Differentials

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Textbook Reference

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Try This First: The Tangent as Approximation

For $f(x) = x^2$, the tangent line at $x = 3$ is $y = 6x - 9$ (slope 6, passing through $(3, 9)$).

Use this tangent line to estimate $f(3.1) = (3.1)^2$.

Predict: the tangent-line estimate and the actual value will be close but not equal. How close?

Tangent-line estimate: $y = 6(3.1) - 9 = 18.6 - 9 = 9.6$.

Actual value: $(3.1)^2 = 9.61$.

Error: $0.01$. This error is the difference $\Delta y - dy$: the actual change in $y$ minus the tangent-line approximation. The differential $dy$ captures the tangent-line part of the change.

Quantity-First Framing

When the input $x$ changes by a small amount $\Delta x$:

• The actual change in output is $\Delta y = f(x + \Delta x) - f(x)$ -- exact but requires computing $f$ at two points.
• The approximated change is $dy = f'(x)\,\Delta x$ -- fast to compute, uses only the slope at $x$.

For small $\Delta x$, the approximation $\Delta y \approx dy$ is accurate. The differential $dy$ is the change you read off the tangent line; $\Delta y$ is the actual change along the curve.

Prerequisite Check

• Find the tangent line equation at a given point
• Compute $f(x + h) - f(x)$ for small $h$ numerically
• State that the tangent line approximates the function near a point

Quick Reference

Differential notation:

Let $x$ change by $dx$ (an arbitrary nonzero number). The differential of $y = f(x)$ is: \[ dy = f'(x)\,dx. \]

This is the vertical change along the tangent line when the horizontal change is $dx$.

Key relationship: \[ \Delta y \approx dy \quad \text{when } dx \text{ is small}. \]

The tangent-line approximation (linearization): \[ f(x + dx) \approx f(x) + f'(x)\,dx. \]

Key Concepts

1. $\Delta y$ vs. $dy$: The Exact and the Approximate

Example 1. For $f(x) = x^3$ at $x = 2$, compute $\Delta y$ and $dy$ when $dx = 0.1$.

$\Delta y = f(2.1) - f(2) = (2.1)^3 - 8 = 9.261 - 8 = 1.261$.

$dy = f'(2)\,dx = 3(4)(0.1) = 1.2$.

Error: $\Delta y - dy = 1.261 - 1.2 = 0.061$.

The approximation $dy = 1.2$ is close to the exact $\Delta y = 1.261$. For smaller $dx$, the approximation improves.

Two representations:

Geometric: On the graph of $f(x) = x^3$, at the point $(2, 8)$:

• $\Delta y = 1.261$ is the vertical rise along the curve from $x = 2$ to $x = 2.1$.
• $dy = 1.2$ is the vertical rise along the tangent line over the same horizontal distance.

Tabular: As $dx$ shrinks, the error $\Delta y - dy$ shrinks faster (proportional to $dx^2$):

The error goes to zero faster than $dx$.

2. Using $dy$ to Estimate Function Values

Example 2. Estimate $\sqrt{4.08}$ using differentials.

Choose $f(x) = \sqrt{x}$, $x = 4$ (a nearby point where we know the exact value), $dx = 0.08$.

$f'(x) = \dfrac{1}{2\sqrt{x}}$. $f'(4) = \dfrac{1}{4}$.

$dy = f'(4)\,dx = \frac{1}{4}(0.08) = 0.02$.

$f(4.08) \approx f(4) + dy = 2 + 0.02 = 2.02$.

Actual: $\sqrt{4.08} \approx 2.0199$. Excellent approximation.

3. Geometric Meaning

On the graph of $y = f(x)$ at point $(x, f(x))$:

• $dx$ is a horizontal displacement.
• $dy = f'(x)\,dx$ is the corresponding vertical displacement ALONG THE TANGENT LINE.
• $\Delta y$ is the vertical displacement ALONG THE CURVE.

The tangent line (with slope $f'(x)$) is the best linear approximation to the curve near $(x, f(x))$. The differential $dy$ captures this linear approximation. The error $\Delta y - dy$ is the deviation of the curve from its tangent line, which is quadratic in $dx$ for smooth $f$.

4. Ask Why: Why Does the Error Go to Zero So Fast?

By Taylor’s theorem (or the definition of differentiability): \[ f(x + dx) = f(x) + f'(x)\,dx + \frac{f''(x)}{2}(dx)^2 + \cdots \]

The differential $dy = f'(x)\,dx$ captures the linear term. The error $\Delta y - dy \approx \frac{f''(x)}{2}(dx)^2$ is quadratic in $dx$: when $dx$ is halved, the error is quartered. This rapid convergence is why linear approximation (the differential) is so useful: even for moderate $dx$, the approximation is quite good if $|f''(x)|$ is small.

5. Multiple Valid Paths: Leibniz Notation

The equation $dy = f'(x)\,dx$ can be rewritten as $\dfrac{dy}{dx} = f'(x)$. This is the Leibniz notation for the derivative -- the ratio of the differential of $y$ to the differential of $x$. The notation makes the chain rule read as a fraction multiplication: \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}. \]

This is “one way to see the chain rule” (as cancellation of differentials); another way is the limit definition. Both views are valid and complement each other.

Named Misconception: limit-as-unreachable-barrier

A differential $dx$ is sometimes thought of as “infinitely small” or as a value that “cannot be reached.” In rigorous analysis, $dx$ is not an infinitesimal; it is a finite, nonzero number -- the horizontal displacement you choose. The approximation $\Delta y \approx dy$ becomes more accurate as $dx$ gets smaller, but $dx$ itself is a specific number you pick.

Thinking of $dx$ as an unreachable barrier (“the limit we approach but never reach”) confuses the differential with the limit concept. A differential is a concrete, computable thing: choose $dx$, compute $dy = f'(x)\,dx$.

Common Errors

Common Misconceptions

Leveled Practice

Level 1 -- Computing Differentials

Problem 1. Find $dy$ for $y = 3x^2 - 5x$ and evaluate when $x = 2$, $dx = 0.1$.

Problem 2. Estimate $(2.03)^5$ using differentials.

Level 2 -- Geometric Meaning

Problem 3. For $f(x) = x^2$, compute $\Delta y$ and $dy$ when $x = 3$ and $dx = 0.5$. Compute the error $\Delta y - dy$.

Level 3 -- Low-Floor-High-Ceiling Extension

Problem 4 (Extension).

(a) (Floor) Use differentials to estimate $\cos(61^\circ)$. Hint: $61^\circ = \pi/3 + \pi/180$ radians; use $x_0 = \pi/3$, $dx = \pi/180$.

(b) (Mid) For $f(x) = e^x$ near $x = 0$: show that $dy = e^x\,dx = dx$ when $x = 0$. This means $e^{dx} \approx 1 + dx$ for small $dx$. Verify numerically with $dx = 0.1$.

(c) (Ceiling) Compute the error $\Delta y - dy$ for $f(x) = x^n$ at a general $x$ with change $dx$, and show the error is $\binom{n}{2}x^{n-2}(dx)^2 + O(dx^3)$. What does this mean for large $n$?

Mastery Checklist

• State the differential $dy = f'(x)\,dx$ and distinguish it from $\Delta y$
• Use the differential to estimate $f(x_0 + dx)$ for a given function and small $dx$
• Choose an appropriate base point $x_0$ when estimating a function value
• Interpret $dx$ and $dy$ geometrically as horizontal and vertical displacements along the tangent line
• Explain why the error $\Delta y - dy$ is proportional to $(dx)^2$

Mental Model

The differential is the tangent-line approximation, expressed in terms of changes rather than absolute values.

When the input changes by $dx$, the tangent line says the output changes by $dy = f'(x)\,dx$ -- this is the slope times the run. The actual output changes by $\Delta y = f(x + dx) - f(x)$, which includes the curvature. For small $dx$, the curvature correction is tiny (quadratic in $dx$), so $dy \approx \Delta y$.

Choosing the base point $x_0$ at a “clean” number (one where $f(x_0)$ and $f'(x_0)$ are easy to compute) makes the estimate accurate and convenient.

Connections

Within MATH161

• Linearization: $f(x_0 + dx) \approx f(x_0) + f'(x_0)\,dx$ is the same as the tangent-line equation -- differentials and linearization are two names for the same approximation.
• Error estimation (next): Using $dy$ to bound the error in a measurement is the direct application of this lesson to physical measurements.
• Integration (Ch. 4): The notation $\int f(x)\,dx$ uses $dx$ as a differential; the integral sums up infinitely many differentials, each contributing $f(x)\,dx$ to the area.

Back to Calculus I Skills | Next: Error Estimation with Differentials