Core Concept
The Big Idea
When x changes by a small amount Δx, the corresponding change in y = f(x) is:
Δy = f(x + Δx) – f(x) (actual change)
But using derivatives, we approximate this as:
dy ≈ Δy when Δx is very small (Δx → 0)
Symbol
Meaning
Δx
Small change in x (actual, can be exact)
Δy
Actual change in y = f(x+Δx) – f(x)
dx
Differential of x = Δx (when small)
dy
Approximate change in y = f'(x)·dx ≈ Δy
Approximation Formula
dy = f'(x) · dx ⟹ f(x + Δx) ≈ f(x) + f'(x)·Δx
Δy vs dy
Δy (Actual Change)
Δy = f(x+Δx) – f(x)
Exact calculation needed
Exact calculation needed
dy (Approximate Change)
dy = f'(x)·dx
Quick, easy estimation
Quick, easy estimation
Key Points
- 1dx = Δx when we use approximation. Both represent the same small change in x.
- 2The approximation is more accurate when Δx is smaller (closer to zero).
- 3Relative error = Δy/y ≈ dy/y; Percentage error = (dy/y) × 100%.
- 4Used to find approximate values like √25.1, ∛8.02, (1.99)³ without a calculator.
- 5Strategy: Pick a nearby perfect value for x, then apply Δx as the difference.
Strategy: To find f(a + Δx) ≈ f(a) + f'(a)·Δx, choose 'a' as the nearest number whose exact value you know.
Solved Examples
Example 01
Approximate value of √25.3
Q. Find approximate value of √25.3.
Let f(x) = √x = x^(1/2) → f'(x) = 1/(2√x)
Take x = 25 (perfect square), Δx = 0.3
f(25) = √25 = 5
f'(25) = 1/(2·5) = 1/10 = 0.1
f(25.3) ≈ f(25) + f'(25)·(0.3) = 5 + 0.1 × 0.3 = 5 + 0.03
∴ √25.3 ≈ 5.03
Example 02
Approximate value of (2.01)⁵
Q. Find approximate value of (2.01)⁵.
Let f(x) = x⁵ → f'(x) = 5x⁴
Take x = 2, Δx = 0.01
f(2) = 2⁵ = 32
f'(2) = 5·(2⁴) = 5·16 = 80
f(2.01) ≈ 32 + 80 × 0.01 = 32 + 0.8 = 32.8
∴ (2.01)⁵ ≈ 32.8
Example 03
Finding Δy and dy for comparison
Q. For y = x² + 2x, find Δy and dy when x = 2 and Δx = 0.1.
Δy = f(2.1) – f(2) = [(2.1)² + 2(2.1)] – [4 + 4]
= [4.41 + 4.2] – 8 = 8.61 – 8 = 0.61
dy = f'(x)·dx = (2x + 2)·Δx = (2·2+2)·0.1 = 6 × 0.1 = 0.6
Δy = 0.61 (actual) | dy = 0.60 (approximate) → Very close!
Example 04
Approximate value of ∛8.05
Q. Find approximate value of ∛8.05.
f(x) = x^(1/3) → f'(x) = (1/3)·x^(–2/3) = 1/(3·x^(2/3))
Take x = 8, Δx = 0.05
f(8) = ∛8 = 2
f'(8) = 1/(3·8^(2/3)) = 1/(3·4) = 1/12
f(8.05) ≈ 2 + (1/12)×0.05 = 2 + 0.00417 ≈ 2.004
∴ ∛8.05 ≈ 2.004