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Differentials

The Notation Behind "dx" and "dy"

You've been writing $\frac{dy}{dx}$ for months, treating it as a single symbol for "the derivative." But what if $dy$ and $dx$ were actually separate quantities that you could manipulate individually?

That's exactly what differentials are. They turn the derivative notation into something you can work with algebraically—and this notation becomes essential when you learn integration by substitution and differential equations.

The differential $dy$ represents how much the tangent line rises when $x$ changes by $dx$. It's the linearization repackaged in a form that's incredibly useful for calculations.

Prerequisite Map

Quick Reference

Property	Value
Concept	Linear Approximations and Differentials
Chapter	2.9
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Definition of Differentials

If $y = f(x)$ where $f$ is differentiable, then:

• $dx$ is an independent variable — it can be any real number (typically small)
• $dy$ is a dependent variable defined by:

$$\boxed{dy = f'(x) \cdot dx}$$

This says: the differential $dy$ equals the derivative times $dx$.

Geometric Meaning: $\Delta y$ vs $dy$

Here's the crucial distinction:

Quantity	Definition	Geometric Meaning
$\Delta y$	$f(x + \Delta x) - f(x)$	Actual change in $f$ along the curve
$dy$	$f'(x) \cdot dx$	Change along the tangent line

When $dx = \Delta x$:

        y
        │           curve y = f(x)
        │               ●─── Q (actual point)
        │              /│
        │             / │ Δy
        │            /  │
        │        ●─●────┤
        │       P  R    │ dy
        │          │    │
        │──────────┼────┼───── x
                   x   x+Δx

  P = (x, f(x))
  Q = (x + Δx, f(x + Δx))
  R = point on tangent line at x + Δx

  dy = height from P to R (tangent line rise)
  Δy = height from P to Q (actual curve rise)

Key insight: As $\Delta x \to 0$, we have $\Delta y \to dy$. The differential $dy$ approximates the actual change $\Delta y$.

Computing Differentials

To find $dy$ for a function:

1. Compute $f'(x)$
2. Write $dy = f'(x) \cdot dx$

Example: For $y = x^3 - 2x$:

• $f'(x) = 3x^2 - 2$
• $dy = (3x^2 - 2)\,dx$

Evaluating Differentials at Specific Values

To find a numerical value for $dy$:

1. Substitute a specific $x$ value
2. Substitute a specific $dx$ value
3. Multiply

Example: For $y = x^3 - 2x$ with $x = 2$ and $dx = 0.1$:

• $dy = (3(2)^2 - 2)(0.1) = (12 - 2)(0.1) = 10(0.1) = 1$

Connection to Linearization

The linear approximation $f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$ can be rewritten as:

$$f(a + dx) \approx f(a) + dy$$

So $dy$ represents the approximate change in $f$ predicted by the linearization.

Practice Problems

CCI-Style Conceptual Questions

Mastery Checklist

• [ ] I can define $dx$ and $dy$ and explain what each represents
• [ ] I can compute the differential $dy = f'(x)\,dx$ for any differentiable function
• [ ] I can evaluate $dy$ at specific values of $x$ and $dx$
• [ ] I can distinguish between $\Delta y$ (actual change) and $dy$ (tangent line change)
• [ ] I can use differentials to estimate changes in function values
• [ ] I understand why $dy \approx \Delta y$ for small $dx$

Mental Model

The "Shadow" Analogy:

Imagine a curved road at night, with a flashlight shining straight ahead at one point. The light beam creates a straight shadow on the ground—that's the tangent line.

• If you drive $dx$ meters along the road, your altitude changes by $\Delta y$
• The shadow (tangent line) would rise/fall by $dy$

For short distances, the shadow closely matches reality. For longer distances, the curve deviates from its shadow.

$dy$ is the "shadow's prediction" of how much you'll climb or descend. It's easy to compute (just multiply slope by distance) and works great for small movements.

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Connections

Looking back:

• Linearization introduced $L(x) = f(a) + f'(a)(x-a)$; differentials rewrite this as $f(a+dx) \approx f(a) + dy$
• The derivative $\frac{dy}{dx}$ literally becomes the ratio of differentials

Looking ahead:

• Error estimation applies $dy$ to propagate measurement uncertainties
• In u-substitution, the relationship $du = g'(x)dx$ is central
• Separable differential equations manipulate $dy$ and $dx$ as separate quantities

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Last updated: 2026-01-22