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⁡Δx→0f(x+Δx)⋅g(x+Δx)−f(x)⋅g(x)Δx\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→0lim​Δxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x)​

=lim⁡Δx→0f(x+Δx)⋅g(x+Δx)−f(x)⋅g(x+Δx)+f(x)⋅g(x+Δx)−f(x)⋅g(x)Δx\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→0lim​Δxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x+Δx)+f(x)⋅g(x+Δx)−f(x)⋅g(x)​

=lim⁡Δx→0f′(x)Δx⋅g(x)+g′(x)Δx⋅f(x)Δx\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→0lim​Δxf′(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)g2(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⁡Δx→0f(x+Δx)g(x+Δx)−f(x)g(x)Δx\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→0lim​Δxg(x+Δx)f(x+Δx)​−g(x)f(x)​​

=lim⁡Δx→0f(x+Δx)g(x)−f(x)g(x+Δx)g(x+Δx)g(x)⋅1Δx\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→0lim​g(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x+Δx)​⋅Δx1​

=lim⁡Δx→0f(x+Δx)g(x)−f(x)g(x)+f(x)g(x)−f(x)g(x+Δx)g(x+Δx)g(x)⋅1Δx\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→0lim​g(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x)+f(x)g(x)−f(x)g(x+Δx)​⋅Δx1​

=lim⁡Δx→0f′(x)Δx⋅g(x)−f(x)⋅g(x)Δxg2(x)⋅Δx\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→0lim​g2(x)⋅Δxf′(x)Δx⋅g(x)−f(x)⋅g(x)Δx​

=f′(x)g(x)−f(x)g′(x)g2(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)​