---
# System prepended metadata

title: 大一奇數微積分共筆 (E6 A204)(Ch.7)

---

# 大一奇數微積分共筆 (E6 A204)(Ch.7)

:::info
授課教師：高華隆
:::
- 輸入數學式的操作說明可以在[這裡](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)找到

- 有的中文沒抄到，所以是找[國家教育研究院的數學名詞中英對照](http://www.mathland.idv.tw/scene/tables/mathchieng.pdf)
- 上為簡易版，[詳細版可按此下載](http://terms.naer.edu.tw/terms/manager_admin/new_file_download.php?Pact=FileDownLoad&button_num=g5&source_id=284&Pval=1585)

- 授課速度有點快，歡迎幫忙校稿或補充

- 方便複製：
	- $\lim \limits_{h \to 0}$
	- $\dfrac {d}{dx}$
	- $\lim \limits_{x \to a} \dfrac {f(x)}{g(x)}$
	- $ [-\frac \pi 2, \frac \pi 2]$

# 微積分 Chapter7：Transcendental Functions

## 7-1 Inverse functions and their derivatives


- 應該是其他版的，正確版本沒問題→~~前面相關單元： 1-6、(2-5)、3-8~~

### One-to-one funtions

- **Defintion**

    ![](https://i.imgur.com/j1KchHD.png)
    - A function $f(x)$ is **one-to-one** on a domain $D$ if $f(x_1) \neq f(x_2)$ whenever $x_1 \neq x_2$ in $D$.


	$g(f(_j)) = x_j, \forall j$
	$f(g(a_j)) = a_j, \forall j$
	$g \equiv f^{-1}$ 
	
	
- **Example**

	![](https://i.imgur.com/MHGKjY6.png)
	![](https://i.imgur.com/1TSHhuR.png)


	:::success
	1. $f(x) = \sqrt x : [0, \infty) \to \Bbb R \Rightarrow f$ is one-to-one on $[0, \infty)$
	2. $f(x) = x^2 : \Bbb R \to \Bbb R \Rightarrow f$ is not one-to-one on $\Bbb R \because f(-1) = f(1) = 1$
	3. $f(x) = \sin x : [0,\pi] \to \Bbb R \Rightarrow f$ is not one-t-one on $[0,\pi] \because f(0) = f(\pi) = 0$
	4. 
	:::
	
	
- **Note:**

	:::warning
    1. Some function are not one-to-one on their natural domain, 
        but we can resrtict the function to a smaller domain 
    	s.t. the function is one-to-one

    2. $f$ is one-to-one on $D \Leftrightarrow$ each horizontal line intersects the grapf $y=f(x)$ at most one point.
	:::
	
	
### Inverse functions

- **Definition:**

	:::info
	$f : D \to f(D)$ is one-to-one on a domain $D$ with range $f(D).$
	The inverse function $f^{-1}:f(D) \to D$ of $f$ is defined by 
	$$f^{-1}(b) = a \quad \mathsf{if} \; f(a) = b, \quad \forall b \in f(D)$$
	
	:::
	
- **Note:**
	:::warning
	1. "$f^{-1}$" reads "$f$ inverse"
	2. $f^{-1} (f(x)) = x,\quad \forall x \in D$
		$f(f^{-1}(y)) = y,\quad \forall y \in f(D)$
	:::
	
### Finding inverses

- 圖形以 $x = y$為對稱軸，將x y對調
- 只有 one-to-one function 會擁有 inverse function

![](https://i.imgur.com/RvfXOCJ.png)



- **Note**

	:::warning
	The graph $y = f^{-1}(x)$ 
	is obtained by refleting the graph $y=f(x)$ about the line $y=x$
	
	:::
	
- **Step**

	:::info
	1. $y=f(x) \Rightarrow x = g(y)$
		ex: $y=f(x) - x^2 \Rightarrow x = \sqrt y =\equiv g(y)$
	
	2. Interchange, $x$ and $y$
		i.e. $y=g(x) \equiv f^{-1}(x)$
		i.e. $y= \sqrt x = f^{-1}(x)$
	:::
	
- **Example**

	![](https://i.imgur.com/fyr4X3u.png)

	
	Find the inverse of $f= \frac x 2 +1$,
	expressed as a function of $x$
	
	:::success
	**Sol:**
	$y = \frac x 2 +1 \Rightarrow x = 2y-2$
	$\Rightarrow x = f^{-1}(y) = 2y-2$
	$\Rightarrow y = f^{-1}(x) = 2x-2$
	
	:::


### Derivatives of inverses of differentiable functions

- **Recall**

	If $f$ and $f^{-1}$ are diff.
	:::info
	$$ f(f^{-1}(x)) = x$$
	:::
	$1 = \frac d {dx}  = \frac d {dx} f(f^{-1}(x)) = f'(f^{-1}(x)) \bullet \frac d {dx} f^{-1}(x)$
	$\Rightarrow \frac d {dx} f^{-1}(x) = \frac a {f'(f^{-1}(x))}$
	
- **Theorem**

	![](https://i.imgur.com/EBBsnQj.png)


	$I \subset \Bbb R$, on interval, $f:I \to f(I)$ is one-to-one
	
	1. If f is conti. on $I \Rightarrow f^{-1}$ is conti. on $I$
	2. If $f$ is diff. on $I$ and $f'(x) \neq 0, \quad \forall x \in I$
	3. $\Rightarrow f^{-1}$ is diff. on I and, 
	:::info
	$$\frac d {dx} f^{-1}(x) = \frac 1 {f'(f^{-1}(x))}$$
	:::
- **Proof**

	![](https://i.imgur.com/vOOSXRK.png)


	1. Check: $\lim \limits_{y \to y_0} f^{-1}(y_0), \quad \forall y_0 \in f(I)$
	:::info
	**Pf:**
	$y_0 = \lim \limits_{y \to y_0} y = \lim \limits_{y \to y_0} f(f^{-1}(y)) = f(\lim \limits_{y \to y_0} f^{-1}(y))$
	$\Rightarrow f^{-1}(y_0) = \lim \limits_{y \to y_0} f^{-1}(y)$
		$\because f$ is one-to-one $\Rightarrow f^{-1}(y_0) = \lim \limits_{y \to y_0} f^{-1}(y)$
	:::
	
	2. Check: $\lim \limits_{y \to y_0} \frac {f^{-1}(y) f^{-1}(y_0)} {y - y_0}$ exist
	:::info
	**Pf:**
	$\forall y, y_0 \in f(I)$
	$\Rightarrow \exists ! x, x_0 \in I$ s.t. $f(x) = y, f(x_0) = y_0$
	$\lim \limits_{y \to y_0} \frac {f^{-1}(y) - f^{-1}(y_0)}{y - y_0} = \lim \limits_{y \to y_0} \frac {x - x_0}{f(x) - f(x_0)}$
	:::
	沒抄到 
	
- **Example**
	:::success
	$f(x) = x^2, x>0 \Rightarrow f^{-1}(x) = \sqrt x, x>0$
	$\Rightarrow f'(x) = 2x$
	$f^{-1}(4) =$
	:::
	
- **Example**
	:::success
	$f(x) = x^3 - 2, x>0$
	Find $f^{-1}(x)$
	**Sol:**
	$f'(x) = 3x^2 \Rightarrow f'(2) = 12$
	$f'(6) =$
	:::
	
## 7-2 Natural logarithms

- 應該是其他版的，正確版本沒問題→~~1-6(定義)、3-8(微分)、8-3(三角函數積分)~~

### logarithmic functions

![](https://i.imgur.com/Zx0iohf.png)

### Natural logarithms

- **Defintion**

	The natural logarithms is the function defined by
	:::info
	$$\ln x = \int_1^x \frac 1 t dt, \quad \forall x >0$$
	:::

![](https://i.imgur.com/BPjqMWW.png)
![](https://i.imgur.com/hfPwU68.png)
![](https://i.imgur.com/dlHpH1h.png)


- **Note**



	:::warning
	1. $\ln 1 = 0$
	2. $\ln x > 0, \quad \forall x >1$
	3. $\ln x <0, \quad \forall 0 < x < 1$
	4. By F.T.O.C. $\Rightarrow \ln x$ is conti.
	   and diff. on $(0, \infty)$, and $\frac d {dx} \ln x = \frac 1 x$
	5. $\ln 2 < 1 < \ln 3$
	   By I.V.T. $\Rightarrow \exists$ a number $e \in (2,3)$ s.t.
	   $1 = \ln e = \int_1^e \frac 1 t dt$
	6. if $f(x) >0$ 
	   By chain rule $\Rightarrow \frac d {dx} \ln f(x) = \frac {f'(x)}{f(x)}$ 
	:::
	   
- **Example**
	:::success
	1. $\frac d {dx} \ln 2x = \frac 2 {2x} = \frac 1 x, x>0$
	2. $\frac d {dx} \ln (x^2 + 3) = \frac{2x}{x^2 +3}$
	3. $\frac d {dx} \ln |x| = \frac 1 {|x|} \cdot \frac x{|x|} = \frac x {x^2} = \frac 1 x$
	4. $\frac d {dx} \ln bx = \frac b {bx} = \frac 1 x, \quad \forall x,b >0$
	:::
	
- **Properties**

	![](https://i.imgur.com/6agf0rx.png)

	:::info
	1. $\ln bx = \ln b + \ln x, \quad \forall x, b >0$
	2. $\ln \frac b x = \ln b - \ln x, \quad \forall x, b >0$
	3. $\ln \frac 1 x = - \ln x, \quad \forall x >0$
	4. $\ln x^r = r \ln x, \quad \forall x >0, \forall r \in \Bbb Q$
	:::
	
### 多項式微分證明


- $$\frac d {dx} x^n = n x^{n-1}$$

- Case 1: $n \in \Bbb N$
    :::success
    **Proof:**(Using M.I.T.)
    $n = 1 \Rightarrow \frac d{dx} x=1 = 1x^{1-1}$
    
    assume 
    $\frac d{dx} x^{n-1} = (n-1) x^{(n-1)-1}$ 
    holds
    
    $\frac d {dx} x^n = \frac d {dx} x \bullet x^{n-1} = x^{n-1} +x(n-1)x^{n-2}$
    $= nx^{n-1}$
- Case 2: $\frac d {dx} x^{-n} = (-n) x^{-n-1}, \quad \forall n \in \Bbb N$
	:::success
	**Proof:**
    
    
	:::
- Case 3: $\frac d {dx} x^{\frac 1 m}, \quad m \in \Bbb N$
	:::success
	**Proof:**
	
	
- $\frac d {dx} x^n = n \cdot x^{n-1}, n \in \Bbb Q$
- Case:
	**Proof**
	
- $\ln x^r = r \ln x,\quad \forall r \in \Bbb Q$
- **Proof**


暫時省略

- **Note**
	:::warning
	$\ln x:(0, \infty) \to (- \infty, \infty)$ is 
	one-to-one, 
	onto, 
	incresing, 
	concave down
	:::
	
	
### Integrals

- $$\frac d {du} \ln |u| = \frac 1 u$$
	$\Rightarrow \int \frac 1 u du = \ln |u| + c$
	
	If $u = f(x),$ diff. and $f(x) \neq 0, \quad \forall x$
	$\Rightarrow du = f'(x) dx$
	$\Rightarrow \int \frac{f'(x)}{f(x)}dx = \ln |f(x)| +c$

- **Example**

	$\int_{-\frac \pi 2}^{\frac \pi 2} \dfrac{4 \cos \theta}{3 + 2 \sin \theta}du$, 
	
	:::success
	$u=3 + 2 \sin \theta$ 
	$du= 2 \cos \theta d \theta$
	
	$\int_1^5 \frac 2 u du = 2 \ln |u| |_1^5 = 2(\ln 5 - \ln 1) = 2 \ln 5$
	:::
	
### The integrals of tan x, cot x, sec x, csc x

- **Prop.**
	
	:::info
	1. $\int \tan u \; du = \ln |\sec u| + C$
	:::
	**Proof:**
		$\int \tan x \; dx = \int \dfrac {\sin x}{\cos x} dx$
		Let $u = \cos x ,\quad du = -\sin x \; dx$
		$= -\ln |u| + C$
		$= \ln \frac 1 {|\cos x|} + C$
		$= \ln |\sec x| +C$
	**Q.E.D**
	:::info
	2. $\int \cot x \; dx = \ln |\sin x| +C$
	:::
	**Proof:**
	$\int \cot x \; dx = \int \dfrac {\cos x}{\sin x} dx$
    Let $u = \sin x, \quad du = \cos x \; dx$
    $=\ln |u| + C$
    $=\ln |\sin x| +C$
	**Q.E.D**
	:::info
	3. $\int \sec x \; dx = \ln |\sec x + \tan x| +C$
	:::
	**Proof:**
	$\int \sec x \; dx$
	$= \int \dfrac {\sec ^2 x + \sec x \tan x}{\sec x + \tan x} dx$
	Let $u = \sec x + \tan x,\quad du = (\sec^2 x + \sec x \tan x)dx$
	$=\ln |u| + C$
    $=\ln |\sec x + \tan x|+C$
	**Q.E.D**
    :::info
	4. $\int \csc x \; dx = - \ln |\csc x + \cot x| + C$
	:::
    **Proof:**
    $\int \csc x \; dx$
    $= \int \dfrac{-\csc^2 x+\csc x \cot x}{\csc x + \cot x}$
    Let $u = \sec x + \cot x,\quad du = (-\csc^2 x + \csc x \cot x)dx$
    $= - \ln |u| +C$
    $= - \ln |\csc x + \cot x| + C$
	**Q.E.D**
- **Note**

	:::warning
	$y = \dfrac {f_1(x) f_2(x) ... f_k(x)}{g_1(x)g_2(x)...g_m(x)}$
	
	$\ln y = \ln f_1(x) + \ln f_2(x) + ... + \ln f_k(x) - \ln g_1(x) - \ln g_2(x) -...- \ln g_m(x)$
	
	$y' = (\dfrac {f_1'(x)}{f_1(x)} + ... + \dfrac {f_k'(x)}{f_k(x)} - \dfrac {g_1'(x)}{g_1(x)} - ... - \dfrac {g_m'(x)}{g_m(x)})y$
	:::
	
## 7-3 Exponential functions


- Recall 
	$\ln x = \int_1^x \dfrac 1 t dt:(0, \infty) \to (-\infty, \infty)$, one-to-one, onto
	
	$\Rightarrow \ln ^{-1} x : \Bbb R \to (0, \infty)$ exists
	Define ==$\exp (x) \equiv \ln ^{-1} x$==, $\quad \forall x \in \Bbb R$
	
	$\Rightarrow$ 
	1. $\ln e = 1 \Rightarrow \exp(1) = e = e^1$

	2. $\ln e^r = r \cdot \ln e = r \Rightarrow \exp(r) = e^r, \quad \forall r \in \Bbb Q$
	
	$\Rightarrow$
	

### e

The inverse function of $\exp (x)$ is $\;\ln (x)$
The inverse function of $\;\;\; e^x \;\;\;$ is $\;\log _e x$

- 
	$$\ln x = \log_e x$$
	
- **Note**
	:::warning
	1. $\ln x = \int_1^x \frac 1 t$
	2. $e^x = \exp (x)$
	3. $\ln x = 10$
	4. $e^{\ln x} = x$
	5. $\ln (e^x) = x$
	:::
	
- **Prop.** (General power rule for derivative)

	For $x \in \Bbb R, \quad n \in \Bbb R \Rightarrow$
	:::info
	$$\dfrac d {dx} x^n = n x ^{n-1}$$
	:::
	wherever $x^n$ and $x^{n-1}$ exist.
	
- **Proof**

	Case 1: $x>0$
	$x^n = e^{n \ln x}$
	
	$\dfrac d {dx} x^n = e^{n \ln x} n \dfrac 1 x = n x^n \dfrac 1 x$
	
	$= nx^{n-1}$
	
	Case 2: $x<0$
	Let $u=-x >0$
	
	$\dfrac d {dx} x^n = \dfrac d {dx} (-1)^n u^n = (-1)^n n u^{n-1} (-1)$
	
	$n(-u)^{n-1} = nx^{n-1}$
	
	Case 3: $x=0$
	
	$\lim \limits_{h \to 0} \frac {(0+h)^n - (0)^n}{h}= \lim \limits_{h \to 0} h^{n-1} = 0 = nx^{n-1}|$
	
	
- **Example**

	For $x>0$
	
	$\dfrac d {dx} x^x$ 
	
	$= \dfrac d {dx} e^{\ln x^x}$ 
	
	$= \dfrac d {dx} e^{x \ln x}$
	
	$= e^{x \ln x}(\ln x +1)$ 
	
	$= x^x(\ln x +1)$
	
	
### The number e

- **Theorem 4**
	:::info
	$e = \lim \limits_{h \to 0}(1+x)^{\dfrac 1 x} \equiv \lim \limits_{h \to \infty}(1 + \dfrac 1 t)^t$
	:::
	
- **Proof**
	Let $f(x) = \ln x \Rightarrow f'(x) = \dfrac 1 x$
	$\ln e = 1 = f'(1) = \lim \limits_{h \to 0} \dfrac{f(1+x) - f(1)} x$
	$= \lim \limits_{h \to 0} \frac 1 x \ln(1+x) = \lim \limits_{h \to 0} \ln (1+x)^{\frac 1 x}$
	$\therefore  \ln x$ is one-to-one
	$\Rightarrow e = \lim \limits_{h \to 0} (1+x)^{\dfrac 1 x}$
	
- **Theorem**

	
	
- **Properties:**

	$\dfrac {d}{dx} a^{f(x)}, \quad a>0$
	:::info
	1. $\dfrac {d}{dx} a^x = (\ln a) \cdot a^x, \quad \forall a >0$
		**Proof:** $\dfrac {d}{dx} a^x = \dfrac {d}{dx} e^{\ln a^x} = \dfrac {d}{dx} e^{x \ln a}$
	2. $\int a^x dx = \dfrac 1 {\ln a} a^x +c, \quad a>0$

	
	3. $\dfrac {d}{dx} a^{f(x)} = \ln a \cdot a^{f(x)} \cdot f'(x)$
	
		**Proof:** $\dfrac {d}{dx} e^{\ln a ^{f(x)}} = \dfrac {d}{dx} e^{(\ln a) \cdot f(x)}$

		$= e^{\ln a \cdot f(x)} = \dfrac {d}{dx} e^{(\ln a) \cdot f(x)}$

		$= e^{(\ln a) \cdot f(x)} \cdot \ln a \cdot f'(x)$

		$= a^{f(x)} \cdot \ln a \cdot f'(x)$
	
	
	4. $\int a^u du = \dfrac 1 {\ln a} a^u + c$

		**i.e.** $\int a^{f(x)} \cdot f'(x)dx = \dfrac 1 {\ln a} a^{f(x)} +c$
	
	
	5. If $a>1$ 
	    $\Rightarrow \ln a > 0 \Rightarrow \dfrac {d}{dx} a^x = \ln a \cdot a^x >0$
		$\Rightarrow a^x$ is increasing, one-to-one
		
	6. If $0<a<1 \Rightarrow \ln a <0$
		$\Rightarrow \dfrac {d}{dx} a^x <0, \dfrac {d^2}{dx^2} a^x >0$
		$\Rightarrow a^x$ is decresing, one-to-one, concave up
	:::
- **Example**

	1. $\dfrac {d}{dx} 3^x = (\ln 3) \cdot 3^x$
	2. $\dfrac {d}{dx} 3^{-x} = -(\ln 3) \cdot 3^{-x}$
	3. $\dfrac {d}{dx} 3^{\sin x} = (\ln 3) \cdot 3^{\sin x} \cdot \cos x$
	4. $\int 2^{\sin x} \cos x dx$
		$(u = \sin x, \quad du = \cos x dx)$
		$= \int 2^u du$
		$= \dfrac 1 {\ln 2} 2^{\sin x} +c$


### Logarithms with base $a: \log_a x$

- **Definition**

	For $a > 0, a \neq 1, \log_a x$ is the inverse function f $a^x : \Bbb R \to (0, \infty)$
	
- **Note**

	$a>0, a \neq 1$
	$a^x : \Bbb R \to (0, \infty)$
	$\log_a x:(0,\infty) \to \Bbb R$ ：
	圖
	
- **Properties**

	:::info
	1. $a^{log_a x} = x, \quad x>0$
	
	2. $\log_a a^x = x, \quad x \in \Bbb R$
	
	3. $\log_a xy = \log_a x + \log_a y$
	
	4. $\log_a \frac x y = \log_a x - \log_a y$
	
	5. $\log_a x^y = y \cdot \log_a x$
	
	6. $\log_a x = \dfrac {\ln x}{\ln a} = \dfrac{\log_e x}{\log_e a}$
	
	7. $\dfrac {d}{dx} \log_a x = \dfrac {d}{dx} \dfrac 1 {\ln a} \cdot \ln x =\dfrac 1 {\ln a} \cdot \dfrac 1 x$
	
	8. $\dfrac {d}{dx} \log_a f(x) = \dfrac {d}{dx} \dfrac{\ln f(x)} {\ln a} = \dfrac 1 {\ln a} \cdot \dfrac {f'(x)}{f(x)}$
	:::
	
- **Example**

	1. $\dfrac {d}{dx} \log_{10} (3x+1) = \dfrac 1 {\ln 10} \cdot \dfrac 3 {3x+1}$
	
	2. $\int \dfrac {\log_2 x } x dx = \int \dfrac {\ln x} {x \cdot \ln 2} dx$
		
		$(u = \ln x, \quad du = \frac {dx} x)$
		
		$= \dfrac 1 {\ln 2} \int udu$
		
		$= \dfrac 1 {2 \ln 2}(\ln x)^2$
	

### 用兩堂課講的四個重點微分公式

$$\mathsf{ 原式 \qquad 微分}$$

$$\ln x \qquad \frac 1 x$$

$$e^x \qquad e^x$$

$$a^{f(x)} \qquad e^{f(x) \ln a}$$

$$\log_a f(x) \qquad \frac {\ln f(x)}{\ln a}
$$


## 7-5 Indeterminate forms and L'Hôpital's rule 羅必達法則


### L'Hôpital's rule 羅必達法則

- Find 
	:::info
	$$\lim \limits_{x \to a} \dfrac {f(x)}{g(x)} = \lim \limits_{x \to a} \dfrac {f'(x)}{g'(x)}$$
    
	:::
	
	Case 1: 
	$$\dfrac {f(a)}{g(a)} = \dfrac 0 0 or \dfrac \infty \infty$$
	
- **Theorem 5 L’Hôpital’s rule**

	$a \in (\alpha, \beta) \subseteq \Bbb R, \quad f, g:(\alpha, \beta) \to \Bbb R$ are diff. on $(\alpha, \beta)$
	
	$f(a) = g(a) = 0$ and $g'(x) \neq 0, \quad \forall x \in (\alpha, \beta)$ \\$\{a\}$
	
	$\lim \limits_{x \to a} \dfrac {f(x)}{g(x)} = \lim \limits_{x \to a} \dfrac {f'(x)}{g'(x)}$ provied the limit of **RHS** exists
	
- **Example**

	1. $\lim \limits_{x \to 0} \dfrac {}{}$
	
	2. $\lim \limits_{x \to 0} \dfrac {}{}$
	
	3. $\lim \limits_{x \to 0} \dfrac{}{}$
	
	4. $\lim \limits_{x \to 0} \dfrac {}{}$
	
	5. $\lim \limits_{x \to 0} \dfrac {}{}$
	
	6. $\lim \limits_{x \to 0} \dfrac {}{}$
	
	7. $\lim \limits_{x \to 0} \dfrac {}{}$

### Cauchy's Mean Value Theorem 柯西均值定理

- **definition**

	$f, g: [a,b] \to \Bbb R$ are conti. on $[a,b]$, diff. on $(a,b)$
	$\Rightarrow \exists c \in (a,b)$ 
	$\qquad s.t. (f(b)-f(a)) g'(c)=(g(b)-g(a))f'(c)$
    $\qquad i.e. \dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
	
- **Proof**

	Let 
	$h(x) = (f(b)-f(a))(g(x)-g(a))-(g(b)-g(a))(f(x)-f(a))$
	
	$\Rightarrow h$ is conti. on $[a,b]$, diff. on $(a,b)$ 
	
	and $h(a) = 0 = h(b)$
	
	By Rolle's thm.
	$\Rightarrow \exists c \in (a,b)$ s.t.
	$0 = h'(c) = (f(b)-f(a))g'(c)-(g(b)-g(a))f'(c)$
	
- **Note**

	$\exists c \in (a,b)$ s.t.
	
	$\dfrac {f(b)-f(a)}{g(b)-g(a)} = \dfrac {f'(c)}{g'(c)}$ 
	
	if $g(a) \neq g(b)$
	
- **Proof of thm 5**

	略?還是現在下面的是在講證明
	
	$\lim \limits_{x \to a} \dfrac {f(x)}{g(x)} = \lim \limits_{x \to a} \dfrac {f'(x)}{g'(x)} \quad$ if $\dfrac {f(a)}{g(a)} = \dfrac 0 0$
	$\dfrac{f(a)}{g(a)} = \dfrac{\pm \infty}{\pm \infty}, \quad f(a) \cdot g(a) = 0 \cdot \infty, \quad f(a) - g(a) = \infty - \infty$
	
- **Note**

	If $f(x), g(x) \to \pm \infty$ as $x \to a$
	$\lim \limits_{x \to a} \dfrac {f(x)}{g(x)} = \lim \limits_{x \to a} \dfrac {\frac 1 {g(x)}}{\frac 1 {f(x)}} = \lim \limits_{x \to a} \dfrac {\frac {+ g'(x)} {g^2(x)}}{\frac {+f'(x)} {f^2(x)}} = \lim \limits_{x \to a} \dfrac {f(x)}{g(x)} \cdot \lim \limits_{x \to a} \dfrac {g'(x)}{ f'(x)}$
	$\lim \limits_{x \to a} \dfrac {g'(x)}{ f'(x)}= \lim \limits_{x \to a} \dfrac {f(x)}{ g(x)}$
	
- **Example**
	1. $\lim \limits_{x \to \frac \pi 2} \dfrac {\sec x}{ 1 +\tan x} = \lim \limits_{x \to \frac \pi 2} \dfrac {1}{ \cos x + \sin x} = \dfrac 1 {1+0} = 1$
	
	2. ($\frac \infty \infty$)
		$\lim \limits_{x \to \frac \pi 2} \dfrac {\sec x}{ 1 +\tan x} = \lim \limits_{x \to \frac \pi 2} \dfrac {1}{ \cos x + \sin x} = \dfrac 1 {1+0} = 1$
	
	3. ($\frac \infty \infty$)
		$\lim \limits_{x \to \infty} \dfrac {\ln x}{ 2 \sqrt x} = \lim \limits_{x \to \infty} \dfrac {\dfrac 1 {x}}{ \frac 1 {\sqrt x}} = \lim \limits_{x \to \infty} \dfrac {1}{ \sqrt x} = 1$
	
	4. ($\infty \cdot 0$) 待補
		$\lim \limits_{x \to \frac \pi 2} \dfrac {\sec x}{ 1 +\tan x} = \lim \limits_{x \to \frac \pi 2} \dfrac {1}{ \cos x + \sin x} = \dfrac 1 {1+0} = 1$
		
	5. ($0 \cdot \pm \infty$) 待補但我發現我沒拍到
		
	6. ($\infty - \infty$) 待補但我發現我沒拍到

- **Note**
	$1^{\infty}, 0^0, \infty^0$
	$\lim \limits_{x \to a} f(x) = \lim \limits_{x \to a} e^{\ln f(x)} = e^{\lim \limits_{x \to a} \ln f(x)}$ 
	
- **Example**
	1. $\lim \limits_{x \to 0^+} (1+x)^{\frac 1 x} = \lim \limits_{x \to 0^+} e^{\frac {\ln (1+x)} x} = e ^{\lim \limits_{x \to 0^+} \frac {\ln (1+x)} x} = e^{\lim \limits_{x \to 0^+} \frac {\frac 1 {1+x}} 1} = e^1 = e$
	
	2. $\lim \limits_{x \to 0^+} x^{\frac 1 x} = \lim \limits_{x \to 0^+} e^{\frac {\ln x} x} = e ^{\lim \limits_{x \to 0^+} \frac {\ln x} x} = e^{\lim \limits_{x \to 0^+} \frac {\ln x} x} = e^{\lim \limits_{x \to 0^+} \frac { \frac 1 x} 1} = e^0 = 1$

## 7-6 Inverse trigonometric functions 反三角函數

- **Definition**

	==反三角函數的值是一個角度==

	$\sin x:[-\frac \pi 2, \frac \pi 2] \to [-1, 1]$ is one-to-one 	
	$\Rightarrow \sin^{-1} x : [-1,1] \to [-\frac \pi 2, \frac \pi 2]$
	
	![](https://i.imgur.com/on36a9N.png) ![](https://i.imgur.com/DQ3wWFD.png)

	![](https://i.imgur.com/661SHBk.png)


### The Arcsine and Arccosine Function

- **Definition**

![](https://i.imgur.com/ZabHzPH.png)

- **Example**

	![](https://i.imgur.com/i6bi3LK.png)


	$\sin ^{-1} (1) = \dfrac \pi 2$
	算了我懶得打我相信大家都知道！！！
	
	$\cos ^{-1} (1) = 0$
	$\cos ^{-1} (\dfrac {\sqrt 3} 2) = \dfrac \pi 6$
	$\cos ^{-1} (\dfrac {\sqrt 2} 2) = \dfrac \pi 4$
	$\cos ^{-1} (\dfrac 1 2) = \dfrac \pi 3$
	$\cos ^{-1} (0) = \dfrac \pi 2$
	$\cos ^{-1} (-1) = \pi$
	$\cos ^{-1} (- \dfrac {\sqrt 3} 2) = \dfrac 5 6 \pi$
	$\cos ^{-1} (- \dfrac {\sqrt 2} 2) = \dfrac 3 4 \pi$
	$\cos ^{-1} (- \dfrac 1 2) = \dfrac  2 3 \pi$
	
> [name=教授] (前略)......然後某一天我就發現我好像能反著背英文字母了，你們知道為什麼我要跟你們講這個小故事了吧
	
- 對sin有夠熟就知道arcsin在幹嘛了
	
- **Note**
	1. $\sin ^{-1} (-x) = -\sin ^{-1} x$,  ==**odd function**==
	
	2. $\cos ^{-1} + cos^{-1} (-x) = \pi$
		
		**Proof**
		Let $\theta = \cos^{-1} (x) \Rightarrow \cos \theta = x$
		$\cos(\pi - \theta) = - \cos \theta = -x$
		$\Rightarrow \cos^{-1} (-x) = \pi - \theta = \pi - cos^{-1} x$
	
	3. $\sin^{-1} x + \cos^{-1} x = \dfrac \pi 2, \quad \forall x \in [0,1]$
	
	4. 
		![](https://i.imgur.com/L68QUae.png)


- **Example**
	![](https://i.imgur.com/ouhEKSr.png)
	
	我要睡著了 所以我要先睡覺

### Inverses of tan x, cot x, sec x, and csc x

![](https://i.imgur.com/T6y1odG.png)
![](https://i.imgur.com/L68QUae.png)

- 睡著了所以tan x, cotx, sec x都沒打

	$\csc x:[- \dfrac \pi 2, 0) \cup (0, \dfrac \pi 2] \to (-\infty, -1] \cup [2, \infty)$ is one-to-one
	$\Rightarrow \csc^{-1} x:(-x,-1] \cup [1, \infty) \to [-\dfrac \pi 2, 0) \cup (0, \dfrac \pi 2]$
	
- **Note**
	$\sec^{-1} x + \csc^{-1} x = \dfrac \pi 2$
	
- **Derivatives**

    $(\dfrac d {dx} f^{-1}(x) = \dfrac 1 {f'(f(x))})$

	1. $\dfrac d {dx} \sin ^{-1} x = \dfrac 1 {\sqrt {1-x^2}}$


### 反三角函數的定義域、值域、微分


- 定義域和值域

	$\sin^{-1} x : [-1, 1] \to [- \dfrac \pi 2, \dfrac \pi 2]$
	
	$\cos^{-1} x : [-1,1] \to [0, \pi]$
	
	$\tan^{-1} x : \Bbb R \to (-\dfrac \pi 2, \dfrac \pi 2)$
	
	$\cot^{-1} x : \Bbb R \to (0, \pi)$
	
	$\sec^{-1} x: \Bbb R \setminus (-1,1) \to [0, \pi] \setminus \{\dfrac \pi 2\}$
	
	$\csc^{-1} x: \Bbb R \setminus (-1,1) \to [- \dfrac \pi 2, \dfrac \pi 2] \setminus \{0\}$

- **Derivatives**

	1. $\dfrac {d}{dx} \sin^{-1} x = \dfrac 1 {\sqrt {1-x^2}}, \quad |x|<1$
	
	2. $\dfrac {d}{dx} \cos^{-1} x = \dfrac {-1}{\sqrt{1-x^2}}, \quad |x|<1$
	
	3. $\dfrac {d}{dx} \tan^{-1} x = \dfrac 1 {1+x^2}, \quad x \in \Bbb R$
		- **Proof:** 
			$\dfrac {d}{dx} \tan^{-1} x = \dfrac 1 {\sec ^2(tan^{-1} x)} = \dfrac 1 {1+\tan^2 (tan^{-1} x)} = \dfrac 1 {1+x^2}$
	
	4. $\dfrac {d}{dx} \cot ^{-1} x = \dfrac {-1}{1+x^2}, \quad x \in \Bbb R$
	
	5. $\dfrac {d}{dx} \sec ^{-1} x = \dfrac 1 {|x| \sqrt{x^2 - 1}}, \quad |x|>1$
		- **Proof:**
			$\dfrac {d}{dx} \sec^{-1} = \dfrac 1 {\sec(\sec ^{-1} x) \cdot \tan(\sec ^{-1} x}$
			$= \dfrac 1 {|x| \sqrt {x^2 -1}}, \quad |x| > 1$
	
	6. $\dfrac {d}{dx} \csc ^{-1} x = \dfrac {-1}{|x| \sqrt {x^2 -1}} , \quad |x| > 1$
	
- **Example**

	1. $\dfrac {d}{dx} \sin ^{-1} x^2 = \dfrac {2x} {\sqrt {1-x^4}}$
	2. $\dfrac {d}{dx} \sec ^{-1} (5x^4) = \dfrac {20x^3} {5x^4 \sqrt {25x^8 -1}} = \dfrac 4 {x \sqrt {25x^8 - 1}}$

- **Integral**

	For $a>0$

	1. $\int \dfrac 1 { \sqrt {a^2 - u^2}} du = \sin ^{-1} ( \dfrac u a) +c \quad$, For $|u|<a$
	2. $\int \dfrac 1 { a^2 + u^2} du = \dfrac 1 a \tan ^{-1} ( \dfrac u a) +c \quad$, $\forall u$
	3. $\int \dfrac 1 { u \sqrt {u^2 - a^2}} du = \dfrac 1 a \sec ^{-1} (  |\dfrac u a|) +c \quad$, $\forall |u| > a$

- **Example**
	1. $\int ^{\frac{\sqrt 3}{2}}_{\frac{\sqrt 2}{2}} \dfrac{1}{\sqrt{1-x^2}}$

	   $=\sin^{-1}x ^{\frac{\sqrt 3}{2}} _{\frac{\sqrt 2}{2}}$
       
       $= \frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}$
       
	2. $\int \dfrac {dx} {\sqrt{3 - 4x^2}}$
		
		($u = 2x, \quad du = 2dx$)
		
		$= \dfrac 1 2 \int \dfrac 1 {\sqrt {3 - u^2}} du$
		
	3. $\int \dfrac 1 {\sqrt {e^{2x} -6}} dx$
		
		$= \dfrac 1 {\sqrt 6} \sec ^{-1} \dfrac {e^x} {\sqrt 6} + c$
	
	4. $\int \dfrac 1 { \sqrt {4x-x^2}}$ 
		
		$= \int \dfrac 1 {\sqrt {4 - (4-4x+x^2)}} dx$
		
		$= \int \dfrac 1 {\sqrt {4 - (x-2)^2}} dx$
		
		($u = x-2, \quad du = dx$)
		
		$= \int \dfrac 1 {\sqrt {4-u^2}} du$
		
		$= \sin ^{-1} (\dfrac {x-2} 2) +c$
		
## 7-8 Relative rates of growth

(衝到 $\infty$ 的速度比較) 
($\ln$最慢，而且是越多越慢)


![](https://i.imgur.com/UdclGaX.png)


- **Definition**

	$f(x), g(x) \to \infty$ as $x \to \infty$
	
	![](https://i.imgur.com/vM9xnu3.png)

	
	1. $f$ grows faster than $g$ ( $g$ is of smaller order than $f$ )
		as $x \to \infty$ (write $g = o(f)$) if $\lim \limits_{x \to \infty} \dfrac {g(x)} {f(x)} = 0$
		(i.e. $\lim \limits_{x \to \infty} \dfrac {f(x)} {g(x)} = \infty$
 	
	2. $f$ and $g$ grow at the same rate as $x \to \infty$ (write $f \sim g$)
 
	3. $f$ is of at most order of $g$ as $x \to \infty$ if $\exists M>0$ s.t. $\dfrac{f(x)}{g(x)} ≤ M$ for large $x$.(write $f = O(g)$)

- **Note**

	1. $h = o(g), g=o(f) \Rightarrow h=o(f)$
	
	2. $f \sim g \Rightarrow g \sim f$
	
	3. $h \sim g, g \sim f \Rightarrow h \sim f$
	
	4. $x^n = o(x^m), m>n$
	
	5. $x^n = o(e^x), \forall n>0$
	
	6. $\ln x = o(x^n), \forall n >0$
	
	7. $p(x) = a_nx^n + ... + a_0, a_n > 0$
		$q(x) = b_m x^m + ... + b_0, b_n > 0$
        $\Rightarrow$ 1. $p = o(q)$ if $m>n$
		2. $p \sim q$ if $m = n$
		
- **Example**
    1. $2^x = o(3^x)$$(since \lim \limits_{x \to \infty} \frac{2^x}{3^x} = 0)$
    2. $a^x = o(b^x)$ if $0<a<b$
    3. $a>1, b>1$ $\Rightarrow log_a x \sim log_b x$

(第八章另開檔案)