Applications of Derivatives · Applied Problems

Newton's Method

xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Newton's method iteratively approximates roots of f(x) = 0. Each step uses the tangent line to get a better approximation.

Conditions. f'(xₙ) ≠ 0 at each step. Convergence depends on the initial guess.

Variables

SymbolNameUnit
xnCurrent approximation
Current x value
fxnf(xₙ)
Function value at xₙ
fpxnf'(xₙ)
Derivative at xₙ

Worked examples

Use Newton's method with x₀ = 2 to approximate √5 (root of x²-5=0).
  1. f(x) = x²-5, f'(x) = 2x
  2. x₁ = 2 - (4-5)/(4) = 2 + 1/4 = 2.25
  3. x₂ = 2.25 - (5.0625-5)/(4.5) = 2.25 - 0.01389 ≈ 2.2361

Answer: x₂ ≈ 2.2361 (actual √5 ≈ 2.23607)

Practice this and 135 more formulas in the CalcRef workspace — quizzes, reference tables, a 16-category unit converter, and an expression evaluator.