函数相乘和相除的导数及证明

函数求导简介

函数相乘求导

[ f ( x ) ⋅ g ( x ) ] ′ = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) [f(x)\cdot g(x)]'=f'(x)g(x)+f(x)g'(x) [f(x)⋅g(x)]′=f′(x)g(x)+f(x)g′(x)

证明：
[ f ( x ) ⋅ g ( x ) ] ′ = lim ⁡ Δ x → 0 f ( x + Δ x ) ⋅ g ( x + Δ x ) − f ( x ) ⋅ g ( x ) Δ x \qquad [f(x)\cdot g(x)]'=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{\Delta x} [f(x)⋅g(x)]′=Δx→0limΔxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x)

= lim ⁡ Δ x → 0 f ( x + Δ x ) ⋅ g ( x + Δ x ) − f ( x ) ⋅ g ( x + Δ x ) + f ( x ) ⋅ g ( x + Δ x ) − f ( x ) ⋅ g ( x ) Δ x \qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x)\cdot g(x+\Delta x)+f(x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{\Delta x} =Δx→0limΔxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x+Δx)+f(x)⋅g(x+Δx)−f(x)⋅g(x)

= lim ⁡ Δ x → 0 f ′ ( x ) Δ x ⋅ g ( x ) + g ′ ( x ) Δ x ⋅ f ( x ) Δ x \qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f'(x)\Delta x\cdot g(x)+g'(x)\Delta x\cdot f(x)}{\Delta x} =Δx→0limΔxf′(x)Δx⋅g(x)+g′(x)Δx⋅f(x)

= f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) \qquad \qquad \qquad \qquad =f'(x)g(x)+f(x)g'(x) =f′(x)g(x)+f(x)g′(x)

函数相除求导

[ f ( x ) g ( x ) ] ′ = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g 2 ( x ) [\dfrac{f(x)}{g(x)}]'=\dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} [g(x)f(x)]′=g2(x)f′(x)g(x)−f(x)g′(x)

证明：
[ f ( x ) g ( x ) ] ′ = lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x \qquad [\dfrac{f(x)}{g(x)}]'=\lim\limits_{\Delta x\rightarrow0}\dfrac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x} [g(x)f(x)]′=Δx→0limΔxg(x+Δx)f(x+Δx)−g(x)f(x)

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x ) − f ( x ) g ( x + Δ x ) g ( x + Δ x ) g ( x ) ⋅ 1 Δ x \qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)}\cdot\dfrac{1}{\Delta x} =Δx→0limg(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x+Δx)⋅Δx1

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x ) − f ( x ) g ( x ) + f ( x ) g ( x ) − f ( x ) g ( x + Δ x ) g ( x + Δ x ) g ( x ) ⋅ 1 Δ x \qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)}\cdot\dfrac{1}{\Delta x} =Δx→0limg(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x)+f(x)g(x)−f(x)g(x+Δx)⋅Δx1

= lim ⁡ Δ x → 0 f ′ ( x ) Δ x ⋅ g ( x ) − f ( x ) ⋅ g ( x ) Δ x g 2 ( x ) ⋅ Δ x \qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f'(x)\Delta x\cdot g(x)-f(x)\cdot g(x)\Delta x}{g^2(x)\cdot \Delta x} =Δx→0limg2(x)⋅Δxf′(x)Δx⋅g(x)−f(x)⋅g(x)Δx

= f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g 2 ( x ) \qquad \qquad \qquad \qquad =\dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} =g2(x)f′(x)g(x)−f(x)g′(x)