初识微积分二

微分和求导

• 导函数的定义：

$$li{m}_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

$$fracddxf(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

%matplotlib inline

def f(x):
    return x**2 + x

x = list(range(0, 11))
y = [f(i) for i in x]
x1 = 3
y1 = f(x1)
h = 3
x2 = x1+h
y2 = f(x2)

plt.xlabel('x')
plt.ylabel('f(x)')
plt.grid()
plt.plot(x,y, color='green')
plt.scatter(x1,y1, c='red')
plt.annotate('(x,f(x))',(x1,y1), xytext=(x1-0.5, y1+3))

plt.scatter(x2,y2, c='red')
plt.annotate('(x+h, f(x+h))',(x2,y2), xytext=(x2+0.5, y2))
plt.show()

导函数也可记为

$$f^{\prime }(\text{a})=\underset{h\to 0}{\lim }\frac{f(\text{a}+h)-f(\text{a})}{h}\iff f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$$

求导数

$$f(x)=x^2+x$$

$$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$

$$f^{\prime }(\text{2})=\underset{h\to 0}{\lim }\frac{f(\text{2}+h)-f(\text{2})}{h}$$

$$f^{\prime }(2)=\underset{h\to 0}{\lim }\frac{((2+h{)}^2+2+h)-(2^2+2)}{h}$$

$$f^{\prime }(2)=\underset{h\to 0}{\lim }\frac{(h^2+5h+6)-6}{h}=\underset{h\to 0}{\lim }\frac{h^2+5h}{h}=\underset{h\to 0}{\lim }h+5=5$$

%matplotlib widget

def f(x):
    return x**2 + x

x = list(range(0, 11))
y = [f(i) for i in x]

x1 = 2
y1 = f(x1)

m = 5

plt.xlabel('x')
plt.ylabel('f(x)')
plt.grid()

plt.plot(x,y, color='green')
plt.scatter(x1,y1, c='red')
plt.annotate('(x,f(x))',(x1,y1), xytext=(x1-0.5, y1+3))

xMin = x1 - 5
yMin = y1 - (5*m)
xMax = x1 + 5
yMax = y1 + (5*m)
plt.plot([xMin,xMax],[yMin,yMax], color='magenta')

plt.show()

FigureCanvasNbAgg()

求导函数

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$
$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{((x+h{)}^2+x+h)-(x^2+x)}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{x^2+h^2+2xh+x+h-x^2-x}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{h^2+2xh+h}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }2x+h+1$$

$$f^{\prime }(x)=2x+1$$

可微

并不是所有的函数都是可以求导函数的。

可微的函数有下面的几何特征

• 连续
• 切线不竖直
• 光滑

$$q(x)=\{\begin{array}{ll}\frac{40,000}{x^2},& \text{if }x<-4,\\ (x^2-2)\cdot (x-1),& \text{if }x\ne 0\text{ and }x\ge -4\text{ and }x<8,\\ (x^2-2),& \text{if }x\ne 0\text{ and }x\ge 8\end{array}$$

%matplotlib inline

def q(x):
    if x != 0:
        if x < -4:
            return 40000 / (x**2)
        elif x < 8:
            return (x**2 - 2) * x - 1
        else:
            return (x**2 - 2)

x = [*range(-10, -5), -4.01]
x2 = [*range(-4, 8), 7.9999, *range(8, 11)]

y = [q(i) for i in x]
y2 = [q(i) for i in x2]

plt.xlabel('x')
plt.ylabel('q(x)')
plt.grid()

plt.plot(x,y, color='purple')
plt.plot(x2,y2, color='purple')
plt.scatter(-4,q(-4), c='red')
plt.annotate('A (x= -4)',(-5,q(-3.9)), xytext=(-7, q(-3.9)))
plt.scatter(0,0, c='red')
plt.annotate('B (x= 0)',(0,0), xytext=(-1, 40))
plt.scatter(8,q(8), c='red')
plt.annotate('C (x= 8)',(8,q(8)), xytext=(8, 100))

plt.show()

• A不连续
• B不连续
• C不光滑

求导规则

基本规则

$$f(x)=\pi \therefore f^{\prime }(x)=0$$

$$f(x)=2g(x)\therefore f^{\prime }(x)=2g^{\prime }(x)$$

$$f(x)=g(x)+h(x)\therefore f^{\prime }(x)=g^{\prime }(x)+h^{\prime }(x)$$
$$f(x)=k(x)-l(x)\therefore f^{\prime }(x)=k^{\prime }(x)-l^{\prime }(x)$$

$$\frac{d}{dx}(2x+6)=\frac{d}{dx}2x+\frac{d}{dx}6=2$$

幂规则

极其常用的规则
\begin{equation}f(x) = x^{n} ;; \therefore ;; f’(x) = nx^{n-1}\end{equation}

例如：

$$f(x)=x^3\therefore f^{\prime }(x)=3x^2$$

$$f(x)=x^{-2}\therefore f^{\prime }(x)=-2x^{-3}$$

$$f(x)=x^2\therefore f^{\prime }(x)=2x$$

幂规则的推导

$$f(x)=x^2$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{(x+h{)}^2-x^2}{h}$$

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{x^2+h^2+2xh-x^2}{h}=\underset{h\to 0}{\lim }\frac{h^2+2xh}{h}=\underset{h\to 0}{\lim }h+2x=2x$$

乘法规则

$$\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$

例如

$$f(x)=2x^2$$

$$g(x)=x+1$$

$$f^{\prime }(x)=4x$$

$$g^{\prime }(x)=1$$

$$\frac{d}{dx}[f(x)g(x)]=(4x\cdot (x+1))+(2x^2\cdot 1)$$

$$\frac{d}{dx}[f(x)g(x)]=6x^2+4x$$

商数规则

$$r(x)=\frac{s(x)}{t(x)}$$
$$r^{\prime }(x)=\frac{s^{\prime }(x)t(x)-s(x)t^{\prime }(x)}{（t(x){）}^2}$$

例如

$$s(x)=3x^2$$

$$t(x)=2x$$

$$r^{\prime }(x)=\frac{(6x\cdot 2x)-(3x^2\cdot 2)}{（2x{）}^2}=\frac{6x^2}{4x^2}=\frac{3}{2}$$

链式规则

$$\frac{d}{dx}[o(i(x))]=o^{\prime }(i(x))\cdot i^{\prime }(x)$$

例如

$$i(x)=x^2$$

$$o(x)=2x$$

$$o^{\prime }(x)=2,i^{\prime }(x)=2x$$

$$\frac{d}{dx}[o(i(x))]=4x$$

极值和优化

对函数$k(x)=-10x^2+100x+3$ 有导函数：

$$k^{\prime }(x)=-20x+100$$

%matplotlib inline

def k(x):
    return -10*(x**2) + (100*x)  + 3

def kd(x):
    return -20*x + 100

x = list(range(0, 11))
y = [k(i) for i in x]

yd = [kd(i) for i in x]

plt.axhline()
plt.axvline()
plt.xlabel('x (time in seconds)')
plt.ylabel('k(x) (height in feet)')
plt.xticks(range(0,15, 1))
plt.yticks(range(-200, 500, 20))
plt.grid()

plt.plot(x,y, color='green')

plt.plot(x,yd, color='purple')

x1 = 2
x2 = 5
x3 = 8
plt.plot([x1-1,x1+1],[k(x1)-(kd(x1)),k(x1)+(kd(x1))], color='r')
plt.plot([x2-1,x2+1],[k(x2)-(kd(x2)),k(x2)+(kd(x2))], color='r')
plt.plot([x3-1,x3+1],[k(x3)-(kd(x3)),k(x3)+(kd(x3))], color='r')

plt.show()

找最大值和最小值

对 $k(x)=-10x^2+100x+3$ 有 $k^{\prime }(x)=-20x+100$

$$-20x+100=0$$

$$x=5$$

如果求二阶导数

$$k^{\prime }(x)=-20x+100\Rightarrow k^{\prime \prime }(x)=-20$$

根据二阶导数为常量，而且是负常量，得知一阶导函数是线性下降的。导函数为0的点就是极大值。

函数
$$w(x)=x^2+2x+7$$

%matplotlib inline

def w(x):
    return (x**2) + (2*x) + 7

def wd(x):
    return 2*x + 2

x = list(range(-10, 11))
y = [w(i) for i in x]

yd = [wd(i) for i in x]

plt.axhline()
plt.axvline()
plt.xlabel('x')
plt.ylabel('w(x)')
plt.xticks(range(-10,15, 1))
plt.yticks(range(-200, 500, 20))
plt.grid()

plt.plot(x,y, color='g')
plt.plot(x,yd, color='m')
plt.show()

极值点

对函数
$$v(x)=x^3-2x+100$$

%matplotlib widget

def v(x):
    return (x**3) - (2*x) + 100

def vd(x):
    return 3*(x**2) - 2

x = list(range(-10, 11))
y = [v(i) for i in x]

yd = [vd(i) for i in x]

plt.axhline()
plt.axvline()
plt.xlabel('x')
plt.ylabel('v(x)')
plt.xticks(range(-10,15, 1))
plt.yticks(range(-1000, 2000, 100))
plt.grid()
plt.plot(x,y, color='g')
plt.plot(x,yd, color='m')
plt.show()

FigureCanvasNbAgg()

%matplotlib inline

def k(x):
    return -10*(x**2) + (100*x)  + 3

def kd(x):
    return -20*x + 100

def k2d(x):
    return -20

plt.axhline()
plt.axvline()
x = list(range(0, 11))
y = [k(i) for i in x]
yd = [kd(i) for i in x]
y2d = [k2d(i) for i in x]

plt.xlabel('x')
plt.ylabel('k(x)')
plt.xticks(range(0,15, 1))
plt.yticks(range(-200, 500, 20))
plt.grid()
plt.plot(x,y, color='g')
plt.plot(x,yd, color='r')
plt.plot(x,y2d, color='b')

plt.show()

练习：

$$w(x)=x^2+2x+7$$

%matplotlib inline

def w(x):
    return (x**2) + (2*x) + 7

def wd(x):
    return 2*x + 2

def w2d(x):
    return 2

x = list(range(-10, 11))
y = [w(i) for i in x]
yd = [wd(i) for i in x]
y2d = [w2d(i) for i in x]

plt.axhline()
plt.axvline()
plt.xlabel('x (time in days)')
plt.ylabel('w(x) (flowers)')
plt.xticks(range(-10,15, 1))
plt.yticks(range(-200, 500, 20))
plt.grid()
plt.plot(x,y, color='green')
plt.plot(x,yd, color='purple')
plt.plot(x,y2d, color='magenta')

plt.show()

极值点不一定是最大值或者最小值

$$v(x)=x^3-6x^2+12x+2$$

$$v^{\prime }(x)=3x^2-12x+12=0$$

$x=2$ 是极值点

%matplotlib inline

def v(x):
    return (x**3) - (6*(x**2)) + (12*x) + 2

def vd(x):
    return (3*(x**2)) - (12*x) + 12

def v2d(x):
    return (3*(2*x)) - 12

from matplotlib import pyplot as plt

x = list(range(-5, 11))
y = [v(i) for i in x]
yd = [vd(i) for i in x]
y2d = [v2d(i) for i in x]


plt.xlabel('x')
plt.ylabel('v(x)')
plt.xticks(range(-10,15, 1))
plt.yticks(range(-2000, 2000, 50))
plt.grid()

plt.plot(x,y, color='green')
plt.plot(x,yd, color='purple')
plt.plot(x,y2d, color='magenta')
plt.show()

print ("v(2) = " + str(v(2)))

print ("v'(2) = " + str(vd(2)))

print ("v''(2) = " + str(v2d(2)))

v(2) = 10
v'(2) = 0
v''(2) = 0