§ 01The Core Idea
The derivative tells us the slope of a function at a single point. If we use that slope to draw a tangent line, that line stays close to the curve for small distances, close enough to use as an approximation.
This comes directly from the definition of the derivative. Writing f′(a) = limx→a [f(x)−f(a)]/(x−a) and rearranging for x near a gives f(x) ≈ f(a) + f′(a)(x−a). That is the entire idea, the rest is application and precision.
In practice this matters because many functions (square roots, cube roots, trigonometric expressions, logarithms) are hard to evaluate by hand for arbitrary inputs. A nearby point where the value is easy to compute (like a = 4 for √x, since √4 = 2) lets us slide along the tangent line to reach the target with almost no work.
§ 02The Linearisation Formula
The tangent line to f at x = a is called the linearisation of f at a, written L(x). It is defined by:
Each piece has a clear job. f(a) is the starting height, the exact value of the function at the base point. f′(a) is the slope of the tangent line: how steeply the function is rising or falling right there. (x − a) is how far we travel horizontally. Multiplying slope by horizontal distance gives the vertical correction on top of f(a).
Small-x Approximations (a = 0)
When a = 0 the formula collapses to L(x) = f(0) + f′(0)·x. Applied to standard functions this yields a short table worth memorising for |x| small:
Valid when |x| ≪ 1. The small-angle approximation sin θ ≈ θ (the foundation of the simple pendulum formula in physics) is just the case n = 1 applied to sin.
§ 03Worked Examples — Building Linearisations
Evaluate at the base point.
f(4) = 2, f′(x) = 1/(2√x) → f′(4) = 1/4Write the linearisation.
L(x) = 2 + (1/4)(x − 4)Substitute x = 4.1 (so x − 4 = 0.1).
L(4.1) = 2 + (1/4)(0.1) = 2.025
- f(8) = 2, f′(x) = (1/3)x^(−2/3) → f′(8) = 1/12
- L(x) = 2 + (1/12)(x − 8)
- L(8.06) = 2 + 0.06/12 = 2 + 0.005 = 2.005
- f(0) = 1, f′(0) = 1 → L(x) = 1 + x
- L(0.1) = 1.1
- f(1) = 0, f′(x) = 1/x → f′(1) = 1
- L(x) = 0 + 1·(x−1) = x − 1
- L(1.08) = 0.08
Closest perfect square: 64. Use a = 64.
- f(64) = 8, f′(64) = 1/16
- L(63) = 8 + (1/16)(63 − 64) = 8 − 1/16 = 7.9375
How Accuracy Falls with Distance from a
The table below shows L(x) = 2 + (x−4)/4 against true √x values as x moves away from a = 4:
| x | True √x | L(x) | Abs. Error | Accuracy |
|---|---|---|---|---|
| 4.1 | 2.02485 | 2.025 | 0.00015 | 99.99% |
| 4.5 | 2.12132 | 2.125 | 0.00368 | 99.8% |
| 5.0 | 2.23607 | 2.250 | 0.01393 | 99.4% |
| 6.0 | 2.44949 | 2.500 | 0.05051 | 97.9% |
| 9.0 | 3.00000 | 3.250 | 0.25000 | 91.7% |
The lesson: linearisation is excellent for small steps and degrades for large ones. Always choose the base point as close as possible to your target.
§ 04Differentials — dy and dx
Differentials reframe linearisation as a language about change rather than value. Instead of asking "what is f(x)?", we ask "if x changes by a small amount, how much does y change?"
For y = f(x) differentiable, define dx as any real number (think of it as a small change in x). Then the differential of y is:
They are equal only when f is linear. For all other functions, Δy − dy is the error, which shrinks faster than dx as dx → 0.
Differential Rules
Every derivative rule has a matching differential rule. Since dy = f′(x) dx, just multiply through by dx:
§ 05Differentials — Worked Examples
- Differentiate: dy/dx = 3x² − 5.
- Multiply by dx: dy = (3x² − 5) dx.
- Chain rule: dy/dx = 2x cos(x²).
- dy = 2x cos(x²) dx.
- a = 2, dx = 0.01. dy = 4x³ dx.
- At x = 2: dy = 4(8)(0.01) = 0.32.
- True change: Δy = (2.01)⁴ − 2⁴ = 16.3224… − 16 = 0.3224.
§ 06Error Estimation and Propagation
In science and engineering, measured quantities always carry uncertainty. If x is measured with error ±Δx, differentials let us estimate the resulting uncertainty in any function y = f(x).
The size of the error in y, in the same units as y. Use when you care about the raw deviation.
Dimensionless, the fractional or percentage error. Usually more meaningful than absolute error.
- V = (4/3)πr³ → dV = 4πr² dr.
- |dV| = 4π(25)(0.02) = 2π ≈ 6.28 cm³.
- Relative error: dV/V = 3 dr/r = 3(0.02)/5 = 1.2%.
§ 07Common Mistakes
§ 08Test Yourself — 10-Question Quiz
What is the linearisation formula for f at x = a?
L(x) = f(a) + f′(a)(x − a). This is the tangent line equation: value f(a) at the base point, plus slope f′(a) times horizontal distance (x − a).
Using the linearisation of √x at a = 9, approximate √9.06. Enter your answer.
- f(9)=3, f′(9)=1/6. L(x)=3+(1/6)(x−9).
- L(9.06)=3+(1/6)(0.06)=3+0.01=3.01.
For small x, eˣ ≈ ?
Linearise eˣ at a=0: f(0)=1, f′(0)=1. L(x)=1+x. This is the first-order Taylor approximation. Option (c) is the second-order approximation.
Find dy for y = x³ − 4x at x = 2 with dx = 0.05.
- dy=(3x²−4)dx. At x=2: dy=(12−4)(0.05)=8×0.05=0.40.
What is the key difference between Δy and dy?
Δy = f(x+Δx)−f(x) is the exact change along the curve. dy = f′(x) dx is the linear approximation: the change along the tangent line. Equal only when f is itself linear.
A cube has side s = 10 cm ± 0.1 cm. Using differentials, the absolute error in V = s³ is:
- dV = 3s² ds = 3(100)(0.1) = 30 cm³.
- Relative error: 3(0.1)/10 = 3%.
To approximate √63, which base point is best?
a = 64. It gives f(64) = 8 (clean integer) and is only 1 unit from the target. a = 49 is 14 units away, much less accurate.
Approximate ln(0.97) using the linearisation of ln x at a = 1.
- L(x) = x − 1 (from Example 4).
- L(0.97) = 0.97 − 1 = −0.03.
- True: ln(0.97) = −0.03046…: error under 2%.
If y = xⁿ and the relative error in x is 2%, what is the relative error in y?
dy/y = n · dx/x. If dx/x = 2%, then dy/y = 2n%. For a sphere (n=3) a 2% radius error gives 6% volume error.
Linearisation is the first term of which series expansion?
L(x) = f(a) + f′(a)(x−a) is the degree-1 Taylor polynomial. Adding f″(a)(x−a)²/2 gives the degree-2 approximation. The full Taylor series includes all higher-order terms.
Quiz complete!