Differentials: dy, Δy and Error Estimates

Differentials estimate how much a function’s output changes when its input changes a little. If y = f(x) and x changes by a small amount dx, the differential of y is $dy = f'(x)\,dx$ . Then the true change Δy is close to dy. Differentials are often used to estimate measurement errors.

Before this lesson: Linear Approximation, Power Rule and Chain Rule

dy versus Δy

When x changes from x to x + dx, the curve rises by Δy = f(x + dx) − f(x). The tangent line rises by dy = f′(x) dx. For a small dx, the two are almost the same, and dy is much easier to compute.

Graph of y = x² (solid blue) with its tangent line at (1, 1) (dashed orange). A horizontal step dx = 0.8 goes from x = 1 to x = 1.8. The rise along the tangent line is dy = 1.6, and the rise along the curve is Δy = 2.24. The step is drawn large so the gap between dy and Δy is visible. — dy is the rise along the tangent line, and Δy is the rise along the curve. For small dx they are nearly equal.

This is the same idea as linear approximation, written in terms of changes instead of values.

Differential formulas

dy = f'(x)\,dx, \qquad \Delta y \approx dy

Term	Meaning	Formula
Absolute error	Estimated change or error in y	$dy = f'(x)\,dx$
Relative error	Error compared to the size of y	$\frac{dy}{y}$
Percent error	Relative error as a percent	$\frac{dy}{y} \times 100\%$

A useful shortcut: if $y = kx^n$ , then $\frac{dy}{y} = n\,\frac{dx}{x}$ . The percent error in y is n times the percent error in x. For a cube’s volume (n = 3), a 1% error in the side gives about a 3% error in the volume.

Differentials examples

Example 1: Finding a differential

Find dy for $y = x^3 - 2x$ .

\frac{dy}{dx} = 3x^2 - 2

Differentiate with the power rule.

dy = (3x^2 - 2)\,dx

Multiply both sides by dx.

Answer: $dy = (3x^2 - 2)\,dx$ .

Example 2: Comparing dy and Δy

Let $y = x^2$ . Find Δy and dy when x changes from 3 to 3.1.

\Delta y = (3.1)^2 - 3^2 = 9.61 - 9 = 0.61

The exact change in y.

dx = 3.1 - 3 = 0.1

The change in x.

dy = 2x\,dx = 2(3)(0.1) = 0.6

The differential at x = 3.

\Delta y - dy = 0.61 - 0.6 = 0.01

The difference is exactly (dx)² for this function.

Answer: Δy = 0.61 and dy = 0.6. The differential misses by only 0.01.

Example 3: Error in the volume of a cube

A cube’s side is measured as 10 cm, with a possible error of ±0.1 cm. Estimate the possible error in its volume, and the percent error.

V = s^3

Volume of a cube.

dV = 3s^2\,ds

Find the differential.

dV = 3(10)^2(0.1) = 30

Substitute s = 10 and ds = 0.1.

\frac{dV}{V} = \frac{30}{1000} = 0.03

Divide by V = 10³ = 1000.

0.03 \times 100\% = 3\%

Write as a percent.

Answer: the volume could be off by about ±30 cm³, a 3% error. The side’s 1% error tripled, matching the shortcut with n = 3.

Example 4: Error in a sphere’s surface area

A ball’s radius is measured as 5 cm ± 0.05 cm. Estimate the error in its surface area $S = 4\pi r^2$ , and the percent error.

dS = 8\pi r\,dr

Differentiate S = 4πr².

dS = 8\pi(5)(0.05) = 2\pi \approx 6.28

Substitute r = 5 and dr = 0.05.

S = 4\pi(5)^2 = 100\pi

Find the surface area.

\frac{dS}{S} = \frac{2\pi}{100\pi} = 0.02 = 2\%

Divide; π cancels.

Answer: the error is about ±6.28 cm², or 2%. The radius had a 1% error, and n = 2 doubles it.

Example 5: Word problem with paint

A dome is a hemisphere with radius 10 m. Estimate the volume of paint needed for a coat 1 mm thick.

V = \frac{2}{3}\pi r^3

Volume of a hemisphere.

dV = 2\pi r^2\,dr

Find the differential.

dr = 1 \text{ mm} = 0.001 \text{ m}

The coat adds 0.001 m to the radius. Use the same units.

dV = 2\pi(10)^2(0.001) = 0.2\pi

Substitute r = 10 and dr = 0.001.

0.2\pi \approx 0.628

Multiply by π.

Answer: about 0.628 m³, or 628 liters of paint. dV is the curved surface area 2πr² times the thickness, which makes sense for a thin layer.

Common mistakes

Mistake	Example of the error	Fix
Leaving off dx	$dy = 3x^2 - 2$	A differential has dx in it: $dy = (3x^2 - 2)\,dx$ .
Substituting the new x	Using x = 3.1 in dy = 2x dx	Use the starting value x = 3.
Mixing units	Using dr = 1 with r in meters for a 1-mm coat	Convert first: dr = 0.001 m.
Calling dy the exact change	“Δy = 0.6” for x = 3 to 3.1	dy is an estimate. The exact change is Δy = 0.61.
Confusing absolute and relative error	Reporting 30 as the percent error	Percent error is (dV/V) × 100% = 3%.

Related methods

Differentials and linear approximation

The estimate f(x + dx) ≈ f(x) + dy is the linearization L(x) written another way. Use whichever form fits the question: L(x) for a value, dy for a change.

Differentials in integration

The notation dy = f′(x) dx comes back in u-substitution for integrals, later in Calculus 1, where du = g′(x) dx is a key step.

Practice problems

Find dy for $y = \sqrt{x}$ .
Find dy for $y = \sin(2x)$ .
Find dy for $y = e^{x^2}$ .
For $y = x^3$ , compare dy and Δy when x changes from 2 to 2.01.
A circle’s radius is measured as 20 cm ± 0.2 cm. Estimate the error in its area and the percent error.
A square tile’s side is 12 cm ± 0.05 cm. Estimate the error in its area and the percent error.
Estimate the change in a sphere’s volume when its radius grows from 6 cm to 6.1 cm.
Use differentials to estimate $\sqrt{4.1}$ .
A pendulum’s period is $T = k\sqrt{L}$ for a constant k. If L is measured with a 2% error, about what percent error results in T?
A can has radius 3 cm and height 10 cm. Estimate the change in volume if the radius grows by 0.1 cm and the height stays the same.

Show answers

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

Differentiate x^(1/2).

dy = \frac{dx}{2\sqrt{x}}

Multiply both sides by dx.

Answer: $dy = \frac{dx}{2\sqrt{x}}$

dy = 2\cos(2x)\,dx

Chain rule: the inside 2x has derivative 2.

Answer: $dy = 2\cos(2x)\,dx$

dy = 2x\,e^{x^2}\,dx

Chain rule on e^(x²).

Answer: $dy = 2xe^{x^2}\,dx$

dy = 3x^2\,dx = 3(4)(0.01) = 0.12

Use x = 2 and dx = 0.01.

\Delta y = (2.01)^3 - 8 = 8.120601 - 8 = 0.120601

Exact change.

Answer: dy = 0.12 and Δy = 0.120601

dA = 2\pi r\,dr = 2\pi(20)(0.2) = 8\pi \approx 25.1

Differentiate A = πr² and substitute.

\frac{dA}{A} = \frac{8\pi}{400\pi} = 0.02

A = π(20)² = 400π.

Answer: about ±25.1 cm², a 2% error

dA = 2s\,ds = 2(12)(0.05) = 1.2

Differentiate A = s² and substitute.

\frac{dA}{A} = \frac{1.2}{144} \approx 0.0083

A = 12² = 144.

Answer: about ±1.2 cm², about 0.83%

dV = 4\pi r^2\,dr = 4\pi(36)(0.1) = 14.4\pi

Differentiate V = (4/3)πr³ and substitute r = 6, dr = 0.1.

14.4\pi \approx 45.2

Multiply by π.

Answer: about 45.2 cm³

dy = \frac{dx}{2\sqrt{x}} = \frac{0.1}{2(2)} = 0.025

Use x = 4 and dx = 0.1.

\sqrt{4.1} \approx 2 + 0.025 = 2.025

Add dy to √4 = 2.

Answer: about 2.025

\frac{dT}{T} = \frac{1}{2}\cdot\frac{dL}{L}

T = kL^(1/2), so n = ½ in the shortcut.

\frac{dT}{T} = \frac{1}{2}(2\%) = 1\%

Substitute the 2% error.

Answer: about 1%

10.

V = \pi r^2 h \Rightarrow dV = 2\pi rh\,dr

Differentiate with h fixed.

dV = 2\pi(3)(10)(0.1) = 6\pi \approx 18.8

Substitute r = 3, h = 10 and dr = 0.1.

Answer: about 18.8 cm³

Differentials FAQ

What is a differential in calculus?

If y = f(x), the differential dy = f′(x) dx is the change in y along the tangent line when x changes by dx. It estimates the true change Δy.

What is the difference between dy and Δy?

Δy is the exact change in the function’s value. dy is the change along the tangent line. They are close when dx is small.

How do you use differentials to estimate error?

Treat the measurement error as dx and compute dy = f′(x) dx. Divide by y for the relative error, and multiply by 100% for the percent error.

Why does the percent error triple for a cube’s volume?

V = s³, so dV/V = 3 ds/s. Each power of s adds the side’s relative error once.

Summary

$dy = f'(x)\,dx$ estimates the change Δy.
Use the starting x, not the new x.
Relative error is dy/y; percent error is that times 100%.
For $y = kx^n$ , the percent error in y is n times the percent error in x.
Keep the units consistent before you substitute.

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