数学分析笔记5
导数
Def 1.
设\(f : I \rightarrow \mathbb{R}\), \(I \subseteq \mathbb{R}\), \(x_0 \in I\), 若极限
\[\lim_{x\rightarrow x_0} \dfrac{f(x) - f(x_0)}{x - x_0}\]
存在, 则称\(f\)在\(x_0\)点可导或可微, 称极限为\(f\)在\(x_0\)点的导数, 记作\(f'(x_0)\).且
\[f'(x_0)=\lim_{x\rightarrow x_0}\dfrac{f(x)-f(x_0)}{x-x_0}=\lim_{h\rightarrow 0}\dfrac{f(x_0+h)-f(x_0)}{h}\]
类似于连续，我们也可以定义左导数和右导数:
\[f'(x_0^+)=\lim_{h\rightarrow 0^+}\dfrac{f(x_0+h)-f(x_0)}{h}\]
\[f'(x_0^-)=\lim_{h\rightarrow 0^-}\dfrac{f(x_0+h)-f(x_0)}{h}\]
Remark 1. 可导：函数在这一点，长得像一次函数。
Statement 1. 设\(f:I\rightarrow \mathbb{R}\), \(I\) 为区间，\(x_0 \in I\) 且\(f\)在\(x_0\)点可导, 则 \(\exists h :I \rightarrow \mathbb{R},h(0)=0\),\(h\)在\(x_0\)连续，使得
\[f(x)=f(x_0)+f'(x-x_0)+h(x)(x-x_0)\]
Remark 2. 上述定理相当于，函数在这一点，长得像一次函数，但是有一个误差项\(h(x)\)，但误差\(h(x)\)是一个一阶的高阶无穷小。
Proof 2. 定义: \[h(x)= \left \{ \begin{array}{rl} 0 & x=x_{0} . \\ \frac{f(x)-f\left(x_{0}\right)-f^{\prime}\left(x_{0}\right)(x-x_0) }{x-x_{0} } & x \neq x_0, x \in I \end{array} \right. \]
由于
\[\begin{aligned}\lim _{x \rightarrow x_{0}} h(x) &=\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)-f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)}{x-x_{0}} \\ &=\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}-f^{\prime}\left(x_{0}\right) \\ &=0\end{aligned}\]
所以\(h(x)\)在\(x_0\)点连续，且\(h(0)=0\).
Remark 3. 由此，我们可以可导函数比连续函数更光滑。下面，我们可以得出可导是一个比连续更强的结论
Statement 2. 设 \(f: I \rightarrow \mathbb{R},I\)为区间，\(x_0 \in I\) 且\(f\)在\(x_0\)点可导, 则\(f\)在\(x_0\)点连续.
Statement 3. 函数的求导法则
- 四则运算
\[\begin{aligned} \left(f+g\right)^{\prime}(x) &=f^{\prime}(x)+g^{\prime}(x) \\ \left(f-g\right)^{\prime}(x) &=f^{\prime}(x)-g^{\prime}(x) \\ \left(fg\right)^{\prime}(x) &=f^{\prime}(x)g(x)+f(x)g^{\prime}(x) \\ \left(\frac{f}{g}\right)^{\prime}(x) &=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)} \end{aligned}\]
- 链式法则
\[\left(f\circ g\right)^{\prime}(x)=f^{\prime}\left(g(x)\right)g^{\prime}(x)\]
- 反函数求导
\[\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}\]
- 隐函数求导
\[\left(f(x,y)\right)^{\prime}(x)=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}\]