设 y = f ( x ) y=f(x) y=f(x)在 x 0 x_0 x0的某个邻域内有定义,自变量增量为 Δ x \Delta x Δx,因变量增量为 Δ y = f ( x 0 + Δ x ) − f ( x 0 ) \Delta y=f(x_0+\Delta x)-f(x_0) Δy=f(x0+Δx)−f(x0),若 lim Δ x → 0 = Δ y Δ x = lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x \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→0limΔxf(x0+Δx)−f(x0)极限存在,则说明 y = f ( x ) y=f(x) y=f(x)在 x 0 x_0 x0处可导。
定义公式: f ′ ( x 0 ) = lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x f'(x_0)=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x} f′(x0)=Δx→0limΔxf(x0+Δx)−f(x0)或 f ′ ( x 0 ) = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 f'(x_0)=\lim\limits_{x\rightarrow x_0}\dfrac{f(x)-f(x_0)}{x-x_0} f′(x0)=x→x0limx−x0f(x)−f(x0)
左导数: f − ′ ( x 0 ) = lim Δ x → 0 − f ( x 0 + Δ x ) − f ( x 0 ) Δ x f'_-(x_0)=\lim\limits_{\Delta x\rightarrow0^-}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x} f−′(x0)=Δx→0−limΔxf(x0+Δx)−f(x0)
右导数: f + ′ ( x 0 ) = lim Δ x → 0 + f ( x 0 + Δ x ) − f ( x 0 ) Δ x f'_+(x_0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x} f+′(x0)=Δx→0+limΔxf(x0+Δx)−f(x0)
导数存在的充要条件: f − ′ ( x 0 ) = f + ′ ( x 0 ) f'_-(x_0)=f'_+(x_0) f−′(x0)=f+′(x0)
y = ∣ x ∣ y=|x| y=∣x∣在 x = 0 x=0 x=0处是否可导?
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\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→0limΔxf(0+Δx)−f(0)=Δx→0limΔx∣Δx∣
f − ′ ( 0 ) = lim Δ x → 0 + − Δ x Δ x = − 1 \qquad f'_-(0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{-\Delta x}{\Delta x}=-1 f−′(0)=Δx→0+limΔx−Δx=−1, f + ′ ( 0 ) = lim Δ x → 0 + Δ x Δ x = 1 f'_+(0)=\lim\limits_{\Delta x\rightarrow0^+}\dfrac{\Delta x}{\Delta x}=1 f+′(0)=Δx→0+limΔxΔx=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 = 0 x=0 x=0处不可导
已知 f ′ ( x 0 ) = 1 f'(x_0)=1 f′(x0)=1,则 lim h → 0 f ( x 0 + h ) − f ( x 0 − h ) h = ‾ \lim\limits_{h\rightarrow 0}\dfrac{f(x_0+h)-f(x_0-h)}{h}=\underline{\qquad\qquad} h→0limhf(x0+h)−f(x0−h)=
解:原式 = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h + f ( x 0 ) − f ( x 0 − h ) h = f ′ ( x 0 ) + f ′ ( x 0 ) = 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→0limhf(x0+h)−f(x0)+hf(x0)−f(x0−h)=f′(x0)+f′(x0)=2