前置知识:微分的定义和介绍
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+11∣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
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limΔx→0Δy−dyΔx=‾\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-dy}{\Delta x}=\underline{\qquad}Δx→0limΔxΔy−dy=.
解:
limΔx→0Δy−dyΔx=limΔx→0Δy−f′(x)ΔxΔx=limΔx→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→0limΔxΔy−dy=Δx→0limΔxΔy−f′(x)Δx=Δx→0limΔxΔy−f′(x)=f′(x)−f′(x)=0
\qquad所以limΔx→0Δy−dyΔx=0\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y-dy}{\Delta x}=0Δx→0limΔxΔy−dy=0