【数学】求导
导数
对于函数 \(f\left(x\right)\)，定义其导数为 \(f'\left(x\right)\)。
\(f'\left(x\right)\) 可以看作 \(f\left(x\right)\) 在相应位置的斜率。
多项式函数的求导
对于一个多项式函数 \(f\left(x\right) = \sum_{i=0} a_i x^i\)，这个函数的导数为 \(f'\left(x\right) = \sum_{i = 0} i \cdot a_i x ^ {i - 1}\)（系数乘上指数，指数上减一）。
特殊函数导数
\[\left(\sin\left(x\right)\right)' = \cos\left(x\right)
\]
\[\left(\cos\left(x\right)\right)' = -\sin\left(x\right)
\]
\[\left(\ln\left(x\right)\right)' = \dfrac{1}{x}
\]
\[\text{指数函数：}\left(a^x\right)' = a^x\ln\left(a\right)\text{（将会在后文给出证明）}
\]
\[\text{特别的：}\left(e^x\right)' = e^x\ln\left(e\right) = e^x \cdot 1 = e^x
\]
一般函数的求导
对于一个一般函数 \(f\left(x\right)\)，其在 \(x\) 处的导数为
\[\lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x}
\]
和积商求导法则
和差
\[\left(f\left(x\right) \pm g\left(x\right)\right)' = f'\left(x\right) \pm g'\left(x\right)
\]
证明略
积
\[\left(f\left(x\right)g\left(x\right)\right)' = f'\left(x\right)g\left(x\right) + f\left(x\right)g'\left(x\right)
\]
证明：
\[\left(f\left(x\right)g\left(x\right)\right)' = \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x\right)}{\Delta x}
\]
\[= \lim_{\Delta x \to 0} \dfrac{\left(f\left(x + \Delta x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x + \Delta x\right)\right) + \left(f\left(x\right)g\left(x + \Delta x\right) - f\left(x\right)g\left(x\right)\right)}{\Delta x}
\]
\[= \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x}g\left(x + \Delta x\right) + f\left(x\right)\lim_{\Delta x \to 0}\dfrac{g\left(x + \Delta x\right) - g\left(x\right)}{\Delta x}
\]
\[= f'\left(x\right)g\left(x\right) + f\left(x\right)g'\left(x\right)
\]
商
\[\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)' = \dfrac{f'\left(x\right)g\left(x\right) - f\left(x\right)g'\left(x\right)}{g^2\left(x\right)}
\]
证明：
\[\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)' = \lim_{\Delta x \to 0} \dfrac{\dfrac{f\left(x + \Delta x\right)}{g\left(x + \Delta x\right)} - \dfrac{f\left(x\right)}{g\left(x\right)}}{\Delta x}
\]
\[= \lim_{\Delta x \to 0} \dfrac{f\left(x + \Delta x\right)g\left(x\right) - g\left(x + \Delta x\right)f\left(x\right)}{g\left(x +\Delta x\right)g\left(x\right)\Delta x}
\]
\[= \lim_{\Delta x \to 0} \dfrac{\left(f\left(x + \Delta x\right)g\left(x\right) - f\left(x\right)g\left(x\right)\right) - \left(g\left(x + \Delta x\right)f\left(x\right) - f\left(x\right)g\left(x\right)\right)}{g\left(x +\Delta x\right)g\left(x\right)\Delta x}
\]
\[= \lim_{\Delta x \to 0} \dfrac{\dfrac{f\left(x + \Delta x\right) - f\left(x\right)}{\Delta x}g\left(x\right) - \dfrac{g\left(x + \Delta x\right) - g\left(x\right)}{\Delta x}f\left(x\right)}{g\left(x +\Delta x\right)g\left(x\right)}
\]
\[= \dfrac{f'\left(x\right)g\left(x\right) - f\left(x\right)g'\left(x\right)}{g^2\left(x\right)}
\]
复合函数
\[\left(f\left(g\left(x\right)\right)\right)' = f'\left(g\left(x\right)\right)g'\left(x\right)
\]
证明：
咕
其他
那么，关于前文“特殊函数求导”的指数函数求导，可以简单给出证明了。
\[\text{设} y = a^x
\]
\[\text{则有} \ln\left(y\right) = x\ln\left(a\right)
\]
\[\left(\ln\left(y\right)\right)' = \left(x\ln\left(a\right)\right)'
\]
\[\dfrac{y'}{y} = 1 \cdot \ln\left(a\right) + 0
\]
（这一步左边使用了复合函数求导法则，右边使用了积的求导法则。）
再将右边的分母上的 \(y\) 乘到右边，并替换回 \(a^x\)，就得到了：
\[y' = a^x\ln\left(a\right)
\]
