导数定义

设y=f(x)y=f(x)y=f(x)在x0x_0x0​的某个领域内有定义，自变量增量为Δx\Delta xΔx，因变量增量为Δy=f(x0+Δx)−f(x0)\Delta y=f(x_0+\Delta x)-f(x_0)Δy=f(x0​+Δx)−f(x0​)，若lim⁡Δx→0=ΔyΔx=lim⁡Δx→0f(x0+Δx)−f(x0)Δx\lim\limits_{\Delta x\rightarrow0}=\dfrac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}Δx→0lim​=ΔxΔy​=Δx→0lim​Δxf(x0​+Δx)−f(x0​)​极限存在，则说明y=f(x)y=f(x)y=f(x)在x0x_0x0​处可导。

定义公式：f′(x0)=lim⁡Δx→0f(x0+Δx)−f(x0)Δxf'(x_0)=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}f′(x0​)=Δx→0lim​Δxf(x0​+Δx)−f(x0​)​或f′(x0)=lim⁡x→x0f(x)−f(x0)x−x0f'(x_0)=\lim\limits_{x\rightarrow x_0}\dfrac{f(x)-f(x_0)}{x-x_0}f′(x0​)=x→x0​lim​x−x0​f(x)−f(x0​)​

左导数：f−′(x0)=lim⁡Δx→0−f(x0+Δx)−f(x0)Δxf'_-(x_0)=\lim\limits_{\Delta x\rightarrow0^-}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}f−′​(x0​)=Δx→0−lim​Δxf(x0​+Δx)−f(x0​)​

右导数：f+′(x0)=lim⁡Δx→0+f(x0+Δx)−f(x0)Δxf'_+(x_0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}f+′​(x0​)=Δx→0+lim​Δxf(x0​+Δx)−f(x0​)​

导数存在的充要条件：f−′(x0)=f+′(x0)f'_-(x_0)=f'_+(x_0)f−′​(x0​)=f+′​(x0​)

例题1

y=∣x∣y=|x|y=∣x∣在x=0x=0x=0处是否可导？

解：
f′(0)=lim⁡Δx→0f(0+Δx)−f(0)Δx=lim⁡Δx→0∣Δx∣Δx\qquad f'(0)=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(0+\Delta x)-f(0)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{|\Delta x|}{\Delta x}f′(0)=Δx→0lim​Δxf(0+Δx)−f(0)​=Δx→0lim​Δx∣Δx∣​

f−′(0)=lim⁡Δx→0−ΔxΔx=−1\qquad f'_-(0)=\lim\limits_{\Delta x\rightarrow0}\dfrac{-\Delta x}{\Delta x}=-1f−′​(0)=Δx→0lim​Δx−Δx​=−1，f+′(0)=lim⁡Δx→0ΔxΔx=1f'_+(0)=\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta x}{\Delta x}=1f+′​(0)=Δx→0lim​ΔxΔx​=1

f−′(0)≠f+′(0)\qquad f'_-(0)\neq f'_+(0)f−′​(0)​=f+′​(0)，所以y=∣x∣y=|x|y=∣x∣在x=0x=0x=0处不可导

例题2

已知f′(x0)=1f'(x_0)=1f′(x0​)=1，则lim⁡h→0f(x0+h)−f(x0−h)h=‾\lim\limits_{h\rightarrow 0}\dfrac{f(x_0+h)-f(x_0-h)}{h}=\underline{\qquad\qquad}h→0lim​hf(x0​+h)−f(x0​−h)​=​

解：原式=lim⁡h→0f(x0+h)−f(x0)h+f(x0)−f(x0−h)h=f′(x0)+f′(x0)=2=\lim\limits_{h\rightarrow 0}\dfrac{f(x_0+h)-f(x_0)}{h}+\dfrac{f(x_0)-f(x_0-h)}{h}=f'(x_0)+f'(x_0)=2=h→0lim​hf(x0​+h)−f(x0​)​+hf(x0​)−f(x0​−h)​=f′(x0​)+f′(x0​)=2

总结

• 求导的充要条件：左导数===右导数
• 可导则一定连续，连续不一定可导