatives of functions where $y$ is neatly isolated on one side of the equation. Functions like $y = x^2 + 3x$ or $y = \sin(x)$ are straightforward because you can see exactly what $y$ equals in terms of $x$. But what happens when you encounter an equation like $x^2 + y^2 = 25$? This describes a circle, and no matter how hard you try, you cannot write $y$ as a single function of $x$ without splitting it into two pieces. This is where implicit differentiation becomes your essential tool. Rather than forcing the equation into a form it does not naturally take, you will learn to differentiate it exactly as it stands. The technique is surprisingly elegant once you understand the underlying idea, and it opens up an entire world of curves that explicit functions cannot easily describe.

Core Concepts

Explicit vs. Implicit Functions

An explicit function gives you $y$ directly in terms of $x$. You can look at it and immediately compute $y$ for any value of $x$:

$$y = x^3 - 4x + 7$$

$$y = e^x \cos(x)$$

$$y = \sqrt{x^2 + 1}$$

An implicit equation describes a relationship between $x$ and $y$ without isolating either variable:

$$x^2 + y^2 = 25$$

$$x^3 + y^3 = 6xy$$

$$e^{xy} = x + y$$

The key distinction is this: with an explicit function, you have a recipe for computing $y$ from $x$. With an implicit equation, you have a condition that $x$ and $y$ must satisfy together. The equation $x^2 + y^2 = 25$ tells you that valid pairs $(x, y)$ lie on a circle of radius 5, but it does not tell you how to compute $y$ from $x$ without additional work.

Why We Need Implicit Differentiation

Consider the circle $x^2 + y^2 = 25$ again. You could solve for $y$ to get:

$$y = \pm\sqrt{25 - x^2}$$

This gives you two separate functions: the upper semicircle $y = \sqrt{25 - x^2}$ and the lower semicircle $y = -\sqrt{25 - x^2}$. You could differentiate each one using the chain rule:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}} \quad \text{or} \quad \frac{dy}{dx} = \frac{x}{\sqrt{25 - x^2}}$$

This works, but it is messy. And for more complicated equations like $x^3 + y^3 = 6xy$, solving for $y$ explicitly is either extremely difficult or impossible. Implicit differentiation lets you find $\frac{dy}{dx}$ without ever solving for $y$. You differentiate the equation as it stands, treating $y$ as a function of $x$, and then solve algebraically for $\frac{dy}{dx}$.

The Key Insight: $y$ Depends on $x$

The central idea of implicit differentiation is deceptively simple: treat $y$ as a function of $x$, even though you do not know what that function is.

When you differentiate $x^2$, you get $2x$. That is straightforward.

When you differentiate $y^2$, you must use the chain rule. Think of $y^2$ as $(y(x))^2$, an outer function (squaring) composed with an inner function ($y$, which depends on $x$). The chain rule gives:

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

The factor $\frac{dy}{dx}$ appears because $y$ is not just a variable; it is a function of $x$ whose derivative we seek.

This same principle applies to any expression involving $y$:

• $\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}$
• $\frac{d}{dx}[\sin(y)] = \cos(y) \cdot \frac{dy}{dx}$
• $\frac{d}{dx}[e^y] = e^y \cdot \frac{dy}{dx}$

Every time you differentiate a term containing $y$, the chain rule contributes a factor of $\frac{dy}{dx}$.

The Technique Step by Step

Here is the systematic procedure for implicit differentiation:

Step 1: Differentiate both sides of the equation with respect to $x$.

Step 2: When differentiating any term involving $y$, apply the chain rule and include the factor $\frac{dy}{dx}$.

Step 3: Collect all terms containing $\frac{dy}{dx}$ on one side of the equation.

Step 4: Factor out $\frac{dy}{dx}$.

Step 5: Solve for $\frac{dy}{dx}$.

The result will typically be an expression involving both $x$ and $y$. This is expected and perfectly acceptable. The derivative $\frac{dy}{dx}$ tells you the slope of the tangent line at any point $(x, y)$ on the curve.

Handling Products: The Product Rule Meets Implicit Differentiation

When $x$ and $y$ appear multiplied together, you need the product rule. For the term $xy$:

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

For the term $x^2 y^3$:

$$\frac{d}{dx}[x^2 y^3] = x^2 \cdot \frac{d}{dx}[y^3] + y^3 \cdot \frac{d}{dx}[x^2]$$

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

Always remember: the product rule applies whenever two factors are both present, and the chain rule applies whenever one of those factors involves $y$.

Finding Slopes at Specific Points

One of the most common applications is finding the slope of the tangent line at a particular point on an implicitly defined curve.

The process is:

1. Use implicit differentiation to find $\frac{dy}{dx}$ as an expression in $x$ and $y$.
2. Substitute the coordinates of the specific point into this expression.
3. Simplify to get the numerical slope.

Since your answer for $\frac{dy}{dx}$ contains both $x$ and $y$, you need both coordinates of the point to evaluate the slope. This makes geometric sense: the same $x$-value might correspond to multiple points on the curve (like the top and bottom of a circle), and those points can have different slopes.

Second Derivatives Implicitly

To find $\frac{d^2y}{dx^2}$ for an implicitly defined curve, you differentiate $\frac{dy}{dx}$ with respect to $x$. This requires implicit differentiation again because $\frac{dy}{dx}$ typically contains $y$, which depends on $x$.

The process often requires substituting your expression for $\frac{dy}{dx}$ back into the result to eliminate it and express $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ alone.

Notation and Terminology

Term	Meaning	Example
Explicit function	$y$ is isolated: $y = f(x)$	$y = x^2 + 1$
Implicit equation	A relation between $x$ and $y$	$x^2 + y^2 = 25$
$\frac{dy}{dx}$	Derivative of $y$ with respect to $x$, treating $y$ as $y(x)$	Found via implicit differentiation
$\frac{d}{dx}[y^n]$	Requires chain rule: $ny^{n-1}\frac{dy}{dx}$	$\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx}$
$y'$	Alternative notation for $\frac{dy}{dx}$	$y’ = -\frac{x}{y}$
$y’'$	Second derivative $\frac{d^2y}{dx^2}$	Found by differentiating $y'$
Level curve	Curve defined by $F(x,y) = c$ for constant $c$	Contour lines on a map

Examples

Key Properties and Rules

The Chain Rule is Central

Every implicit differentiation problem ultimately relies on the chain rule. When you see $y$ or any function of $y$, remember that $y$ depends on $x$:

$$\frac{d}{dx}[f(y)] = f’(y) \cdot \frac{dy}{dx}$$

The Result Contains Both Variables

Unlike explicit differentiation where $\frac{dy}{dx}$ is purely a function of $x$, implicit differentiation typically yields $\frac{dy}{dx}$ as a function of both $x$ and $y$. This is not a flaw but a feature: it reflects the geometric reality that the same $x$-value may correspond to multiple points with different slopes.

Horizontal Tangents

For $\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}$, horizontal tangents occur where:

• The numerator $N(x,y) = 0$
• The denominator $D(x,y) \neq 0$

Vertical Tangents

Vertical tangents occur where:

• The denominator $D(x,y) = 0$
• The numerator $N(x,y) \neq 0$

Singular Points

If both numerator and denominator are zero simultaneously, you have a singular point where the tangent may not be well-defined or multiple tangent lines may exist.

Differentiating Both Sides

You can differentiate both sides of an equation because if two expressions are equal for all values in some interval, their derivatives are also equal. This is the foundation that makes implicit differentiation valid.

Multiple Implicit Derivatives

Higher derivatives can be found by differentiating lower derivatives implicitly, then substituting back to eliminate $\frac{dy}{dx}$ in favor of $x$ and $y$.

Real-World Applications

Curves in Nature and Design

Many natural and designed curves are more easily described implicitly than explicitly. The equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ describes planetary orbits and the cross-sections of whispering galleries. While you could solve for $y$ explicitly, implicit differentiation provides a unified formula for the slope at any point without needing to consider upper and lower halves separately.

Level Curves and Contour Maps

Topographic maps show elevation through contour lines, each representing a constant height $h$. If elevation is given by $z = F(x, y)$, then each contour line satisfies $F(x, y) = c$ for some constant $c$. Implicit differentiation tells you the slope of these contour lines, which is essential for understanding terrain steepness and water flow patterns.

Economic Indifference Curves

In economics, an indifference curve shows all combinations of two goods that give a consumer equal satisfaction. If utility is $U(x, y) = c$, implicit differentiation gives the marginal rate of substitution:

$$\frac{dy}{dx} = -\frac{\partial U / \partial x}{\partial U / \partial y}$$

This tells economists how much of good $y$ a consumer would give up for one more unit of good $x$ while maintaining the same satisfaction.

Engineering Constraint Equations

Many engineering problems involve constraints that relate multiple variables. A gear system might satisfy $xy = k$ (constant angular momentum), or a structural element might satisfy a stress equilibrium equation. Implicit differentiation helps engineers understand how changing one variable affects another while maintaining the constraint.

Related Rates Problems

Implicit differentiation underlies related rates problems where multiple quantities change with time. If $x$, $y$, and $z$ are related by an equation and all depend on time $t$, differentiating implicitly with respect to $t$ reveals how their rates of change connect.

Thermodynamics and Phase Diagrams

Phase boundaries between solid, liquid, and gas states are often described by implicit equations relating pressure, volume, and temperature. Implicit differentiation helps determine how these variables must change together to stay on a phase boundary, which is crucial for understanding phase transitions.

Self-Test Problems

Test your understanding with these practice problems. Try each one before revealing the answer.

Summary

• Explicit functions give $y$ directly as $y = f(x)$, while implicit equations describe a relationship between $x$ and $y$ without isolating either variable.

• Implicit differentiation allows you to find $\frac{dy}{dx}$ without solving for $y$ explicitly. This is essential for curves like circles, ellipses, and more exotic shapes that cannot be written as single functions.

• The key technique is to differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Every time you differentiate an expression involving $y$, apply the chain rule and include the factor $\frac{dy}{dx}$.

• After differentiating, collect all terms with $\frac{dy}{dx}$ on one side, factor it out, and solve algebraically.

• The result typically contains both $x$ and $y$, which is expected. To find the slope at a specific point, substitute both coordinates into your expression for $\frac{dy}{dx}$.

• Horizontal tangents occur where the numerator of $\frac{dy}{dx}$ is zero and the denominator is nonzero. Vertical tangents occur where the denominator is zero and the numerator is nonzero.

• Second derivatives can be found by differentiating $\frac{dy}{dx}$ implicitly, then substituting the first derivative expression to simplify.

• Real-world applications include analyzing curves in design and nature, understanding contour maps, economic modeling with indifference curves, and solving engineering constraint problems.

The elegance of implicit differentiation lies in its directness. Instead of forcing an equation into a form it does not naturally take, you work with it as it stands. This technique will serve you well not only in calculus but in multivariable calculus, differential equations, and many areas of applied mathematics where relationships between variables are naturally implicit.