除法定则

除法定则或商定則（英語：Quotient rule）是数学中关于两个函数的商的导数的一个计算定则。 可用口訣：分子纖維（分子先微分） 若已知两个可導函数g,h及其导数g',h'，且h(x)≠0，则它们的商 f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)={\frac {g(x)}{h(x)}}} 的导数为： f ′ ( x ) = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) [ h ( x ) ] 2 {\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}} 4 x − 2 x 2 + 1 {\displaystyle {\frac {4x-2}{x^{2}+1}}} 的导数为： f ( x ) = 2 x 2 x 3 {\displaystyle f(x)={\frac {2x^{2}}{x^{3}}}} 的导数为： 设 f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)={\tfrac {g(x)}{h(x)}}} ， h ( x ) ≠ 0 {\displaystyle h(x)\neq 0} ，且 g {\displaystyle g} 和 h {\displaystyle h} 均可导。 f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim Δ x → 0 g ( x + Δ x ) h ( x + Δ x ) − g ( x ) h ( x ) Δ x {\displaystyle f'(x)=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)}{h(x+\Delta x)}}-{\frac {g(x)}{h(x)}}}{\Delta x}}} = lim Δ x → 0 1 Δ x ⋅ g ( x + Δ x ) h ( x ) − g ( x ) h ( x + Δ x ) h ( x ) h ( x + Δ x ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot {\frac {g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)}}} = lim Δ x → 0 1 Δ x ⋅ ( g ( x + Δ x ) h ( x ) − g ( x ) h ( x ) ) − ( g ( x ) h ( x + Δ x ) − g ( x ) h ( x ) ) h ( x ) h ( x + Δ x ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot {\frac {(g(x+\Delta x)h(x)-g(x)h(x))-(g(x)h(x+\Delta x)-g(x)h(x))}{h(x)h(x+\Delta x)}}} = lim Δ x → 0 1 Δ x ⋅ h ( x ) ( g ( x + Δ x ) − g ( x ) ) − g ( x ) ( h ( x + Δ x ) − h ( x ) ) h ( x ) h ( x + Δ x ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot {\frac {h(x)(g(x+\Delta x)-g(x))-g(x)(h(x+\Delta x)-h(x))}{h(x)h(x+\Delta x)}}} = lim Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x h ( x ) − g ( x ) h ( x + Δ x ) − h ( x ) Δ x h ( x ) h ( x + Δ x ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)-g(x)}{\Delta x}}h(x)-g(x){\frac {h(x+\Delta x)-h(x)}{\Delta x}}}{h(x)h(x+\Delta x)}}} = lim Δ x → 0 ( g ( x + Δ x ) − g ( x ) Δ x ) h ( x ) − g ( x ) lim Δ x → 0 ( h ( x + Δ x ) − h ( x ) Δ x ) h ( x ) h ( lim Δ x → 0 ( x + Δ x ) ) {\displaystyle ={\frac {\lim _{\Delta x\to 0}\left({\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right)h(x)-g(x)\lim _{\Delta x\to 0}\left({\frac {h(x+\Delta x)-h(x)}{\Delta x}}\right)}{h(x)h(\lim _{\Delta x\to 0}(x+\Delta x))}}} = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) [ h ( x ) ] 2 {\displaystyle ={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}} 假设 f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)={\frac {g(x)}{h(x)}}} 。 那么 g ( x ) = f ( x ) h ( x )   {\displaystyle g(x)=f(x)h(x){\mbox{ }}\,} g ′ ( x ) = f ′ ( x ) h ( x ) + f ( x ) h ′ ( x )   {\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,} f ′ ( x ) = g ′ ( x ) − f ( x ) h ′ ( x ) h ( x ) = g ′ ( x ) − g ( x ) h ( x ) ⋅ h ′ ( x ) h ( x ) {\displaystyle f'(x)={\frac {g'(x)-f(x)h'(x)}{h(x)}}={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}} f ′ ( x ) = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) ( h ( x ) ) 2 {\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{\left(h(x)\right)^{2}}}} 考虑恒等式，v≠0 u v = 1 4 [ ( u + 1 v ) 2 − ( u − 1 v ) 2 ] {\displaystyle {\frac {u}{v}}\;=\;{\frac {1}{4}}\left[\left(u+{\frac {1}{v}}\right)^{2}-\;\left(u-{\frac {1}{v}}\right)^{2}\right]} 那么： d ( u v ) d x = 1 4 d d x [ ( u + 1 v ) 2 − ( u − 1 v ) 2 ] {\displaystyle {\frac {d\left({\frac {u}{v}}\right)}{dx}}\;=\;{\frac {1}{4}}{\frac {d}{dx}}\left[\left(u+{\frac {1}{v}}\right)^{2}-\;\left(u-{\frac {1}{v}}\right)^{2}\right]} 于是： d ( u v ) d x = 1 4 [ 2 ( u + 1 v ) ( d u d x − d v v 2 d x ) − 2 ( u − 1 v ) ( d u d x + d v v 2 d x ) ] {\displaystyle {\frac {d\left({\frac {u}{v}}\right)}{dx}}\;=\;{\frac {1}{4}}\left[2\left(u+{\frac {1}{v}}\right)\left({\frac {du}{dx}}-{\frac {dv}{v^{2}dx}}\right)-\;2\left(u-{\frac {1}{v}}\right)\left({\frac {du}{dx}}+{\frac {dv}{v^{2}dx}}\right)\right]} 展开，得： d ( u v ) d x = 1 4 [ 4 v d u d x − 4 u v 2 d v d x ] {\displaystyle {\frac {d\left({\frac {u}{v}}\right)}{dx}}\;=\;{\frac {1}{4}}\left[{\frac {4}{v}}{\frac {du}{dx}}-{\frac {4u}{v^{2}}}{\frac {dv}{dx}}\right]} 最后，把分子和分母同除以4，便得： d ( u v ) d x = [ v d u d x − u d v d x ] v 2 {\displaystyle {\frac {d\left({\frac {u}{v}}\right)}{dx}}\;=\;{\frac {\left[v{\frac {du}{dx}}-u{\frac {dv}{dx}}\right]}{v^{2}}}} 乘法定则 鏈式法則