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:
- Compute $f'(x)$
- 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 $dy$:
- Substitute a specific $x$ value
- Substitute a specific $dx$ value
- 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
Find the differential $dy$ for $y = 5x^2 - 3x + 7$.
Find $dy$ for $y = \sqrt{1 + x^2}$.
For $y = x^2 + 2x$, compute both $\Delta y$ and $dy$ when $x = 1$ and $\Delta x = dx = 0.1$.
How close is $dy$ to $\Delta y$?
Use differentials to estimate the value of $\cos(62°)$.
Hint: Work in radians. Use $\pi/3 = 60°$ as your base point.
Prove the following differential rules, where $u$ and $v$ are differentiable functions of $x$:
(a) $d(u + v) = du + dv$
(b) $d(uv) = u\,dv + v\,du$
(c) $d\left(\frac{u}{v}\right) = \frac{v\,du - u\,dv}{v^2}$
CCI-Style Conceptual Questions
A student computes $dy = 2.4$ and $\Delta y = 2.53$ for some function.
Which statement is true?
(A) The tangent line rose by $2.4$ and the curve rose by $2.53$
(B) The tangent line rose by $2.53$ and the curve rose by $2.4$
(C) Both values represent the same quantity with rounding error
(D) The values cannot both be correct since $dy$ should equal $\Delta y$
If $f'(3) = -2$ and $dx = 0.5$, what does the sign of $dy$ tell you?
(A) The function is increasing at $x = 3$
(B) The tangent line rises as we move right from $x = 3$
(C) The tangent line falls as we move right from $x = 3$
(D) The concavity is negative at $x = 3$
Common Misconceptions
$dy$ and $\Delta y$ represent the same change in output and can be used interchangeably.
This is the height-vs-slope error. The differential $dy = f'(x)\,dx$ measures the vertical change along the tangent line, while $\Delta y = f(x + \Delta x) - f(x)$ measures the actual vertical change along the curve. For $y = x^2 + 2x$ at $x = 1$ with $\Delta x = 0.1$, the actual change is $\Delta y = 0.41$ and the differential gives $dy = 0.4$; these are close but not equal, and the difference grows as $\Delta x$ increases.
the differential $dx$ is an infinitely small quantity that cannot be given a specific value.
This is the concept-image-conflicts-definition error. In formal calculus, $dx$ is not an infinitesimal; it is an ordinary real number chosen by the user. Writing $dy = f'(x)\,dx$ with $x = 2$ and $dx = 0.1$ gives $dy = (3(4) - 2)(0.1) = 1.0$ for $y = x^3 - 2x$, a concrete computation with a concrete number. The confusion between $dx$ as a “variable that approaches zero” (from limits) and $dx$ as a “specific chosen increment” (from differentials) arises from conflating two distinct uses of the notation.
Mastery Checklist
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 shadow is 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.
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
| Previous | Up | Next |
|---|---|---|
| Linearization | Skills Index | Error Estimation |
Last updated: 2026-01-22