Calculus I › Applications of derivatives › full formula sheet
Differentials
Turn the derivative into a measuring instrument — estimate tiny changes and bound measurement errors.
Notation: dx is a small change you choose in x; dy is the resulting change in y along the tangent line. The true change along the curve is Δy.
Before this lesson: Definition of the derivative · Linear approximation
Where it comes from
The problem: a ball’s radius is measured as 10 cm, but the ruler might be off by 0.1 cm. How wrong could the computed x²-style quantity be? Take the simpler y = x² at x = 10 with a wobble of dx = 0.1. The naive approach: compute f twice — 10.1² − 10² = 102.01 − 100 = 2.01. That works, but for a messy f it means two painful evaluations. The differential shortcut reads the change straight off the tangent line:
Picture it: dx is a horizontal step; dy is how far the tangent line rises over that step, while Δy is how far the curve rises. For small steps the two rises nearly agree — and dy is far cheaper to compute.
Before reading on: dy = f′(x) dx. What happens to the gap between dy and Δy as dx shrinks — and why does that make dy a measuring instrument?
Derivation
We start from the derivative as a limit, then define two new quantities — dx and dy — so that dy/dx becomes an honest ratio.
So is dy/dx “really” a fraction? After these definitions, yes: dy and dx are ordinary quantities and dy/dx = f′(x) exactly. The limit in Step 1 was the motivation; the definitions in Steps 2–3 make the ratio honest. (Purists note: this doesn’t make every “cancel the d’s” manipulation legal — but for differentials it is exactly what the notation was built for.)
Before reading on: the radius has a 1% measurement error. Predict the percentage error in the sphere’s volume before you compute it.
How to use it
The procedure, every time:
- Identify y = f(x), the base x, and the small change dx. In error problems dx is the measurement uncertainty (e.g. ±0.1). Keep its sign if direction matters.
- Compute dy = f′(x)·dx. Differentiate first, then substitute x and dx — never substitute before differentiating.
- Interpret. dy ≈ Δy: the estimated change in the quantity. For errors, report ±|dy|.
- Relative error: divide by y. dy/y estimates the fractional error — often cleaner than the absolute one (see Example 3).
When to reach for it
Any “small change in x, find the change in y” or “measurement error propagates” problem. It replaces two evaluations of f with one derivative and one multiplication.
dy is not Δy
dy rides the tangent; Δy rides the curve. They differ by about ½f″(c)(dx)². For small dx the gap is negligible — for large dx, compute Δy directly.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: y = x², x = 10, dx = 0.1
- Differentiate first. dy = 2x·dx. (Why 2x? Power rule on x² — before any numbers go in.)
- Substitute. dy = 2(10)(0.1) = 2.
- Compare with truth. Δy = 10.1² − 10² = 2.01. The estimate 2 is off by 0.01 ✓.
Your turn: y = x³, x = 5, dx = 0.2 — estimate Δy.
Answer: dy = 15.
dy = 3x² dx = 3(25)(0.2) = 15. True Δy = 5.2³ − 125 = 15.608 ✓.
Example 2 — error propagation: sphere volume, r = 10 ± 0.1 cm
- Set up. V = (4/3)πr³, r = 10, dr = 0.1. (Why dr? The radius is the measured quantity; its uncertainty is the dx.)
- Differentiate first. dV = 4πr²·dr.
- Substitute. dV = 4π(100)(0.1) = 40π ≈ 125.7 cm³.
- Compare. True ΔV = (4/3)π(10.1³ − 1000) = (4/3)π(30.301) ≈ 126.9 cm³. Error of the estimate: about 1% ✓.
Your turn: A cube’s side is s = 20 ± 0.2 cm — bound the volume error.
Answer: dV = 240 cm³.
V = s³, dV = 3s² ds = 3(400)(0.2) = 240. True ΔV = 20.2³ − 8000 = 242.4 ✓.
Example 3 — relative error: square side s = 5 ± 0.05
- Set up. A = s², s = 5, ds = 0.05.
- Differentiate. dA = 2s·ds = 2(5)(0.05) = 0.5.
- Relative error. dA/A = 0.5/25 = 0.02 = 2%, while ds/s = 0.05/5 = 1%. The area’s percentage error is double the side’s.
- The pattern. For y = xn: dy/y = n·dx/x. (Why? dy = nxn−1dx, divide by xn.) Powers multiply relative error by the exponent.
Your turn: A circle’s radius is r = 10 ± 0.1 — the relative error in the area?
Answer: 2%.
A = πr², dA = 2πr dr, so dA/A = 2 dr/r = 2(0.1)/10 = 0.02 = 2% ✓.
Example 4 — estimating a change: √49.5 − √49
- Set up. y = √x, x = 49, dx = 0.5. We want Δy without computing √49.5.
- Differentiate. dy = (1/(2√x))·dx.
- Substitute. dy = (1/(2·7))(0.5) = 0.5/14 ≈ 0.0357.
- Compare. True Δy = √49.5 − 7 ≈ 0.0356. The estimate matches to 4 decimal places ✓ — one division replaced a hard root.
Your turn: Estimate ∛√64.4 − ∛√64 without a calculator.
Answer: ≈ 0.00833.
y = x1/3, x = 64, dx = 0.4: dy = (1/(3·16))(0.4) = 0.4/48 ≈ 0.00833. Check: 4.00833³ ≈ 64.40 ✓.
Memorization tips
- Say it: “dee-why equals f-prime-of-x dee-ex.” Slope times run equals rise — along the tangent.
- Picture the triangle: run = dx, hypotenuse = tangent line, rise = dy. The curve’s rise Δy is a hair above or below.
- dx is the step, never the destination: for “x goes from 49 to 49.5”, dx = 0.5.
- Differentiate, then substitute. Numbers go in last — the #1 error is substituting first and differentiating a constant.
- Relative error shortcut: for y = xn, dy/y = n·dx/x. Exponents multiply percentage errors.
- dy ≈ Δy needs small dx. If dx isn’t small, dy is still defined — it’s just not a good estimate anymore.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and differentials are yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is a differential?
For y = f(x), the differential dy = f′(x)·dx is the change in y along the tangent line when x changes by dx — the linear part of the true change Δy.
What is the difference between dy and Δy?
Δy = f(x+dx) − f(x) is the exact change along the curve; dy = f′(x)·dx is the change along the tangent line. They agree closely for small dx: for y = x² at x = 10, dx = 0.1, dy = 2 and Δy = 2.01.
Why can dy/dx be treated like a fraction?
We define dx = Δx and dy = f′(x)·dx, so dy/dx = f′(x) holds as an honest ratio. The limit definition motivated the notation; the definitions make it exact.
When is dy a good estimate of Δy?
When dx is small and f is smooth near x. The gap is about ½f″(c)(dx)² — quadratic in dx, so tiny steps give tiny gaps.
What is relative error via differentials?
dy/y estimates the fractional error. For y = xn, dy/y = n·dx/x — a 1% error in a square’s side gives about a 2% error in its area.
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