總之一個因緣際會複習微積分，所以寫下這篇文章。
本篇文章部分要仰賴一些先備知識：
$\displaystyle{ \lim_{\theta \to 0} \frac{sin(\theta)}{\theta} = 1 }$
（證明可以參考：sinx/xについて覚えておくべき２つのこと）
題外話，在 tutor 板上出現過為什麼不可以用羅必達法則（l’Hôpital’s Rule）證明的問題。
有興趣可以自己前去觀看：[討論] limsinx/x＝1 when x→0
另一個部分是：
$\displaystyle{ \lim_{\theta \to 0} \frac{cos(\theta) - 1}{\theta} = 0 }$
推倒如下。
$\displaystyle{ \lim_{\theta \to 0} \frac{cos(\theta) - 1}{\theta} = \lim_{\theta \to 0} \frac{cos(\theta) - 1}{\theta} \cdot \frac{cos(\theta) + 1}{cos(\theta) + 1}}$
$\displaystyle{ = \lim_{\theta \to 0} \frac{cos^{2}(\theta) - 1^{2}}{\theta [cos(\theta) + 1]} = \lim_{\theta \to 0} \frac{-sin^{2}(\theta)}{\theta [cos(\theta) + 1]}}$
$\displaystyle{ = - \lim_{\theta \to 0} \frac{sin(\theta)}{\theta}\frac{sin(\theta)}{cos(\theta) + 1} = - { \lim_{\theta \to 0} \frac{sin(\theta)}{\theta} \cdot \lim_{\theta \to 0} \frac{sin(\theta)}{cos(\theta) + 1} }}$
$\displaystyle{ = - { 1 \cdot \frac{0}{1 + 1} } = 0 }$
sin(x) 正弦
利用微分定義推導：
$\displaystyle{ \frac{d}{dx} sin(x) = \lim_{h \to 0} \frac{sin(x + h) - sin(x)}{h} }$
$\displaystyle{ = \lim_{h \to 0} \frac{sin(x)cos(h) + cos(x)sin(h) - sin(x)}{h} \text{（※ 正弦和角公式）} }$
$\displaystyle{ = \lim_{h \to 0} \frac{sin(x)cos(h) - sin(x) + cos(x)sin(h)}{h} }$
$\displaystyle{ = \lim_{h \to 0} \frac{sin(x) [cos(h) - 1]}{h} + \lim_{h \to 0} \frac{cos(x)sin(h)}{h} }$
$\displaystyle{ = sin(x) \cdot \lim_{h \to 0} \frac{cos(h) - 1}{h} + cos(x) \cdot \lim_{h \to 0} \frac{sin(h)}{h} }$
$\displaystyle{ = sin(x) \cdot 0 + cos(x) \cdot 1 = cos(x) }$
cos(x) 餘弦
利用微分定義推導：
$\displaystyle{ \frac{d}{dx} cos(x) = \lim_{h \to 0} \frac{cos(x + h) - cos(x)}{h} }$
$\displaystyle{ = \lim_{h \to 0} \frac{cos(x)cos(h) - sin(x)sin(h) - cos(x)}{h} \text{（※ 餘弦和角公式）} }$
$\displaystyle{ = \lim_{h \to 0} \frac{cos(x)cos(h) - cos(x) - sin(x)sin(h)}{h} }$
$\displaystyle{ = \lim_{h \to 0} \frac{cos(x) [cos(h) - 1]}{h} - \lim_{h \to 0} \frac{sin(x)sin(h)}{h} }$
$\displaystyle{ = cos(x) \cdot \lim_{h \to 0} \frac{cos(h) - 1}{h} - sin(x) \cdot \lim_{h \to 0} \frac{sin(h)}{h} }$
$\displaystyle{ = cos(x) \cdot 0 - sin(x) \cdot 1 = - sin(x) }$
tan(x) 正切
由 $tan(x) = \frac{sin(x)}{cos(x)}$ 開始：
$\frac{d}{dx} tan(x) = \frac{d}{dx} \frac{sin(x)}{cos(x)}$
$ = \frac{\frac{d}{dx}sin(x) \cdot cos(x) - sin(x) \cdot \frac{d}{dx}cos(x)}{cos^{2}(x)} \text{（※ 微分除法公式）}$
$ = \frac{cos(x) \cdot cos(x) - sin(x) \cdot [-sin(x)]}{cos^{2}(x)} = \frac{cos^{2}(x) + sin^{2}(x)}{cos^{2}(x)}$
$ = \frac{1}{cos^{2}(x)} = sec^{2}(x)$
（※ 使用了 $sin^{2}(x) + cos^{2}(x) = 1$ 這條式子，證明請參考：sin^2+cos^2=1）
csc(x) 餘割
由 $csc(x) = \frac{1}{sin(x)}$ 開始：
$\frac{d}{dx} csc(x) = \frac{d}{dx} \frac{1}{sin(x)}$
$ = \frac{\frac{d}{dx} 1 \cdot sin(x) - 1 \cdot \frac{d}{dx} sin(x)}{sin^{2}(x)} \text{（※ 微分除法公式）}$
$ = \frac{0 \cdot sin(x) - 1 \cdot cos(x)}{sin^{2}(x)} = - \frac{cos(x)}{sin^{2}(x)}$
$ = - \frac{cos(x)}{sin(x)} \cdot \frac{1}{sin(x)} = - cot(x) csc(x)$
sec(x) 正割
由 $sec(x) = \frac{1}{cos(x)}$ 開始：
$\frac{d}{dx} sec(x) = \frac{d}{dx} \frac{1}{cos(x)}$
$ = \frac{\frac{d}{dx} 1 \cdot cos(x) - 1 \cdot \frac{d}{dx} cos(x)}{cos^{2}(x)} \text{（※ 微分除法公式）}$
$ = \frac{0 \cdot cos(x) - 1 \cdot [-sin(x)] }{cos^{2}(x)} = \frac{sin(x)}{cos^{2}(x)}$
$ = \frac{sin(x)}{cos(x)} \cdot \frac{1}{cos(x)} = tan(x) sec(x)$
cot(x) 餘切
由 $cot(x) = \frac{cos(x)}{sin(x)}$ 開始：
$\frac{d}{dx} cot(x) = \frac{d}{dx} \frac{cos(x)}{sin(x)}$
$ = \frac{\frac{d}{dx}cos(x) \cdot sin(x) - cos(x) \cdot \frac{d}{dx} sin(x)}{sin^{2}(x)} \text{（※ 微分除法公式）}$
$ = \frac{[-sin(x)] \cdot sin(x) - cos(x) \cdot cos(x)}{sin^{2}(x)} = \frac{ - sin^{2}(x) - cos^{2}(x)}{sin^{2}(x)}$
$ = - {\frac{sin^{2}(x) + cos^{2}(x)}{sin^{2}(x)}} = - \frac{1}{sin^{2}(x)} = -csc^{2}(x)$
（※ 跟 $tan(x)$ 一樣使用了 $sin^{2}(x) + cos^{2}(x) = 1$ 這條式子，證明請參考上面的資料。）
其他
可以發現一些手法，除了 $sin(x)$ 以及 $cos(x)$ 要從定義來之外，
其他的有兩兩對應的步驟，像是 $tan(x)$ 跟 $cot(x)$ 就很類似、而 $csc(x)$ 跟 $sec(x)$ 也很相像。