导数的定义和介绍

例题

若f(x)={eax,x<0b+sin⁡2x,x≥0f(x)=\left\{\begin{matrix}e^{ax},\qquad\quad x<0\\b+\sin 2x, \ x\geq 0\end{matrix}\right.f(x)={eax,x<0b+sin2x, x≥0​在x=0x=0x=0处可导，求a,ba,ba,b的值。

解：
f−′(0)=lim⁡Δx→0−f(0+Δx)−f(0)Δx=lim⁡Δx→0−eaΔx−1Δx=lim⁡Δx→0−eaΔx−1aΔx×a=a\qquad f'_-(0)=\lim\limits_{\Delta x\rightarrow0^-}\dfrac{f(0+\Delta x)-f(0)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0^-}\dfrac{e^{a\Delta x}-1}{\Delta x}=\lim\limits_{\Delta x\rightarrow0^-}\dfrac{e^{a\Delta x}-1}{a\Delta x}\times a=af−′​(0)=Δx→0−lim​Δxf(0+Δx)−f(0)​=Δx→0−lim​ΔxeaΔx−1​=Δx→0−lim​aΔxeaΔx−1​×a=a

f+′(0)=lim⁡Δx→0+f(0+Δx)−f(Δx)Δx=lim⁡Δx→0+(b+sin⁡2Δx)−(b+sin⁡0)Δx=lim⁡Δx→0+sin⁡2Δx2Δx×2=2\qquad f'_+(0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{f(0+\Delta x)-f(\Delta x)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{(b+\sin 2\Delta x)-(b+\sin 0)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{\sin 2\Delta x}{2\Delta x}\times 2=2f+′​(0)=Δx→0+lim​Δxf(0+Δx)−f(Δx)​=Δx→0+lim​Δx(b+sin2Δx)−(b+sin0)​=Δx→0+lim​2Δxsin2Δx​×2=2

∵f(x)\qquad\because f(x)∵f(x)在x=0x=0x=0处可导

∴f−′(0)=f+′(0)\qquad \therefore f'_-(0)=f'_+(0)∴f−′​(0)=f+′​(0)，即a=2a=2a=2

∵lim⁡Δx→0−eaΔx=f(0)=b+sin⁡0=b\qquad \because \lim\limits_{\Delta x\rightarrow0^-}e^{a\Delta x}=f(0)=b+\sin 0=b∵Δx→0−lim​eaΔx=f(0)=b+sin0=b

∴b=lim⁡Δx→0−eaΔx=1\qquad \therefore b=\lim\limits_{\Delta x\rightarrow0^-}e^{a\Delta x}=1∴b=Δx→0−lim​eaΔx=1

\qquad综上所述，{a=2b=1\left\{\begin{matrix}a=2\\b=1\end{matrix}\right.{a=2b=1​

总结

可导的充要条件是左导数===右导数。解题过程中用到了无穷小替换，这个要用熟。