除法定则除法定则或商定则(英语:Quotient rule)是数学中关于两个函数的商的导数的一个计算定则。 可用口诀:分子纤维(分子先微分) 若已知两个可导函数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}}}} 目录 1 例子 2 证明 2.1 从牛顿差商推出 2.2 从乘积法则推出 2.3 从复合函数求导法则推出 3 参见 例子 4 x − 2 x 2 + 1 {\displaystyle {\frac {4x-2}{x^{2}+1}}} 的导数为: d d x ( 4 x − 2 x 2 + 1 ) {\displaystyle {\frac {d}{dx}}\left({\frac {4x-2}{x^{2}+1}}\right)} = ( x 2 + 1 ) ( 4 ) − ( 4 x − 2 ) ( 2 x ) ( x 2 + 1 ) 2 {\displaystyle ={\frac {(x^{2}+1)(4)-(4x-2)(2x)}{(x^{2}+1)^{2}}}} = ( 4 x 2 + 4 ) − ( 8 x 2 − 4 x ) ( x 2 + 1 ) 2 {\displaystyle ={\frac {(4x^{2}+4)-(8x^{2}-4x)}{(x^{2}+1)^{2}}}} = − 4 x 2 + 4 x + 4 ( x 2 + 1 ) 2 {\displaystyle ={\frac {-4x^{2}+4x+4}{(x^{2}+1)^{2}}}} f ( x ) = 2 x 2 x 3 {\displaystyle f(x)={\frac {2x^{2}}{x^{3}}}} 的导数为: f ′ ( x ) {\displaystyle f'(x)\,} = ( 4 x ⋅ x 3 ) − ( 2 x 2 ⋅ 3 x 2 ) ( x 3 ) 2 {\displaystyle ={\frac {\left(4x\cdot x^{3}\right)-\left(2x^{2}\cdot 3x^{2}\right)}{\left(x^{3}\right)^{2}}}} = 4 x 4 − 6 x 4 x 6 {\displaystyle ={\frac {4x^{4}-6x^{4}}{x^{6}}}} = − 2 x 4 x 6 {\displaystyle ={\frac {-2x^{4}}{x^{6}}}} = − 2 x 2 {\displaystyle =-{\frac {2}{x^{2}}}} 证明 从牛顿差商推出 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}}}} 从乘积法则推出 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}}}} 从复合函数求导法则推出 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}}}} 参见 乘法定则 链式法则