前置知识：微分的定义和介绍

例1

函数y=x+1y=\sqrt{x+1}y=x+1​在点x=0x=0x=0处，当自变量改变量Δx=0.04\Delta x=0.04Δx=0.04时，dy=‾dy=\underline{\qquad}dy=​.

解：
f′(x)=12x+1∣x=0=0.5\qquad f'(x)=\dfrac{1}{2\sqrt{x+1}}|_{x=0}=0.5f′(x)=2x+1​1​∣x=0​=0.5

dy=f′(x)Δx=0.5×0.04=0.02\qquad dy=f'(x)\Delta x=0.5\times 0.04=0.02dy=f′(x)Δx=0.5×0.04=0.02

例2

设函数y=f(x)y=f(x)y=f(x)在x0x_0x0​处可导，且f′(x0)≠0f'(x_0)\neq0f′(x0​)​=0，则lim⁡Δx→0Δy−dyΔx=‾\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-dy}{\Delta x}=\underline{\qquad}Δx→0lim​ΔxΔy−dy​=​.

解：
lim⁡Δx→0Δy−dyΔx=lim⁡Δx→0Δy−f′(x)ΔxΔx=lim⁡Δx→0ΔyΔx−f′(x)=f′(x)−f′(x)=0\qquad \lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-dy}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-f'(x)\Delta x}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y}{\Delta x}-f'(x)=f'(x)-f'(x)=0Δx→0lim​ΔxΔy−dy​=Δx→0lim​ΔxΔy−f′(x)Δx​=Δx→0lim​ΔxΔy​−f′(x)=f′(x)−f′(x)=0

\qquad所以lim⁡Δx→0Δy−dyΔx=0\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-dy}{\Delta x}=0Δx→0lim​ΔxΔy−dy​=0