前置知识：导数的定义和介绍
若
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sin
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f(x)=\left\{eax,x<0b+sin2x, x≥0
解：
∵
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\qquad \because \lim\limits_{\Delta x\rightarrow0^-}e^{a\Delta x}=f(0)=b+\sin 0=b
∵Δx→0−limeaΔx=f(0)=b+sin0=b
∴ b = lim Δ x → 0 − e a Δ x = 1 \qquad \therefore b=\lim\limits_{\Delta x\rightarrow0^-}e^{a\Delta x}=1 ∴b=Δx→0−limeaΔx=1
f − ′ ( 0 ) = lim Δ x → 0 − f ( 0 + Δ x ) − f ( 0 ) Δ x = lim Δ x → 0 − e a Δ x − 1 Δ x = lim Δ x → 0 − e a Δ x − 1 a Δ x × 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=a f−′(0)=Δx→0−limΔxf(0+Δx)−f(0)=Δx→0−limΔxeaΔx−1=Δx→0−limaΔxeaΔx−1×a=a
f + ′ ( 0 ) = lim Δ x → 0 + f ( 0 + Δ x ) − f ( 0 ) Δ x = lim Δ x → 0 + ( b + sin 2 Δ x ) − ( b + sin 0 ) Δ x = lim Δ x → 0 + sin 2 Δ x 2 Δ x × 2 = 2 \qquad f'_+(0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{f(0+\Delta x)-f(0)}{\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=2 f+′(0)=Δx→0+limΔxf(0+Δx)−f(0)=Δx→0+limΔx(b+sin2Δx)−(b+sin0)=Δx→0+lim2Δxsin2Δx×2=2
∵ f ( x ) \qquad\because f(x) ∵f(x)在 x = 0 x=0 x=0处可导
∴ f − ′ ( 0 ) = f + ′ ( 0 ) \qquad \therefore f'_-(0)=f'_+(0) ∴f−′(0)=f+′(0)，即 a = 2 a=2 a=2
\qquad
综上所述，
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\left\{a=2b=1
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。