第一单元用python学习微积分（七）第一单元总结_用python学习微积分(七)-优快云博客

本文链接：https://blog.youkuaiyun.com/bullseye/article/details/121892942

本文回顾了麻省理工学院的单变量微积分课程，讲解了e底法和对数微分的证明，强调了自然对数在变化率表示中的应用，并提供了第一单元的公式总结，包括通用公式、特殊公式及习题解答。讨论了函数连续性的判断方法和对特定函数的求导过程。

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本文内容来自学习麻省理工学院公开课：单变量微积分-第一次考试复习-网易公开课

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(4) 函数在a是否连续，要看a点处函数的左极限切线斜率是否等于右极限切线斜率

(5) 对求导

一、对 $\frac{d}{dx}x^r = rx^{r-1}$ 的证明

1、e底法

$\frac{d}{dx}x^r = \frac{d}{dx}(e^{lnx})^r = \frac{d}{dx}e^{rlnx} = \frac{d}{dx}rlnx\frac{d}{dx}e^{rlnx} = \frac{r}{x} e^{rlnx}$

$=\frac{r}{x} x^r =rx^{r-1}$

2、对数微分法

$ln(x^r) = rln(x)$

$(ln(x^r))' =(rln(x))' = \frac{r}{x}$ (由 $(ln(u))' = \frac{u'}{u}$ )

$(x^r)' = \frac{r}{x}x^r = rx^{r-1}$

二、自然对数是自然的

当考虑变化量时，通常是用变化量除以总量，当FTSE在一天中跌掉27.9，我们考虑一天跌掉的百分比

$\frac{\Delta p}p = \frac{27.9}{6432} =$ .43% ，而当考虑瞬时变化时则有 $\frac{p'}p = (ln(p))'$

三、第一单元复习

1、通用的公式

$(u+v)' = u' +v'; (cu)' = cu';(uv)' = u'v + v'u; (\frac{u}v)' = \frac{uv' -v'u}{v^2}$

$\frac{d}{dx}f(u) = f'(u)u'(x)...... [u = u(x)]$

2、特殊的公式

$(x^r )'= rx^{r-1}$

$(sin(x))' = cos(x)$

$(cos(x))' = -sin(x)$

$tan(x) = \frac{1}{(cos(x))^2} = (sec(x))^2$

$(sec(x))' = sec(x) tan(x)$

$(ex)' = ex$

$(ln(x))' = \frac{1}{x}$

$(tan^{-1}(x))' = \frac{1}{x^2 +1}......tan^{}-1 =>atan$

$(sin^{-1}(x))' = \frac{1}{(1-x^2)^{\frac{1}{2}}}......sin^-1=>asin$

3、习题：

(1) $(sec(x))' = ((cos(x))^-1)' = -(cos(x))^{-2}(-sin(x)) = \frac{sin(x)}{(cos(x))^2}$

$=sec(x)tan(x)$

(2) $(ln(sec(x)))' = \frac{1}{sec(x)}(sec(x))' = \frac{sec(x)tan(x)}{sec(x)} = tan(x)$

(3) $((x^{10}+8x)^6)' =6(x^{10}+8x)^5(10x^9+8)$

(4) $(e^{xtan^{-1}x})' = e^{xtan^{-1}x}(xtan^{-1}x)' = e^{xtan^{-1}x}(tan^{-1}x + \frac{x}{x^2+1})$

4、其他公式

(1) $f(x)' = \lim_{\Delta x \rightarrow 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}}$

(2) $\lim_{u \rightarrow 0}{\frac{e^u-1}{u}} = \lim_{u \rightarrow 0}{\frac{e^{u+\Delta u}-e^u}{\Delta u}} = \lim_{u \rightarrow 0}{\frac{e^{u+\Delta u}-e^u}{\Delta u}} =\frac{d}{dx}e^u|_{u=0} = 1$

验证下：

from sympy import *
import numpy as np 
import matplotlib.pyplot as plt 

u = symbols('u')
y = (np.e**u - 1)/u
limit_expr = limit(y, u, 0) 
limit_expr

(3) $y = sin^{-1}(x)$ 求导

$x = sin(y)$

$\frac{d}{dx}(x = sin(y))$

$1 = cos(y)y'$

$y' = \frac{1}{cos(y)} = \frac{1}{(1-(sin(y))^2)^{\frac{1}{2}}} = \frac{1}{(1-x^2)^{\frac{1}{2}}}$

(4) 函数在a是否连续，要看a点处函数的左极限切线斜率是否等于右极限切线斜率

(5) 对 $y = tan^{-1}(x)$ 求导

$\frac{d}{dx}(tan(y) = x)$

$(sec(y))^2y' = 1$

$y' = \frac{1}{(sec(y))^2} = (cos(y))^2$

辅助作图：

from sympy import *
import numpy as np 
import matplotlib.pyplot as plt 

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.spines['left'].set_position('zero')
ax.spines['bottom'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
ax.set_aspect( 1 ) 
def plotText(plt,x,y, text):
plt.text(x+0.1, y+0.01, text, fontsize=12)
    
plt.plot([0,1],[0,0], c='b')    
plt.plot([1,1],[0,tan(45/180*np.pi)], c='c',label='tan(y)....== x')    
plotText(plt, 1, 0.5, 'x')
plt.plot([0,1],[0,tan(45/180*np.pi)],c='r')
plotText(plt, 0.2,0.6,'(1+x^2)^(1/2)')
plotText(plt, 0,0.02,'y')
plt.legend(loc='upper right')
plt.show()