3.1 The Derivative as a Function

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这一节继续上一章关于切线和斜率的讨论，我们知道对于函数f(x)，在x0x_0的切线的斜率可以由：

m=lim⁡h→0f(x0+h)−f(x0)hm = \lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)}{h}

求得，称m为f(x)在x0x_0处的导数(derivative)。如果我们考察函数每一个x处的导数，那么可以得到：

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

f′(x)f'(x)称为f(x)的导函数(derivative function)。


f(x)在x0x_0处有导数，我们称为f(x)在x0x_0处可微(differentiable)，如果f(x)在定义域内处处可微，那么可以称f(x)可微。

如果设z=x+hz = x + h，那么导函数可以表达为：

f′(x)=lim⁡z→xf(z)−f(x)z−xf'(x) = \lim_{z \to x}\frac{f(z) - f(x)}{z -x}

求导数的计算过程称之为微分(differentiation)，上面这种表达方式在求某些函数的导数的时候比较方便。


一些表达微分的常见符号：

f′(c)=y′=dydx=dfdx=ddxf(x)=D(f)(x)=Dxf(x)f'(c) = y' = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}f(x) = D(f)(x) = D_xf(x)



可微的判断

可微的定义是通过极限来描述的，所以可微的判断等价于判断对应极限是否存在。由于极限有单向极限的存在，可微也有相同的概念，另外极限为无穷，也等同于不可微。


区间内的可微判断

f(x)在一个开区间内可微，当且仅当它在区间内各个点都可微。f(x)在一个闭区间内可微，当且仅当它在区间内给个点都可微，且在端点a和b处分别存在左和右导数。



可微和连续的关系

这是很重要的概念，它指的是如果函数f在x=c处可微，那么它在x=c处连续。

证明：

如果f在x=c处连续，那么根据连续定义，需要证明

lim⁡x→cf(x)=f(c)\lim_{x \to c}f(x) = f(c)

我们将其转换成另一种形式(思维突破点)

lim⁡x→c[f(x)−f(c)]=0\lim_{x \to c}[f(x) - f(c)] = 0

要证明以上式子成立，我们根据已知即f在x=c处可微，存在

lim⁡x→cf(x)−f(c)x−c\lim_{x \to c}\frac{f(x) - f(c)}{x - c}

我们观察f(x)−f(c)x−c\frac{f(x) - f(c)}{x - c}和f(x)−f(c)f(x) - f(c)，比较相似，要将f(x)−f(c)x−c\frac{f(x) - f(c)}{x - c}转换为f(x)−f(c)f(x) - f(c)，可以写成

f(x)−f(c)x−c⋅(x−c)\frac{f(x) - f(c)}{x - c}\cdot (x - c)

变成要证明

lim⁡x→c[f(x)−f(c)x−c⋅(x−c)]=0\lim_{x \to c}[\frac{f(x) - f(c)}{x - c}\cdot (x - c)] = 0

首先，f(x)−f(c)x−c⋅(x−c)\frac{f(x) - f(c)}{x - c}\cdot (x - c)在x=c处的极限是否存在？我们知道存在

lim⁡x→c(x−c)=0\lim_{x \to c}(x - c) = 0

因为f在x=c处可微，那么也存在

lim⁡x→cf(x)−f(c)x−c\lim_{x \to c}\frac{f(x) - f(c)}{x - c}

这里不需要知道x=c处的导数是多少，我们知道它存在即可。那么应用极限的乘法定理可得(这里要特别注意，极限乘法定理成立的前提是乘法的两项的极限都要存在，所以没有上面两个前提的成立，无法应用乘法定理)

lim⁡x→c[f(x)−f(c)x−c⋅(x−c)]=lim⁡x→c[f(x)−f(c)x−c]⋅lim⁡x→c(x−c)=lim⁡x→c[f(x)−f(c)]=0\lim_{x \to c}[\frac{f(x) - f(c)}{x - c}\cdot (x - c)] = \lim_{x \to c}[\frac{f(x) - f(c)}{x - c}]\cdot \lim_{x \to c}(x - c) = \lim_{x \to c}[f(x) - f(c)] = 0

证毕。


要注意f在x=c处连续，不能推导出f在x=c处可微。因为要证明此推论成立的话，就要证明存在

lim⁡x→cf(x)−f(c)x−c\lim_{x \to c}\frac{f(x) - f(c)}{x - c}

我们观察f(x)−f(c)x−c\frac{f(x) - f(c)}{x - c}，对于x=c

lim⁡x→c[f(x)−f(c)]=0\lim_{x \to c}[f(x) - f(c)] = 0

并且

lim⁡x→c(x−c)=0\lim_{x \to c}(x - c) = 0

似乎可以应用极限的除法定理，但是除法定理要求分母不能为零，所以无法证明

lim⁡x→cf(x)−f(c)x−c\lim_{x \to c}\frac{f(x) - f(c)}{x - c}

存在



练习

  1. f(x)=4−x2f(x) = 4 - x^2；求f′(−3),f′(0),f′(1)f'(-3), f'(0), f'(1)
lim⁡h→0f(x)=f(x+h)−f(x)h=4−(x+h)2−4+x2h=x2−(x2+h2+2xh)h=−h2−2xhh=−h−2x=−2x\begin{aligned} \lim_{h \to 0}f(x) &= \frac{f(x + h) - f(x)}{h} \\ &= \frac{4 - (x + h)^2 - 4 + x^2}{h} \\ &= \frac{x^2 - (x^2 + h^2 + 2xh)}{h} \\ &= \frac{-h^2 - 2xh}{h} \\ &= - h - 2x \\ &= -2x \end{aligned}

∴f′(−3)=6f′(0)=0f′(1)=−2\begin{aligned} \therefore f'(-3) &= 6 \\ f'(0) &= 0 \\ f'(1) &= -2 \end{aligned}

  1. F(x)=(x−1)2+1F(x) = (x - 1)^2 + 1
lim⁡h→0F(x)=F(x+h)−F(x)h=(x+h−1)2+1−(x−1)2−1h=h2+(x−1)2+2h(x−1)−(x−1)2h=h+2x−2=2x−2\begin{aligned} \lim_{h \to 0}F(x) &= \frac{F(x + h) - F(x)}{h} \\ &= \frac{(x + h - 1)^2 + 1 - (x - 1)^2 - 1}{h} \\ &= \frac{h^2 + (x - 1)^2 + 2h(x - 1) - (x - 1)^2}{h} \\ &= h + 2x - 2 \\ &= 2x - 2 \end{aligned}

  1. g(t)=1t2g(t) = \frac{1}{t^2}
lim⁡h→0g(t)=g(t+h)−g(t)h=1(t+h)2−1t2h=t2−(t+h)2(t+h)2⋅t2h=−h2−2th(t+h)2⋅t2h=−h−2t(t+h)2⋅t2=−2tt4=−2t3\begin{aligned} \lim_{h \to 0}g(t) &= \frac{g(t + h) - g(t)}{h} \\ &= \frac{\frac{1}{(t + h)^2} - \frac{1}{t^2}}{h} \\ &= \frac{\frac{t^2 - (t + h)^2}{(t + h)^2\cdot t^2}}{h} \\ &= \frac{\frac{-h^2 - 2th}{(t + h)^2 \cdot t^2}}{h} \\ &= \frac{-h - 2t}{(t + h)^2 \cdot t^2} \\ &= \frac{-2t}{t^4} \\ &= \frac{-2}{t^3} \end{aligned}

  1. k(z)=1−z2zk(z) = \frac{1 - z}{2z}
lim⁡h→0k(z)=k(z+h)−k(z)h=1−z−h2(z+h)−1−z2zh=−2h4z2+4zh=0\begin{aligned} \lim_{h \to 0}k(z) &= \frac{k(z + h) - k(z)}{h} \\ &= \frac{\frac{1 - z - h}{2(z + h)} - \frac{1 - z}{2z}}{h} \\ &= \frac{-2h}{4z^2 + 4zh} \\ &= 0 \end{aligned}

  1. p(ϕ)=3ϕp(\phi) = \sqrt{3\phi}
lim⁡h→0p(ϕ)=p(ϕ+h)−p(ϕ)h=3(ϕ+h)−3ϕh=(3(ϕ+h)−3ϕ)h⋅3(ϕ+h)+3ϕ3(ϕ+h)+3ϕ=3(ϕ+h)−3ϕh⋅(3(ϕ+h)+3ϕ)=33(ϕ+h)+3ϕ=323ϕ\begin{aligned} \lim_{h \to 0}p(\phi) &= \frac{p(\phi + h) - p(\phi)}{h} \\ &= \frac{\sqrt{3(\phi + h)} - \sqrt{3\phi}}{h} \\ &= \frac{(\sqrt{3(\phi + h)} - \sqrt{3\phi})}{h} \cdot \frac{\sqrt{3(\phi + h)} + \sqrt{3\phi}}{\sqrt{3(\phi + h)} + \sqrt{3\phi}} \\ &= \frac{3(\phi + h) - 3\phi}{h \cdot (\sqrt{3(\phi + h)} + \sqrt{3\phi})} \\ &= \frac{3}{\sqrt{3(\phi + h)} + \sqrt{3\phi}} \\ &= \frac{3}{2\sqrt{3\phi}} \end{aligned}

  1. r(s)=2s+1r(s) = \sqrt{2s + 1}
lim⁡h→0r(s)=r(s+h)−r(s)h=2(s+h)+1−2s+1h⋅2(s+h)+1+2s+12(s+h)+1+2s+1=2s+2h+1−2s−1h⋅(2(s+h)+1+2s+1)=22(s+h)+1+2s+1=12s+1\begin{aligned} \lim_{h \to 0}r(s) &= \frac{r(s + h) - r(s)}{h} \\ &= \frac{\sqrt{2(s + h) + 1} - \sqrt{2s + 1}}{h} \cdot \frac{\sqrt{2(s + h) + 1} + \sqrt{2s + 1}}{\sqrt{2(s + h) + 1} + \sqrt{2s + 1}} \\ &= \frac{2s + 2h + 1 - 2s - 1}{h \cdot (\sqrt{2(s + h) + 1} + \sqrt{2s + 1})} \\ &= \frac{2}{\sqrt{2(s + h) + 1} + \sqrt{2s + 1}} \\ &= \frac{1}{\sqrt{2s + 1}} \end{aligned}

  1. dydx\frac{dy}{dx} if y=2x3y = 2x^3
lim⁡h→0y=2(x+h)3−2x3h=2x2+2xh2+4x2h+2hx2+2h3+4xh2−2x3h=2xh+4x2+2x2+2h2+4xh=6x2\begin{aligned} \lim_{h \to 0}y &= \frac{2(x + h)^3 - 2x^3}{h} \\ &= \frac{2x^2 + 2xh^2 + 4x^2h + 2hx^2 + 2h^3 + 4xh^2 - 2x^3}{h} \\ &= 2xh + 4x^2 + 2x^2 + 2h^2 + 4xh \\ &= 6x^2 \end{aligned}

  1. drds\frac{dr}{ds} if r=s32+1r = \frac{s^3}{2} + 1
lim⁡h→0r=(s+h)3−s32h=3sh2+3hs2+h32h=3sh+3s2+h22=32s2\begin{aligned} \lim_{h \to 0}r &= \frac{(s + h)^3 - s^3}{2h} \\ &= \frac{3sh^2 + 3hs^2 + h^3}{2h} \\ &= \frac{3sh + 3s^2 + h^2}{2} \\ &= \frac{3}{2}s^2 \end{aligned}
  1. dsdt\frac{ds}{dt} if s=t2t+1s = \frac{t}{2t + 1}
lim⁡h→0s=t+h2t+2h+1−t2t+1h=(t+h)(2t+1)−t(2t+2h+1)h(2t+2h+1)(2t+1)=2t2+t+2th+h−2t2−2th−th(2t+2h+1)(2t+1)=1(2t+1)2\begin{aligned} \lim_{h \to 0}s &= \frac{\frac{t + h}{2t + 2h + 1} - \frac{t}{2t + 1}}{h} \\ &= \frac{(t + h)(2t + 1) - t(2t + 2h + 1)}{h(2t + 2h + 1)(2t + 1)} \\ &= \frac{2t^2 + t + 2th + h -2t^2 -2th -t}{h(2t + 2h + 1)(2t + 1)} \\ &= \frac{1}{(2t + 1)^2} \end{aligned}

  1. dvdt\frac{dv}{dt} if v=t−1tv = t - \frac{1}{t}
lim⁡h→0v=t+h−1t+h−t+1th=h−(1t+h−1t)h=1−−1t(t+h)=1+1t2\begin{aligned} \lim_{h \to 0}v &= \frac{t + h - \frac{1}{t + h} - t + \frac{1}{t}}{h} \\ &= \frac{h - (\frac{1}{t + h} - \frac{1}{t})}{h} \\ &= 1 - \frac{-1}{t(t + h)} \\ &= 1 + \frac{1}{t^2} \end{aligned}

  1. dpdq\frac{dp}{dq} if p=1q+1p = \frac{1}{\sqrt{q + 1}}
lim⁡h→0p=1q+h+1−1q+1h=q+1−q+h+1h⋅q+h+1⋅q+1⋅q+1+q+h+1q+1+q+h+1=−hh⋅2(q+1)3=−12(q+1)3\begin{aligned} \lim_{h \to 0}p &= \frac{\frac{1}{\sqrt{q + h + 1}} - \frac{1}{\sqrt{q + 1}}}{h} \\ &= \frac{\sqrt{q + 1} - \sqrt{q + h + 1}}{h \cdot \sqrt{q + h + 1} \cdot \sqrt{q + 1}} \cdot \frac{\sqrt{q + 1} + \sqrt{q + h + 1}}{\sqrt{q + 1} + \sqrt{q + h + 1}} \\ &= \frac{-h}{h \cdot 2(\sqrt{q + 1})^3} \\ &= \frac{-1}{2(\sqrt{q + 1})^3} \end{aligned}

  1. dzdw\frac{dz}{dw} if z=13w−2z = \frac{1}{\sqrt{3w - 2}}
lim⁡h→0z=13(w+h)−2−13w−2h=3w−2−3(w+h)−2h⋅3(w+h)−2⋅3w−2⋅3w−2+3(w+h)−23w−2+3(w+h)−2=−3h2h⋅(3w−2)3=−32⋅(3w−2)3\begin{aligned} \lim_{h \to 0}z &= \frac{\frac{1}{\sqrt{3(w + h) - 2}} - \frac{1}{\sqrt{3w - 2}}}{h} \\ &= \frac{\sqrt{3w - 2} - \sqrt{3(w + h) - 2}}{h \cdot \sqrt{3(w + h) - 2} \cdot \sqrt{3w - 2}} \cdot \frac{\sqrt{3w - 2} + \sqrt{3(w + h) - 2}}{\sqrt{3w - 2} + \sqrt{3(w + h) - 2}} \\ &= \frac{-3h}{2h \cdot (\sqrt{3w -2})^3} \\ &= -\frac{3}{2 \cdot (\sqrt{3w -2})^3} \end{aligned}

13 - 22 略...


  1. f(x)=1x+2f(x) = \frac{1}{x + 2}
lim⁡z→xf(z)−f(x)z−x=1z+2−1x+2z−x=x+2−z−2(z−x)(z+2)(x+2)=−(z−x)(z−x)(z+2)(x+2)=−1(x−2)2\begin{aligned} \lim_{z \to x}\frac{f(z) - f(x)}{z - x} &= \frac{\frac{1}{z + 2} - \frac{1}{x + 2}}{z - x} \\ &= \frac{x + 2 - z - 2}{(z - x)(z + 2)(x + 2)} \\ &= \frac{-(z - x)}{(z - x)(z + 2)(x +2)} \\ &= \frac{-1}{(x - 2)^2} \end{aligned}

  1. f(x)=1(x−1)2f(x) = \frac{1}{(x - 1)^2}
lim⁡z→xf(z)−f(x)z−x=1(z−1)2−1(x−1)2z−x=(x−1)2−(z−1)2(z−x)(z−1)2(x−1)2=[(x−1)−(z−1)][(x−1)+(z−1)](z−x)(z−1)2(x−1)2=−(z−x)(x+z−2)(z−x)(z−1)2(x−1)2=−2(x−1)(x−1)4=−2(x−1)3\begin{aligned} \lim_{z \to x}\frac{f(z) - f(x)}{z - x} &= \frac{\frac{1}{(z - 1)^2} - \frac{1}{(x - 1)^2}}{z - x} \\ &= \frac{(x - 1)^2 - (z - 1)^2}{(z - x)(z - 1)^2(x - 1)^2} \\ &= \frac{[(x - 1) - (z - 1)][(x - 1) + (z - 1)]}{(z - x)(z - 1)^2(x - 1)^2} \\ &= \frac{-(z - x)(x + z - 2)}{(z - x)(z - 1)^2(x - 1)^2} \\ &= \frac{-2(x - 1)}{(x - 1)^4} \\ &= \frac{-2}{(x - 1)^3} \end{aligned}

  1. g(x)=xx−1g(x) = \frac{x}{x - 1}
lim⁡z→xg(z)−g(x)z−x=zz−1−xx−1z−x=z(x−1)−x(z−1)(z−x)(z−1)(x−1)=−(z−x)(z−x)(z−1)(x−1)=−1(x−1)2\begin{aligned} \lim_{z \to x}\frac{g(z) - g(x)}{z - x} &= \frac{\frac{z}{z - 1} - \frac{x}{x - 1}}{z - x} \\ &= \frac{z(x -1) - x(z - 1)}{(z - x)(z - 1)(x - 1)} \\ &= \frac{-(z - x)}{(z - x)(z - 1)(x - 1)} \\ &= \frac{-1}{(x - 1)^2} \end{aligned}
  1. g(x)=1+xg(x) = 1 + \sqrt{x}
lim⁡z→xg(z)−g(x)z−x=z−xz−x=z−xz−x⋅z+xz+x=1z+x=12x\begin{aligned} \lim_{z \to x}\frac{g(z) - g(x)}{z - x} &= \frac{\sqrt{z} - \sqrt{x}}{z - x} \\ &= \frac{\sqrt{z} - \sqrt{x}}{z - x} \cdot \frac{\sqrt{z} + \sqrt{x}}{\sqrt{z} + \sqrt{x}} \\ &= \frac{1}{\sqrt{z} + \sqrt{x}} \\ &= \frac{1}{2\sqrt{x}} \end{aligned}

f(x)={x2,if x<0x,if x>00,if x=0f(x) = \begin{cases} x^2, & \text{if } x < 0 \\ x, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \end{cases}

xp=0x_p = 0

lim⁡h→0−(xp+h)2−xp2h=h+2xp=0\lim_{h \to 0^-}\frac{(x_p + h)^2 - x_p^2}{h} = h + 2x_p = 0
lim⁡h→0+xp+h−xph=1\lim_{h \to 0^+}\frac{x_p + h - x_p}{h} = 1

∴f(x)在P点不可微\therefore f(x)\text{在P点不可微}


f(x)={2,if x≤12x,if x>1f(x) = \begin{cases} 2, & \text{if } x \le 1 \\ 2x, & \text{if } x > 1 \end{cases}

xp=1x_p = 1

lim⁡h→1−2−2h=0\lim_{h \to 1^-}\frac{2 - 2}{h} = 0
lim⁡h→1+2(xp+h)−2xph=2\lim_{h \to 1^+}\frac{2(x_p + h) - 2x_p}{h} = 2

∴f(x)在P点不可微\therefore f(x)\text{在P点不可微}


f(x)={x,if 0<x<12x−1,if x≥1f(x) = \begin{cases} \sqrt{x}, & \text{if } 0 < x < 1 \\ 2x - 1, & \text{if } x \ge 1 \end{cases}

xp=1x_p = 1

lim⁡h→1−xp+h−xph=12xp=1/2\lim_{h \to 1^-}\frac{\sqrt{x_p + h} - \sqrt{x_p}}{h} = \frac{1}{2\sqrt{x_p}} = 1/2
lim⁡h→1+2(xp+h)−1−2xp+1h=2\lim_{h \to 1^+}\frac{2(x_p + h) - 1 - 2x_p + 1}{h} = 2

∴f(x)在P点不可微\therefore f(x)\text{在P点不可微}


f(x)={x,if x<11/x,if x≥1f(x) = \begin{cases} x, & \text{if } x < 1 \\ 1/x, & \text{if } x \ge 1 \end{cases}

xp=1x_p = 1

lim⁡h→1−xp+h−xph=1\lim_{h \to 1^-}\frac{x_p + h - x_p}{h} = 1
lim⁡h→1+1/(xp+h)−1/xph=−1xp2=−1\lim_{h \to 1^+}\frac{1/(x_p + h) - 1/x_p}{h} = -\frac{1}{x_p^2} = -1

∴f(x)在P点不可微\therefore f(x)\text{在P点不可微}

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