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4.2 Linear Approximations and Differentials

4.2 Linear Approximations and Differentials

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus I
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Linear Approximations and Differentials

Linear approximations and differentials let you estimate function values without computing them exactly. The core idea: zoom in close enough on any smooth curve, and it starts to look like a straight line. That straight line (the tangent line) becomes your estimation tool.

These techniques show up constantly in physics, engineering, and later math courses whenever exact answers are impractical and a close estimate will do.

Linear Approximation at a Point

If a function is differentiable at a point, its tangent line hugs the curve closely in a small neighborhood around that point. This means you can swap out the actual (possibly complicated) function for the much simpler tangent line and get a solid estimate, as long as you stay near the point of tangency.

The further you move from that point, the worse the approximation gets. This is strictly a local tool.

Linear approximation at a point, Calculus - Wikipedia

Construction of Function Linearization

The linearization of f(x)f(x) at x=ax = a is just the equation of the tangent line at that point:

L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x - a)

Each piece has a clear role:

  • f(a)f(a): the known function value at the base point (your starting height)
  • f(a)f'(a): the slope of the tangent line (how fast ff is changing at aa)
  • (xa)(x - a): how far your input is from the base point

How to use it (step by step):

  1. Choose a base point aa where f(a)f(a) and f(a)f'(a) are easy to compute.

  2. Calculate f(a)f(a) and f(a)f'(a).

  3. Plug into L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x - a).

  4. Substitute the xx-value you want to estimate.

Example: Estimate sin(0.1)\sin(0.1) using linearization at a=0a = 0.

  1. f(x)=sin(x)f(x) = \sin(x), so f(0)=sin(0)=0f(0) = \sin(0) = 0

  2. f(x)=cos(x)f'(x) = \cos(x), so f(0)=cos(0)=1f'(0) = \cos(0) = 1

  3. L(x)=0+1(x0)=xL(x) = 0 + 1 \cdot (x - 0) = x

  4. L(0.1)=0.1L(0.1) = 0.1

The actual value is sin(0.1)0.0998\sin(0.1) \approx 0.0998, so the estimate of 0.10.1 is off by only about 0.00020.0002. That's the power of staying close to the base point.

Example: Estimate 4.1\sqrt{4.1} using linearization at a=4a = 4.

  1. f(x)=xf(x) = \sqrt{x}, so f(4)=2f(4) = 2

  2. f(x)=12xf'(x) = \frac{1}{2\sqrt{x}}, so f(4)=14f'(4) = \frac{1}{4}

  3. L(x)=2+14(x4)L(x) = 2 + \frac{1}{4}(x - 4)

  4. L(4.1)=2+14(0.1)=2.025L(4.1) = 2 + \frac{1}{4}(0.1) = 2.025

The actual value is 4.12.0248\sqrt{4.1} \approx 2.0248, so the estimate is very close.

Linear approximation at a point, Linear Approximations and Differentials · Calculus

Graphical Representation of Differentials

Differentials give you a way to talk about small changes in input and output separately.

For y=f(x)y = f(x), define:

  • dxdx: a small change in xx (this is an independent quantity you choose)
  • dydy: the corresponding estimated change in yy, given by dy=f(x)dxdy = f'(x) \, dx

The key distinction to keep straight:

  • dydy is the change along the tangent line (your estimate)
  • Δy=f(x+dx)f(x)\Delta y = f(x + dx) - f(x) is the actual change along the curve

Graphically, if you move dxdx units to the right from a point on the curve, dydy is the vertical distance traveled along the tangent line, while Δy\Delta y is the vertical distance traveled along the actual curve. The gap between dydy and Δy\Delta y is the approximation error, and it shrinks as dxdx gets smaller.

Example: For f(x)=x2f(x) = x^2 at x=3x = 3 with dx=0.01dx = 0.01:

  • dy=f(3)dx=6(0.01)=0.06dy = f'(3) \cdot dx = 6(0.01) = 0.06
  • Δy=f(3.01)f(3)=9.06019=0.0601\Delta y = f(3.01) - f(3) = 9.0601 - 9 = 0.0601

The differential dy=0.06dy = 0.06 is a quick, close estimate of the actual change 0.06010.0601.

Error Analysis in Differential Approximations

Since differentials give estimates, you need ways to measure how good those estimates are.

  • Absolute error = Δydy|\Delta y - dy|, the raw size of the mistake
  • Relative error = ΔydyΔy\frac{|\Delta y - dy|}{|\Delta y|}, the error as a fraction of the true change
  • Percentage error = relative error ×\times 100%

A small percentage error means the linear approximation did its job well. What counts as "acceptable" depends on context: engineering tolerances differ from back-of-the-envelope physics estimates.

The general pattern: the smaller dxdx is, the smaller the error. And functions with less curvature near the base point produce better approximations, because the tangent line stays closer to the curve.

Continuity and Differentiability

Linear approximation only works where the function is differentiable. Differentiability requires continuity, but goes further: the function must also have a well-defined derivative (no sharp corners, cusps, or vertical tangents).

If ff is differentiable at aa, then ff can be approximated by its tangent line near aa. If ff is not differentiable at aa, there's no tangent line to use, and the whole method breaks down at that point.

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