函数的微分

单变量函数的微分学

张瑞
中国科学技术大学数学科学学院

函数的微分

微分

定义 1. (可微)
设$y=f(x)$在区间$I$内有定义，$x_0,x_0+\Delta x$在$I$内，$\Delta y=f(x_0+\Delta x)-f(x_0)$为函数的增量，若

\[\Delta y=A\Delta x+o(\Delta x),\quad \Delta x\to0 \]

其中$A=A(x)$不依赖$\Delta x$。 则称$f(x)$在$x_0$处可微，而$A\Delta x$则为函数$f(x)$的在$x_0$处的微分，记为$dy$或$df$，

\[dy=A\Delta x , \ \ \ \mbox{or}\ \ \ df=A\Delta x \]

定理 1.
$f(x)$在$x_0$处可微，当且仅当$f(x)$在$x_0$可导，且

\[dy=f'(x_0)dx \]

注： 对于一元函数，可微与可导等价。

证： ($\Rightarrow$)

\[\begin{aligned} \lim_{\Delta x\to0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}&=\lim_{\Delta x\to0}\dfrac{A\Delta x+o(\Delta x)}{\Delta x} \\ &=\lim_{\Delta x\to0}\left(A+\dfrac{o(\Delta x)}{\Delta x}\right)=A \end{aligned} \]

所以，$f(x)$在$x_0$处可导。

($\Leftarrow$)

\[\lim_{\Delta x\to0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}=f'(x_0) \]

则有

\[\lim_{\Delta x\to0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}-f'(x_0)=0 \]

这样，

\[\begin{aligned} &\lim_{\Delta x\to0}\dfrac{\Delta y-f'(x_0)\Delta x}{\Delta x} \\ =&\lim_{\Delta x\to0}\dfrac{f(x_0+\Delta x)-f(x_0)-f'(x_0)\Delta x}{\Delta x} \\ =&0 \end{aligned} \]

即

\[\Delta y=f'(x_0)\Delta x+o(\Delta x) \]

所以$f(x)$可微，且

\[dy=f'(x_0)\Delta x \]

取$f(x)=x$，则有

\[dx=(x)' \Delta x=\Delta x \]

这样，有

\[dy=f'(x_0)\Delta x=f'(x_0) dx \]

导数：强调的是变化率，

微分：强调的是从$\Delta y$中，取出$\Delta x$的线性部分

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.8] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm % \draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \path[name path=axis x] (-0.4,0) -- (2.8,0); \draw[->] (-0.4,0) -- (2.8,0) node[right] {$x$}; \draw[->] (0,-0.2) -- (0,2.6) node[left] {$y$}; \draw[blue, name path=func] (-0.2,0.3) .. controls (1,0.3) and (2,0.6) .. (2.5,1.9); \path[name path=l1] (1.1,0) -- (1.1,2.9); \path[name path=l2] ($ (1,0)+(0.01,0) $) -- ($ (1,2.9)+(0.01,0) $); \path[name intersections={of=func and l1, name=o}]; \coordinate (m) at (o-1); \path[name path=slant] ($(m)!-0.5!(2.6,1.7)$) -- (2.6, 1.7); \path[name intersections={of=func and slant}]; \coordinate (n) at (intersection-2); %\node at (n) {$N$}; \fill (m) circle(1pt) node[above] {$M$}; \fill (n) circle(1pt) node[above] {$N$}; % \node (a) {$M$}; \coordinate (xe) at (2.7, 0); \coordinate (orig) at (0,0); \draw[dashed] (n-|orig) -- (n -| xe); \draw[dashed] (m) -- (m -| xe); \draw[dashed] (m) -- (m|-orig) node[below] {$x$}; \draw[dashed, name path=line dx] (n|-orig) node[below] {$x+\Delta x$}--(n); \node[below] at ($ (m)!0.5!(m-|n) $) {$dx=\Delta x$}; \node[right] at ($ (n)!0.5!(m-|n)+(0.2,0) $) {$\Delta y$}; \draw[purple, <->] ($ (n)+(0.3,0) $) -- ($(m-|n)+(0.3,0) $); \path[name intersections={of=func and l2, name=p}]; \draw[red, name path=tangent] ($ (o-1)!-15!(p-1) $)--($ (o-1)!10!(p-1) $); \path[name intersections={of=tangent and line dx}]; \coordinate (dy) at (intersection-1); \node[right] at ($ (dy)!0.5!(m-|n) $) {$dy$}; \draw[red, <->] (dy) -- (m-|n); \draw (dy)--($ (dy)+(0.2,0) $); \draw (m-|n) -- (m -| xe); \draw (n) -- (n -| xe); \end{tikzpicture}

定义 2.
若函数$f(x)$在区间$I$内的每一点都可微，则称函数$f(x)$在区间$I$上可微。记为

\[dy=f'(x)dx \]

这样

\[\dfrac{dy}{dx}=f'(x), \mbox{ or } \dfrac{df(x)}{dx}=f'(x) \]

所以，导数是微分$dy$与自变量微分$dx$的商；导数又称为微商

初等函数微分公式

\[\begin{aligned} &d(\sin(x))=\cos(x)dx, &&d(\cos(x))=-\sin(x)dx, \\ &d(\tan(x))=\sec^2(x)dx=\dfrac{dx}{\cos^2(x)}, &&d(\cot(x))=\dfrac{-dx}{\sin^2(x)}, \\ &d(\arcsin(x))=\dfrac{1}{\sqrt{1-x^2}}dx, &&d(\arccos(x))=\dfrac{-1}{\sqrt{1-x^2}}dx, \\ &d(\arctan(x))=\dfrac{1}{1-x^2}dx, \\ &d(e^x)=e^xdx, &&d(a^x)=a^x \ln adx, \\ &d(\ln(|x|))=\dfrac{1}{x}dx, &&d(\log_a x)=\dfrac{1}{x\ln a}dx, \\ &d(x^\mu)=\mu x^{\mu-1}dx, \end{aligned} \]
\[\begin{aligned} d(\sinh(x))&=d\left(\frac{e^x-e^{-x}}2\right)=\frac{e^x+e^{-x}}2dx=\cosh(x)dx , \\ d(\cosh(x))&=\sinh(x)dx \\ d(\tanh(x))&=\displaystyle\frac1{\cosh^2(x)}dx \\ d(\mbox{arcsinh}(x))&=d(\ln(x+\sqrt{1+x^2}))=\dfrac{1}{\sqrt{1+x^2}}dx \\ d(\mbox{arccosh}(x))&=d(\ln(x+\sqrt{x^2-1}))=\dfrac{1}{\sqrt{x^2-1}}dx \\ d(\mbox{arctanh}(x))&=d\left(\frac12 \ln\dfrac{1+x}{1-x}\right)=\dfrac{1}{1-x^2}dx \end{aligned} \]

微分的四则运算

$u(x)$,$v(x)$可微，则有

\[\begin{aligned} d(cu)=& cdu,(c=const) \\ d(u\pm v)=&du\pm dv \\ d(uv)=&u'dv+v'du \\ d(\dfrac{u}{v})=&\dfrac{vdu-udv}{v^2} \\ \end{aligned} \]

复合函数微分

$y=f(u),u=\phi(x)$可微，则$f(\phi(x))$可微，且

\[d(f(\phi(x)))=(f(\phi(x)))'dx=f'(\phi(x))\phi'(x)dx \]

由$u=\phi(x)$，则$du=\phi'(x)dx$，所以，上式又可以写成

\[\begin{aligned} d(f(\phi(x)))=&f'(\phi(x))\phi'(x)dx=f'(u)du=d(f(u)) \end{aligned} \]

若$f$是$u$的函数，$u$是$x$的函数，若$x$是$t$的函数，均可微，则

\[\begin{aligned} d(f(u))&=f'(u)du=f'(u)\phi'(x)dx \\ &=f'(u)\phi'(x)x'(t)dt \end{aligned} \]

称为，一阶微分的形式不变性

例 1. 求$f(x)$的微分，其中

\[f(x)=\ln(x+\sqrt{1+x^2}) \]

例 2. $u,v$是$x$的可微函数，求$y=\arctan\dfrac{u}{v}$的微分

例 3. 求$\dfrac{d}{d(x^3)}(x^3-2x^6-x^9)$

\[\begin{aligned} df(x)&=d(\ln(x+\sqrt{1+x^2})) \\ &=\dfrac{1}{x+\sqrt{1+x^2}}d(x+\sqrt{1+x^2}) \\ &=\dfrac{1}{x+\sqrt{1+x^2}}(dx+d\sqrt{1+x^2}) \\ &=\dfrac{1}{x+\sqrt{1+x^2}}\left(dx+\dfrac{1}{2\sqrt{1+x^2}}d(1+x^2)\right) \\ &=\dfrac{1}{x+\sqrt{1+x^2}}\left(dx+\dfrac{1}{2\sqrt{1+x^2}}2xdx\right) \\ &=\dfrac{1}{x+\sqrt{1+x^2}}\left(dx+\dfrac{x}{\sqrt{1+x^2}}dx\right) \\ \end{aligned} \]
\[df=\dfrac{1}{\sqrt{1+x^2}}dx \]

类似，

\[d(\ln(x+\sqrt{x^2+a^2}))=\dfrac{1}{\sqrt{x^2+a^2}}dx \]

2.

\[\begin{aligned} dy&=(\arctan\dfrac{u}{v})'dx=\dfrac{1}{1+\dfrac{u^2}{v^2}}\cdot(\dfrac{u}{v})'dx \\ &=\dfrac{v^2}{u^2+v^2}\dfrac{u'v-v'u}{v^2}dx =\dfrac{u'v-v'u}{u^2+v^2}dx \end{aligned} \]

或者

\[\begin{aligned} dy&=\dfrac{1}{1+\dfrac{u^2}{v^2}}\cdot d(\dfrac{u}{v}) \\ &=\dfrac{v^2}{u^2+v^2}\dfrac{vdu-udv}{v^2} =\dfrac{vdu-udv}{u^2+v^2} \end{aligned} \]

3.

本题的意思是，以$u=x^3$为自变量，求$y$的微分。由$y=u-2u^2-u^3$知，

\[\dfrac{dy}{d(x^3)}=\left.\dfrac{d(u-2u^2-u^3)}{du}\right|_{u=x^3}=1-4x^3-3x^6 \]

或者，由“一元微分的形式不变性”，

\[\begin{aligned} \dfrac{dy}{d(x^3)}=&\dfrac{d(x^3-2x^6-x^9)}{d(x^3)} \\ =&\dfrac{(3x^2-13x^5-9x^8)dx}{3x^2dx}=1-4x^3-3x^6 \end{aligned} \]

含参变量函数的微分

$\left\{\begin{aligned} x=\phi(t) \\ y=\psi(t) \end{aligned} \right. , t\in(a,b)$，则

\[\dfrac{dy}{dx}=\dfrac{\psi'(t)dt}{\phi'(t)dt}=\dfrac{\psi'(t)}{\phi'(t)} \]

目录

本节读完

例 4. Thanks

4.

高阶微分

定义 3.
$dy=f'(x)dx$仍然是$x$的函数，若它还可微，则称$f(x)$二阶可微。记为

\[d^2y=d(dy), d^2f=d(df) \]

类似，可以定义$n-1$阶微分和微分为$n$阶微分，

\[d^ny=d(d^{n-1}y), d^nf=d(d^{n-1}f) \]

运算中$dx=\Delta x$表示是$x$增量，与$x$无关。所以有$d(dx)=0$

\[\begin{aligned} d^2y&=d(dy)=d(f'(x)dx)=f''(x)dx\cdot dx \\ &=f''(x)(dx)^2=f''(x)dx^2 \end{aligned} \]

这样，有

\[d^2y=f''(x)dx^2, \dfrac{d^2y}{dx^2}=f''(x) \]

更一般地，

\[d^ny=f^{(n)}(x)dx^n, \dfrac{d^ny}{dx^n}=f^{(n)}(x) \]

记$dx^n=(dx)^n$，而$d(x^n)=nx^{n-1}dx$

  • 当$x$是自变量时，$dx$为与$x$无关的常量，$d(dx)=d^2x=0$。

  • 当$x$是$t$的函数时，$dx=x'(t)dt$为与$t$相关的变量，

    $d(dx)=d^2x=x''(t)dt$。

这样，

\[\begin{aligned} d(df(x))&=d(f'(x)dx)=f''(x)dx^2+f'(x)d(dx) \\ &=f''(x)dx^2+f'(x)d^2x \end{aligned} \]

高阶微分不具有形式不变性

例 5. (例3.2.4) $y=\ln(x)$，其中$x$为$t$的函数。求$dy$, $d^2y$, $d^3y$

解. :

\[\begin{aligned} dy&=\dfrac{1}{x}dx \\ \\ d^2y&=d(\dfrac{1}{x}dx)=d(\dfrac{1}{x})dx+\dfrac{1}{x}d(dx) \\ &=\dfrac{-1}{x^2}dx^2+\dfrac{1}{x}d^2x \end{aligned} \]
\[\begin{aligned} d^3y&=d(\dfrac{-1}{x^2}dx^2+\dfrac{1}{x}d^2x) \\ &=d(\dfrac{-1}{x^2}dx^2)+d(\dfrac{1}{x}d^2x) \\ &=d(\dfrac{-1}{x^2})dx^2+\dfrac{-1}{x^2}d(dx^2)+d(\dfrac{1}{x})d^2x+\dfrac{1}{x}d(d^2x) \\ &=\dfrac{2}{x^3}dx dx^2+\dfrac{-1}{x^2}2dx d(dx)+\dfrac{-1}{x^2}dx d^2x+\dfrac{1}{x}d(d^2x) \\ &=\dfrac{2}{x^3}dx^3+\dfrac{-3}{x^2}dx d^2x+\dfrac{1}{x}d^3x \end{aligned} \]

例 6. 设$x=a\sin\theta$, $y=b\cos\theta$，指出如下计算中的错误。

$dx=a\cos\theta d\theta$, $dy=-b\sin\theta d\theta$, $d^2y=-b\cos\theta d\theta^2$, 则

\[\frac{d^2y}{dx^2}=\frac{-b\cos\theta d\theta^2}{(a\cos\theta d\theta)^2}=\frac{-b}{a^2\cos\theta} \]

例 7. 设$x=\sin t$，将关系式：

\[(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+a^2y=0 \]

换成以$t$为自变量的关系式

解. :

\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\cos t\frac{dy}{dx} \]
\[\begin{aligned} \frac{dy^2}{dt^2}=\frac{d}{dt}(\cos t\frac{dy}{dx}) =-\sin t\frac{dy}{dx}+\cos t\frac{d}{dt}(\frac{dy}{dx}) \\ =-\sin t\frac{dy}{dx}+\cos t\frac{d^2y}{dx^2}\frac{dx}{dt} =-\sin t\frac{dy}{dx}+\cos^2t\frac{d^2y}{dx^2} \\ =-\sin t\frac{dy}{dx}+(1-\sin^2t)\frac{d^2y}{dx^2} =-x\frac{dy}{dx}+(1-x^2)\frac{d^2y}{dx^2} \end{aligned} \]

即

\[\frac{d^2y}{dt^2}+a^2y=0 \]

解. :

\[\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac1{\cos t}\frac{dy}{dt} \]
\[\begin{aligned} \frac{dy^2}{dx^2}=\frac{d}{dt}(\frac{dy}{dx})\frac{dt}{dx} =\frac{d}{dt}(\frac{1}{\cos t}\frac{dy}{dt})\frac1{\cos t} \\ =(\frac{\sin t}{\cos^2t}\frac{dy}{dt}+\frac1{\cos t}\frac{d^2y}{dt^2})\frac1{\cos t} \end{aligned} \]

含参变量函数$\left\{\begin{aligned} x=\phi(t) \\ y=\psi(t) \end{aligned} \right. , t\in(a,b)$，则

\[\begin{aligned} \dfrac{d^2y}{dx^2}=&\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) \\ =&\dfrac{d\left(\dfrac{\psi'(t)}{\phi'(t)}\right)}{dx} \\ =&\dfrac{\psi''(t)\phi'(t)dt-\psi'(t)\phi''(t)dt}{(\phi'(t))^2}\dfrac{1}{\phi'(t)dt} \\ =&\dfrac{\psi''(t)\phi'(t)-\psi'(t)\phi''(t)}{(\phi'(t))^2}\dfrac{1}{\phi'(t)} \\ \end{aligned} \]