导数的基本公式推导
主要推导了人教版A版数学选择性必修二上直接给出的基本的导数公式
本文写于作者初三暑假，更新于高一暑假
可能含有很多不足，如果您方便的话可以联系我修改 awa
大概率会在高二暑假再更新一次吧
一、导数的四则运算法则
设 \(f(x),~g(x)\) 均为可导函数
命题：\({[f(x) \pm g(x)]}^{\prime} = f^{\prime}(x) \pm g^{\prime}(x)\)
证明：
\[ \begin{aligned} {[f(x) \pm g(x)]}^{\prime}&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x) \pm\left[g(x+\Delta x)-g(x)\right]}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}} \pm \lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}} \\\\&=f^{\prime}(x)\pm g^{\prime}(x) \end{aligned} \]
命题：\([f(x)g(x)]^{\prime}=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)\)
证明：
\[ \begin{aligned} {[f(x)g(x)]}^{\prime} &=\lim\limits_{\Delta x \to 0}{\dfrac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}} \\\\&=\lim\limits_{\Delta x \to 0}{\dfrac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{g(x+\Delta x)}+f(x)\lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}} \\\\&=f^{\prime}(x)g(x)+f(x)g^{\prime}(x) \end{aligned} \]
命题：\(\left[\dfrac{f(x)}{g(x)}\right]^{\prime} = \dfrac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{\left[g(x)\right]^2}\)
证明：
\[ \begin{aligned} {\left[\dfrac{f(x)}{g(x)}\right]}^{\prime}&=\lim\limits_{\Delta x\to 0}{\dfrac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)g(x)-g(x+\Delta x)f(x)}{\Delta xg(x+\Delta x)g(x)}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)g(x)-f(x)g(x)}{\Delta xg(x+\Delta x)g(x)}}-\lim\limits_{\Delta x \to 0}{\dfrac{g(x+\Delta x)f(x)-f(x)g(x)}{\Delta xg(x+\Delta x)g(x)}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{\dfrac{g(x)}{g(x+\Delta x)g(x)}} - \lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{\dfrac{f(x)}{g(x+\Delta x)g(x)}} \\\\&=f^{\prime}(x)\dfrac{g(x)}{\left[g(x)\right]^2} - g^{\prime}(x)\dfrac{f(x)}{\left[g(x)\right]^2} \\\\&=\dfrac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{\left[g(x)\right]^2} \end{aligned} \]
二、复合函数的求导法则（链式法则）
命题：设函数 \(y=f(u)\) 在 \(u=g(x)\) 处可导，而 \(u=g(x)\) 在 \(x\) 处可导，
则函数 \(y=f(g(x))\) 在 \(x\) 处可导，且 \[ {\left[f(g(x))\right]}^{\prime}=f^{\prime}(g(x))\cdot g^{\prime}(x) \]
证明：
\[ \begin{aligned} {\left[f(g(x))\right]}^{\prime}&=\lim\limits_{\Delta x \to 0}\dfrac{\Delta y}{\Delta g(x)} \\\\&=\lim\limits_{\Delta x \to 0}\dfrac{\Delta y}{\Delta u}\cdot\dfrac{\Delta u}{\Delta x} \end{aligned} \] 因为 \(u=g(x)\) 在 \(x\) 处可导，所以它在这一点也是连续的
也就是 \(\Delta u \to 0\) 当 \(\Delta x \to 0\) ，故 \[ \begin{aligned} {\left[f(g(x))\right]}^{\prime}&=\lim\limits_{\Delta u \to 0}\dfrac{\Delta y}{\Delta u}\lim\limits_{\Delta x \to 0}\dfrac{\Delta u}{\Delta x} \\\\&=f^{\prime}(u)g^{\prime}(x) \\\\&=f^{\prime}(g(x))\cdot g^{\prime}(x) \end{aligned} \]
三、基本初等函数的导数公式
命题：若 \(f(x)=x^a(a\in \mathbb{N})\) ，则 \(f^{\prime}(x) = a x^{a-1}\)
证明：
\[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x\to 0}{\dfrac{(x+\Delta x)^{a}-(x)^{a}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\sum\limits_{r=0}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r}}-C_{a}^{0}x^{a}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\sum\limits_{r=1}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r}}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\sum\limits_{r=2}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r-1}+C_{a}^{1}x^{a-1}}} \\\\&=C_a^{1}x^{a-1} \\\\&=a x^{a-1} \end{aligned} \]
可以推广到 \(a\in \mathbb{R}\) 的情况，即
命题：若 \(f(x)=x^a(a\in \mathbb{R})\) ，则 \(f^{\prime}(x) = a x^{a-1}\)
证明：
\[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x \to 0}\dfrac{(x+\Delta x)^a-x^a}{\Delta x} \\\\&=\lim\limits_{\Delta x \to 0}x^a\dfrac{\left(1+\frac{\Delta x}{x}\right)^a-1}{\Delta x} \\\\&=x^a\lim\limits_{\Delta x \to 0}\dfrac{\left(1+\frac{\Delta x}{x}\right)^a-1}{\ln\left(1+\frac{\Delta x}{x}\right)^a}\cdot\dfrac{\ln\left(1+\frac{\Delta x}{x}\right)^a}{\Delta x} \end{aligned} \] 令 \(t=(1+\frac{\Delta x}{x})^a-1\) ，则 \[ \begin{aligned} f^{\prime}(x)&=x^a\lim\limits_{\Delta x \to 0}\dfrac{t}{\ln\left(1+t\right)} \cdot \dfrac{a\ln\left(1+\frac{\Delta x}{x}\right)}{\Delta x} \\\\&=x^a\lim\limits_{\Delta x \to 0}\dfrac{t}{\ln\left(1+t\right)} \cdot \dfrac{\ln\left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}} \cdot\dfrac{a}{x} \\\\&=ax^{a-1} \end{aligned} \]
命题：若 \(f(x)=\sin x\) ，则 \(f^{\prime}(x)=\cos x\)
证明：
\[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x\to 0}{\dfrac{\sin (x+\Delta x)-\sin x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\sin x \cos\Delta x+\sin \Delta x \cos x-\sin x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\sin x(\cos\Delta x-1)+\sin\Delta x\cos x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\sin x\left[\left(1-2\sin^2\frac{\Delta x}{2}\right)-1\right]+\cos x\left(2\sin\frac{\Delta x}{2}\cos \frac{\Delta x}{2}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\frac{\Delta x}{2}\cos\frac{\Delta x}{2}\cos x-2\sin^2\frac{\Delta x}{2}\sin x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\frac{\Delta x}{2}\left(\cos\frac{\Delta x}{2}\cos x-\sin x\sin \frac{\Delta x}{2}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\frac{\Delta x}{2}\cos\left(\frac{\Delta x}{2}+x\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\cos\left(\dfrac{\Delta x}{2}+x\right)\dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}} \\\\&=\lim\limits_{\Delta x\to 0}\cos\left(\dfrac{\Delta x}{2}+x\right) \\\\&=\cos x \end{aligned} \]
命题：若 \(f(x)=\cos x\) ，则 \(f^{\prime}(x) = -\sin x\)
证明：
\[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x\to 0}{\dfrac{\cos(x+\Delta x)-\cos x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}\dfrac{\cos x\cos \Delta x- \sin \Delta x\sin x-\cos x}{\Delta x} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\cos x(\cos\Delta x-1)-\sin \Delta x\sin x}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\cos x \left[\left(1-2\sin^2\frac{\Delta x}{2}\right)-1\right]-\sin x\left(2\sin\frac{\Delta x}{2}\cos\frac{\Delta x}{2}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin^2\frac{\Delta x}{2}\cos x-2\sin\frac{\Delta x}{2}\sin x \cos\frac{\Delta x}{2}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin\frac{\Delta x}{2}\left(\sin \frac{\Delta x}{2}\cos x+\sin x \cos \frac{\Delta x}{2}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin \frac{\Delta x}{2}\sin\left(x+\frac{\Delta x}{2}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{-\sin\left(x+\dfrac{\Delta x}{2}\right)\dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}} \\\\&=\lim\limits_{\Delta x\to 0}{-\sin\left(x+\dfrac{\Delta x}{2}\right)} \\\\&=-\sin x \end{aligned} \]
命题：若 \(f(x)=a^x (a>0,\,\texttt{且}\,a \ne 1)\) ，则 \(f^{\prime}(x)=a^x\ln a\)
证明： \[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x\to 0}{\dfrac{a^{x+\Delta x}-a^{x}}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{a^x(a^{\Delta x}-1)}{\Delta x}} \end{aligned} \] 令 \(t=a^{\Delta x}-1\) ，则\(\Delta x = \log_a(t+1)\) \[ \begin{aligned} &\because\Delta x\to 0 \\\\&\therefore a^{\Delta x}\to 1 \\\\&\therefore t+1\to1 \end{aligned} \] 即 \[ t \to 0 \]
则原极限可化为 \[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{t\to 0}{\dfrac{a^xt}{\log_a(t+1)}} \\\\&=\lim\limits_{t\to 0}{\dfrac{a^x}{\frac{1}{t}\log_a(t+1)}} \\\\&=\lim\limits_{t\to 0}{\dfrac{a^x}{\log_a(1+t)^{\frac{1}{t}}}} \\\\&=a^x\dfrac{1}{\log_ae} \\\\&=a^x\dfrac{\ln a}{\ln e} \\\\&=a^x\ln a \end{aligned} \]
特别地，当 \(a=e\) 时，即 \(f(x)=e^x\) ，有 \(f^{\prime}(x)=e^x\)
命题：若 \(f(x)=\log_ax(a>0,\,\texttt{且}\,a \ne 1)\) ，则 \(f^{\prime}(x) = \dfrac{1}{x\ln a}\)
证明： \[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{\Delta x\to 0}{\dfrac{\log_a(x+\Delta x)-\log_ax}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\dfrac{\log_a\left(1+\frac{\Delta x}{x}\right)}{\Delta x}} \\\\&=\lim\limits_{\Delta x\to 0}{\log_a{\left(1+\dfrac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}}} \end{aligned} \] 令 \(t=\dfrac{\Delta x}{x}\) ，则 \(\dfrac{1}{\Delta x} = \dfrac{1}{tx}\) \[ \begin{aligned} &\because \Delta x\to 0 \\\\&\therefore \dfrac{\Delta x}{x}\to 0 \\\\&\therefore t\to 0 \end{aligned} \] \(\therefore\) 原极限可化为 \[ \begin{aligned} f^{\prime}(x)&=\lim\limits_{t\to 0}{\log_a\left(1+t\right)^{\frac{1}{tx}}} \\\\&=\lim\limits_{t\to 0}{\dfrac{1}{x}\log_a\left(1+t\right)^{\frac{1}{t}}} \\\\&=\dfrac{1}{x}\log_ae \\\\&=\dfrac{1\ln e}{x\ln a} \\\\&=\dfrac{1}{x\ln a} \end{aligned} \]
特别地，当 \(a=e\) 时，即 \(f(x)=\ln x\) ，有 \(f^{\prime}(x)=\dfrac{1}{x}\)
命题：若 \(f(x)=\varphi ( \varphi \texttt{为常数})\) ，则 \(f^{\prime}(x)=0\)
证明： \[ \begin{aligned} f^{\prime}(x) &=\lim\limits_{\Delta x\to 0}{\dfrac{\varphi-\varphi}{\Delta x}} \\&=0 \end{aligned} \]
参考文献
[1] 复合函数的导数（链式法则）
[2] 基本初等函数的导数
