牛頓法者，導數之法也，以解方程之近似值。
方程 f ( x ) {\displaystyle f(x)} ，導數 f ′ ( x ) {\displaystyle f'(x)} 且
f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x {\displaystyle f'(x)=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}
則有
Δ x = lim Δ x → 0 f ( x + Δ x ) − f ( x ) f ′ ( x ) {\displaystyle \Delta x=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{f'(x)}}}
又方程一側值爲〇，則
x 1 = x 0 + Δ x = x 0 + 0 − f ( x 0 ) f ′ ( x 0 ) = x 0 − f ( x 0 ) f ′ ( x 0 ) {\displaystyle x_{1}=x_{0}+\Delta x=x_{0}+{\frac {0-f(x_{0})}{f'(x_{0})}}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}}
得
x n = x n − 1 − f ( x n − 1 ) f ′ ( x n − 1 ) {\displaystyle x_{n}=x_{n-1}-{\frac {f(x_{n-1})}{f'(x_{n-1})}}}
求值 2 {\displaystyle {\sqrt {2}}}
設 y = x 2 − 2 {\displaystyle y=x^{2}-2} ，則 y ′ = 2 x {\displaystyle y'=2x}
x 0 = 1.4 {\displaystyle x_{0}=1.4}
x 1 = 1.4 − ( − 0.04 ÷ 2.8 ) = 1.4142857 ⋯ ⋯ ≈ 1.4142 {\displaystyle x_{1}=1.4-(-0.04\div 2.8)=1.4142857\cdots \cdots \thickapprox 1.4142}
x 2 ≈ 1.41421356 {\displaystyle x_{2}\approx 1.41421356}
⋯ ⋯ {\displaystyle \cdots \cdots }
故 2 ≈ 1.4142135623 {\displaystyle {\sqrt {2}}\approx 1.4142135623}
，導數
且
，則
