18.01学习笔记——导数

导数的定义

几何解释:函数图像在某点 \((x_0, y_0)\) 的切线斜率,

物理角度: 瞬时变化率,\(\frac{\Delta y}{\Delta x}, \Delta x \to 0\)\(\frac{\Delta y}{\Delta y} \to {f}'(x)\)

由此可得,\(f(x)\) 在点 \((x_0, f(x_0))\) 的导数为:

\[{f}'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]

\(f(x)\)\(x = x_0\) 处的极限与 \(f(x_0)\) 的取值无关

连续

  • \(\lim_{x \to {x_0^+}}f(x)\) 为 f(x) 在 \(x = x_0\) 处的右极限

  • \(\lim_{x \to {x_0^-}}f(x)\) 为 f(x) 在 \(x = x_0\) 处的左极限

间断点:

  • 跳跃间断点:左右极限均存在,但不相等
  • 可去间断点:左右极限存在且相等,但是 \(f(x)\)\(x=x_0\) 处不存在
  • 无穷间断点:左右极限有一个为 \(\infty\)

定理:\(f(x)\)\(x=x_0\) 可导,则 \(f(x)\)\(x=x_0\) 处连续

\( \lim_{x \to x_0} (f(x) - f(x_0) ) = \lim_{x \to x_0} \frac{f(x) - f(x_0) }{x - x_0} {(x - x_0)} = {f}'(x_0)(x-x_0) = 0\)

求导

特定函数求导

\[\sin{x}' = \cos{x} \]

\[\cos{x}' = -\sin{x} \]

\[\lim_{\Delta x \to 0} \frac{\cos{x} - 1}{x} = \frac{d}{dx}\cos{x}|_{x=0} = 0 \]

\[\lim_{\Delta x \to 0} \frac{\sin{x}}{x} = \frac{d}{dx}\sin{x}|_{x=0} = 1 \]

通用求导公式:

\[({u + v})' = {u}' + {v}' \]

\[{cu}' = c{u}' \]

\[{uv}' = {u}'v + u{v}' \]

\[{\frac{u}{v}}' = \frac{{u}'v - u{v}'}{v^2}, (v != 0) \]

证明:

\(\begin{equation} \notag ({u + v})' = \lim_{\Delta x \to 0}\frac{u(x+\Delta x) + v(x | \Delta x)}{\Delta x} = \lim_{\Delta x \to 0}\frac{u(x+\Delta x) - u(x)}{\Delta x} + \lim_{\Delta x \to 0}\frac{v(x+\Delta x) - v(x)}{\Delta x} = {u}' + {v}' \end{equation}\)

\(\begin{equation} \notag ({cu})' = \lim_{\Delta x \to 0} \frac{(cu(x+\Delta x) - (cu(x)))}{\Delta x} = c \cdot \lim_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} = c{u}'\end{equation}\)

\(\begin{equation} \notag ({uv})' = \lim_{\Delta x \to 0} \frac{(u(x+\Delta x)v(x+\Delta x) - (u(x)v(x)))}{\Delta x} = \lim_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x}v + \frac{v(x + \Delta x) - v(x)}{\Delta x}u = {u}'v + u{v}'\end{equation}\)

链式求导法则

\[ {f(g(x))}' = {f}'(g(x)){g}'(x) \]

\(v = g(x)\),则原式可以写为 \(f(g(x)) = f(v), v = g(x)\),则

\[ {f(g(x))}' = \frac{d}{dx}f(g(x)) = \frac{d}{dv}f(v) \cdot \frac{dv}{dx} \]

posted @ 2022-05-29 16:13  crwen  阅读(361)  评论(0编辑  收藏  举报