Implicit Differentiation: When y Hides Inside the Equation

Sometimes y isn't isolated. Sometimes you can't solve for it.

x² + y² = 25 defines a circle. You could solve for y = ±√(25 - x²), but then you have two functions, and square roots complicate things.

Here's the unlock: you don't need to solve for y. Differentiate both sides with respect to x, treating y as a function of x. Every time you differentiate y, a dy/dx pops out (chain rule). Then solve for dy/dx.

The equation stays implicit. The derivative becomes explicit.

The Basic Idea

When y depends on x but isn't explicitly solved for:

1. Differentiate both sides with respect to x
2. When you differentiate y (or any function of y), apply the chain rule: dy/dx appears
3. Solve for dy/dx

Example: Circle x² + y² = 25

Differentiate both sides with respect to x:

d/dx (x²) + d/dx (y²) = d/dx (25)

2x + 2y · dy/dx = 0

The d/dx(y²) = 2y · dy/dx comes from the chain rule: y² is a function of y, and y is a function of x.

Solve for dy/dx:

2y · dy/dx = -2x dy/dx = -x/y

At the point (3, 4) on the circle: dy/dx = -3/4

The slope of the circle at (3, 4) is -3/4.

Why the Chain Rule?

y is a function of x, even when we can't write it explicitly.

d/dx(y²) = d/dy(y²) · dy/dx = 2y · dy/dx

d/dx(y³) = 3y² · dy/dx

d/dx(sin(y)) = cos(y) · dy/dx

Anything involving y picks up a dy/dx factor when differentiated with respect to x.

Example: x³ + y³ = 6xy

This equation defines the "folium of Descartes."

Differentiate both sides:

3x² + 3y² · dy/dx = 6(1·y + x · dy/dx)

3x² + 3y² · dy/dx = 6y + 6x · dy/dx

Gather dy/dx terms:

3y² · dy/dx - 6x · dy/dx = 6y - 3x²

dy/dx (3y² - 6x) = 6y - 3x²

dy/dx = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x)

Example: Inverse Functions

Find d/dx(arcsin(x)).

Let y = arcsin(x), so sin(y) = x.

Differentiate implicitly:

cos(y) · dy/dx = 1

dy/dx = 1/cos(y)

Now express in terms of x. Since sin(y) = x, cos(y) = √(1 - sin²(y)) = √(1 - x²).

dy/dx = 1/√(1 - x²)

So d/dx(arcsin(x)) = 1/√(1 - x²).

Implicit differentiation derived the inverse trig derivative.

Example: Exponential Implicit

e^(xy) = x + y

Differentiate:

e^(xy) · d/dx(xy) = 1 + dy/dx

e^(xy) · (1·y + x · dy/dx) = 1 + dy/dx

e^(xy) · y + e^(xy) · x · dy/dx = 1 + dy/dx

Gather dy/dx:

e^(xy) · x · dy/dx - dy/dx = 1 - e^(xy) · y

dy/dx (x · e^(xy) - 1) = 1 - y · e^(xy)

dy/dx = (1 - y · e^(xy)) / (x · e^(xy) - 1)

Finding Second Derivatives

Differentiate dy/dx again, treating y as a function of x.

From x² + y² = 25, we found dy/dx = -x/y.

d²y/dx² = d/dx(-x/y)

Using quotient rule:

= (-1·y - (-x)·dy/dx) / y² = (-y + x·dy/dx) / y²

Substitute dy/dx = -x/y:

= (-y + x·(-x/y)) / y² = (-y - x²/y) / y² = (-y² - x²) / y³ = -25/y³ (since x² + y² = 25)

Horizontal and Vertical Tangents

Horizontal tangent: dy/dx = 0

For the circle: -x/y = 0, so x = 0. Points: (0, 5) and (0, -5).

Vertical tangent: dy/dx undefined (denominator = 0)

For the circle: y = 0. Points: (5, 0) and (-5, 0).

When to Use Implicit Differentiation

• The equation can't easily be solved for y
• You want dy/dx at a specific point without finding y(x) everywhere
• The curve has multiple y-values for some x-values
• Deriving inverse function derivatives

Logarithmic Differentiation

A powerful technique: take ln of both sides, then differentiate.

Find d/dx(x^x).

Let y = x^x. Then ln(y) = x·ln(x).

Differentiate:

(1/y) · dy/dx = 1·ln(x) + x·(1/x) = ln(x) + 1

dy/dx = y·(ln(x) + 1) = x^x(ln(x) + 1)

This handles cases where both base and exponent vary.

Common Mistakes

Forgetting dy/dx on y terms: Wrong: d/dx(y²) = 2y Right: d/dx(y²) = 2y · dy/dx

Not applying chain rule to nested functions: d/dx(sin(y)) = cos(y) · dy/dx, not just cos(y)

The Core Insight

Implicit differentiation lets you find dy/dx without isolating y.

Treat y as a function of x, even when you can't write it explicitly. The chain rule handles the dependency: every differentiation of y spawns a dy/dx.

This works because derivatives respect algebraic relationships. If x and y satisfy an equation, their rates of change satisfy the differentiated equation.

The y stays hidden. The derivative emerges anyway.

Part 7 of the Calculus Derivatives series.

Previous: The Quotient Rule: Derivatives of Fractions Next: Related Rates: How Changes Propagate Through Systems