導數
對於函數 \(f\left(x\right)\)，定義其導數為 \(f'\left(x\right)\)。
\(f'\left(x\right)\) 可以看作 \(f\left(x\right)\) 在相應位置的斜率。
多項式函數的求導
對於一個多項式函數 \(f\left(x\right) = \sum_{i=0} a_i x^i\)，這個函數的導數為 \(f'\left(x\right) = \sum_{i = 0} i \cdot a_i x ^ {i - 1}\)（系數乘上指數，指數上減一）。
特殊函數導數
\[\left(\sin\left(x\right)\right)' = \cos\left(x\right) \]
\[\left(\cos\left(x\right)\right)' = -\sin\left(x\right) \]
\[\left(\ln\left(x\right)\right)' = \dfrac{1}{x} \]
\[\text{指數函數：}\left(a^x\right)' = a^x\ln\left(a\right)\text{（將會在后文給出證明）} \]
\[\text{特別的：}\left(e^x\right)' = e^x\ln\left(e\right) = e^x \cdot 1 = e^x \]
一般函數的求導
對於一個一般函數 \(f\left(x\right)\)，其在 \(x\) 處的導數為
\[\lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x} \]
和積商求導法則
和差
\[\left(f\left(x\right) \pm g\left(x\right)\right)' = f'\left(x\right) \pm g'\left(x\right) \]
證明略
積
\[\left(f\left(x\right)g\left(x\right)\right)' = f'\left(x\right)g\left(x\right) + f\left(x\right)g'\left(x\right) \]
證明：
\[\left(f\left(x\right)g\left(x\right)\right)' = \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x\right)}{\Delta x} \]
\[= \lim_{\Delta x \to 0} \dfrac{\left(f\left(x + \Delta x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x + \Delta x\right)\right) + \left(f\left(x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x\right)\right)}{\Delta x} \]
\[= \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x}g\left(x + \Delta x\right) + f\left(x\right)\lim_{\Delta x \to 0}\dfrac{g\left(x + \Delta x\right) - g\left(x\right)}{\Delta x} \]
\[= f'\left(x\right)g\left(x\right) + f\left(x\right)g'\left(x\right) \]
商
\[\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)' = \dfrac{f'\left(x\right)g\left(x\right) - f\left(x\right)g'\left(x\right)}{g^2\left(x\right)} \]
證明：
\[\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)' = \lim_{\Delta x \to 0} \dfrac{\dfrac{f\left(x + \Delta x\right)}{g\left(x + \Delta x\right)} - \dfrac{f\left(x\right)}{g\left(x\right)}}{\Delta x} \]
\[= \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right)g\left(x\right) - g\left(x + \Delta x\right)f\left(x\right)}{g\left(x +\Delta x\right)g\left(x\right)\Delta x} \]
\[= \lim_{\Delta x \to 0} \dfrac{\left(f\left(x + \Delta x\right)g\left(x\right) - f\left(x\right)g\left(x\right)\right) - \left(g\left(x + \Delta x\right)f\left(x\right) - f\left(x\right)g\left(x\right)\right)}{g\left(x +\Delta x\right)g\left(x\right)\Delta x} \]
\[= \lim_{\Delta x \to 0} \dfrac{\dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x}g\left(x\right) - \dfrac{g\left(x + \Delta x\right) - g\left(x\right)}{\Delta x}f\left(x\right)}{g\left(x +\Delta x\right)g\left(x\right)} \]
\[= \dfrac{f'\left(x\right)g\left(x\right) - f\left(x\right)g'\left(x\right)}{g^2\left(x\right)} \]
復合函數
\[\left(f\left(g\left(x\right)\right)\right)' = f'\left(g\left(x\right)\right)g'\left(x\right) \]
證明：
咕
其他
那么，關於前文“特殊函數求導”的指數函數求導，可以簡單給出證明了。
\[\text{設} y = a^x \]
\[\text{則有} \ln\left(y\right) = x\ln\left(a\right) \]
\[\left(\ln\left(y\right)\right)' = \left(x\ln\left(a\right)\right)' \]
\[\dfrac{y'}{y} = 1 \cdot \ln\left(a\right) + 0 \]
（這一步左邊使用了復合函數求導法則，右邊使用了積的求導法則。）
再將右邊的分母上的 \(y\) 乘到右邊，並替換回 \(a^x\)，就得到了：
\[y' = a^x\ln\left(a\right) \]