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Applications of Differentiation

3.9 Differentials

Focus on differentials, linear approximation, propagating small input changes.

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Ideas in this lesson

01DifferentialsWorked steps · five questions02Linear approximationWorked steps · five questions03Propagating small input changesWorked steps · five questions
Idea 1 of 3

Differentials

The linear change dy=f′(x)dxdy = f'(x) dx approximates the actual output change for small dxdx.

Worked example 1

For y=x2y = x^{2} at x=2x = 2 with dx=0.01dx = 0.01, find dydy.
  1. dy=f′(x)dxdy = f'(x) dx; f′(x)=2xf'(x) = 2x
  2. dy=2(2)(0.01)=125≈0.04dy = 2(2)(0.01) = \dfrac{1}{25} \approx 0.04

Answer: 125≈0.04\dfrac{1}{25} \approx 0.04

Worked example 2

For y=xy = \sqrt{x} at x=3x = 3 with dx=0.02dx = 0.02, find dydy.
  1. dy=f′(x)dxdy = f'(x) dx; f′(x)=12 xf'(x) = \dfrac{1}{2\,\sqrt{x}}
  2. dy=12 3(0.02)=2200 3≈0.00577dy = \dfrac{1}{2\,\sqrt{3}}(0.02) = \dfrac{2}{200\,\sqrt{3}} \approx 0.00577

Answer: 2200 3≈0.00577\dfrac{2}{200\,\sqrt{3}} \approx 0.00577

Idea 2 of 3

Linear approximation

Use L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a) near a convenient base point aa.

Worked example 1

Use linearization at x=4x = 4 to approximate 4.02\sqrt{4.02}.
  1. f(4)=2f(4) = 2; f′(4)=14f'(4) = \dfrac{1}{4}
  2. L(x)=2+x−44L(x) = 2 + \dfrac{x - 4}{4}
  3. L(4.02)=2+0.024=401200≈2L(4.02) = 2 + \dfrac{0.02}{4} = \dfrac{401}{200} \approx 2

Answer: 401200≈2\dfrac{401}{200} \approx 2

Worked example 2

Use linearization at x=9x = 9 to approximate 9.04\sqrt{9.04}.
  1. f(9)=3f(9) = 3; f′(9)=16f'(9) = \dfrac{1}{6}
  2. L(x)=3+x−96L(x) = 3 + \dfrac{x - 9}{6}
  3. L(9.04)=3+0.046=451150≈3.01L(9.04) = 3 + \dfrac{0.04}{6} = \dfrac{451}{150} \approx 3.01

Answer: 451150≈3.01\dfrac{451}{150} \approx 3.01

Idea 3 of 3

Propagating small input changes

Keep the original input in the derivative. The differential approximates error; it is not the exact difference.

Worked example 1

For y=x2y = x^{2} at x=2x = 2 with dx=0.01dx = 0.01, find dydy.
  1. dy=f′(x)dxdy = f'(x) dx; f′(x)=2xf'(x) = 2x
  2. dy=2(2)(0.01)=125≈0.04dy = 2(2)(0.01) = \dfrac{1}{25} \approx 0.04

Answer: 125≈0.04\dfrac{1}{25} \approx 0.04

Worked example 2

For y=xy = \sqrt{x} at x=3x = 3 with dx=0.02dx = 0.02, find dydy.
  1. dy=f′(x)dxdy = f'(x) dx; f′(x)=12 xf'(x) = \dfrac{1}{2\,\sqrt{x}}
  2. dy=12 3(0.02)=2200 3≈0.00577dy = \dfrac{1}{2\,\sqrt{3}}(0.02) = \dfrac{2}{200\,\sqrt{3}} \approx 0.00577

Answer: 2200 3≈0.00577\dfrac{2}{200\,\sqrt{3}} \approx 0.00577

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.9,p3.9, p. 231231. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.