unlockable/MathematicalAnalysis
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity
Files
main
MathematicalAnalysis/03函数的导数.tex
unlockable 58f36b9bb1 qedhere是个好东西
2022-11-07 11:08:25 +08:00

670 lines
30 KiB
TeX
Raw Permalink Blame History

This file contains ambiguous Unicode characters
This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.
\chapter{函数的导数}
\section{导数的概念}
\begin{definition}[导数]
设函数$f(x)$在$x_0$附近有定义（包括$x_0$点）。定义$f$在$x_0$点的导数
\[\deriv{f}(x_0) = \tolim{x}{x_0}\frac{f(x) - f(x_0)}{x - x_0}\]
若极限存在，则称$f$在$x_0$点可导。
\end{definition}
\begin{remark}
应用代换$\delx = x - x_0$，导数可以等价地表示为
\[\deriv{f} = \tolim{\delx}{0} \frac{(x_0 + \delx) - f(x_0)}{\delx}\eqper\]
\end{remark}
Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
\[\frac{\dif f}{\dif x} = \tolim{\delx}{0} \frac{\Delta f}{\delx}\]
或记为
\[\frac{\dif f}{\dif x}(x_0) = \frac{\dif f}{\dif x}\bigg|_{x=x_0} = \deriv{f}(x_0) \eqper\]
\begin{definition}[单侧导数]
左导数：$\deriv{f}_{-} (x_0) = \tolim{\delx}{0^-} \dfrac{(x_0 + \delx) - f(x_0)}{\delx}$
右导数：$\deriv{f}_{+} (x_0) = \tolim{\delx}{0^+} \dfrac{(x_0 + \delx) - f(x_0)}{\delx}$
\end{definition}
\begin{corollary}
$\deriv{f}$存在等价于$\deriv{f}_{-}(x_0)$和$\deriv{f}_{+}(x_0)$都存在且相等。
\end{corollary}
\begin{theorem}[函数可导与连续的关系]
若$f$在$x_0$点可导，则$f$在$x_0$点连续。
\end{theorem}
\begin{proof}
注意到有
\begin{align*}
f(x) & = f(x_0) + (f(x) - f(x_0))\\
& = f(x_0) + \frac{f(x) - f(x_0)}{x-x_0}(x-x_0)
\end{align*}
对上面等式两边求极限，应用极限的四则运算性质：
\begin{align*}
\tolim{x}{x_0}f(x) & = \tolim{x}{x_0}f(x_0) + \tolim{x}{x_0}\frac{f(x) - f(x_0)}{x-x_0}(x-x_0)\\
& = f(x_0) + \deriv{f}(x_0) \cdot 0\\
& = f(x_0)
\end{align*}
因此$f$连续。
\end{proof}
\section{导数的计算}
\subsection{利用定义直接计算}
\begin{example}
$f(x) = c$，$c$为常数，求$\deriv{f}(x)$。
\end{example}
\begin{proof}[解]
对于任意的$x$，有
\[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper \qedhere\]
\end{proof}
\begin{example}
$f(x) = x^n \quad (n \in \naturalnum)$，求$\deriv{f}(x)$。
\end{example}
\begin{proof}[解]
应用二项式展开
\begin{align*}
\frac{(x + \delx)^n - x^n}{\delx} & = \frac{nx^{n-1} \delx + \frac{n(n-1)}{2}x^{n-2}\delx^2 + \cdots + \delx^n}{\delx}\\
& = nx^{n-1} + \frac{n(n-1)}{2}x^{n-1}\delx + \cdots + \delx^{n-1} \to nx^{n-1}
\end{align*}
因此$\deriv{f}(x) = nx^{n-1}$，也即$\deriv{(x^n)} = nx^{n-1}$。
\end{proof}
\begin{example}
$f(x) = \sin x$，求$\deriv{f}(x)$。
\end{example}
\begin{proof}[解]
应用和差化积公式有
\[\frac{\sin (x + \delx) - \sin x}{\delx} = \frac{1}{\delx}\cos \left(x + \frac{\Delta}{2}\right) \sin \left(\frac{\delx}{2}\right) \to \cos x\]
因此$\deriv{(\sin x)} = \cos x$，类似有$\deriv{(\cos x)} = -\sin x$。
\end{proof}
\begin{example}
$f(x) = \ln x,\ x > 0$，求$\deriv{f}(x)$。
\end{example}
\begin{proof}[解]
利用对数的性质
\[\frac{\ln(x + \delx) - \ln x}{\delx} = \frac{1}{\delx} \ln \frac{x + \delx}{x} = \frac{1}{x}\ln \left(1 + \frac{\delx}{x}\right)^\frac{x}{\delx}\]
注意$\tolim{\delx}{0} \left(1 + \frac{\delx}{x}\right)\frac{x}{\delx} = 3$，在利用对数函数的连续性得到
\[\frac{\ln(x + \delx) - \ln x}{\delx} \to \frac{1}{x}\ln e = \frac{1}{x}\]
因此$\deriv{(\ln x)} = \dfrac{1}{x}, \ x>0$。
\end{proof}
\subsection{导数的四则运算}
\begin{theorem}[导数的四则运算]
设函数$f$，$g$在$x$处可导，那么有
\begin{enumerate}
\item $\deriv{(f \pm g)}(x) = \deriv{f}(x) \pm \deriv{g}(x)$；
\item $\deriv{(fg)} = \deriv{f}(x)g(x) + f(x)\deriv{g}(x)$；
\item $\deriv{\left(\dfrac{f}{g}\right)} = \dfrac{\deriv{f}(x)g(x) - f(x)\deriv{g}(x)}{g^2(x)}$。
\end{enumerate}
\end{theorem}
\begin{proof}
只证2，3。
对于2，有
\begin{align*}
& \tolim{\delx}{0} \frac{f(x+\delx)g(x+\delx) - f(x)g(x)}{\delx}\\
= & \tolim{\delx}{0}\frac{\left[f(x+\delx) - f(x)\right]g(x+\delx)}{\delx} + \frac{f(x)\left[g(x+\delx) - g(x)\right]}{\delx}\\
= & \deriv{f}(x)g(x) + f(x)\deriv{g}(x)
\end{align*}
对于3，可先证$f = 1$的情况：
\[\deriv{\left(\frac{f}{g}\right)}(x) = \frac{\deriv{f}(x)g(x) - f(x)\deriv{g}(x)}{g^2(x)}\text{简化为}\deriv{\left(\frac{1}{g}\right)}(x) = - \frac{\deriv{g}(x)}{g^2(x)}\]
为此，考虑
\[\tolim{\delx}{0}\frac{1}{\delx}\left[\frac{1}{g(x+\delx)}-\frac{1}{g(x)}\right] = \tolim{\delx}{0} -\frac{g(x+\delx)g(x)}{\delx} \cdot \frac{1}{g(x+\delx)g(x)} = - \frac{\deriv{g}(x)}{g^2(x)}\]
对于一般的$f$可由乘法求导公式得到
\[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper \qedhere\]
\end{proof}
\begin{example}
\begin{align*}
\deriv{(\tan x)} & = \deriv{\left(\frac{\sin x}{\cos x}\right)} = \frac{\deriv{(\sin x)}\cos x - \sin x \deriv{(\cos x)}}{\cos^2 x}\\
& = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x\\
\deriv{(\cot x)} & = \deriv{\left(\frac{1}{\tan x}\right)} = -\frac{1}{\sin^2 x} = -\csc^2 x
\end{align*}
\end{example}
\subsection{复合函数的求导}
对于复合函数导数的计算，有链式法则：
\begin{theorem}[链式法则]
设函数$x = \varphi(t)$在$t$处可导。$y = f(x)$在$x = \varphi(t)$处可导。则复合函数$(f \circ \varphi)(t)$在$t$处可导，且
\[\deriv{(f(\varphi(t)))} = \deriv{(f\circ \varphi)}(t) = \deriv{f}(\varphi(t)) \cdot \deriv{\varphi}(t)\eqper\]
\end{theorem}
\begin{remark}
用Leibniz记号上式可记为
\[\frac{\dif y}{\dif t} = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t}\]
\end{remark}
\begin{proof}
$\deriv{f}(x) = \tolim{\delx}{0}\dfrac{\Delta y}{\Delta x}$。
思路：
\begin{align*}
\tolim{\Delta t}{0} \frac{\Delta y}{\Delta t} & = \tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}\\
\intertext{注意：这里不能保证$\Delta x \neq 0$}
& = \tolim{\Delta t}{0} \frac{\Delta x}{\Delta t} \cdot \tolim{\Delta x}{0} \frac{\Delta y}{\Delta x}\\
& = \deriv{\varphi}(t) \cdot \deriv{f}(x)\\
& = \deriv{\varphi}(t) \cdot \deriv{f}(\varphi(t))
\end{align*}
因此证法为：$\Delta x \neq 0$时，引入记号
\[\alpha (\Delta x) = \frac{f(x+\Delta) - f(x)}{\Delta x} - \deriv{f}(x) \to 0 \quad (\Delta x \to 0)\]
则
\begin{equation*}
f(x+\Delta x) - f(x) = [\deriv{f}(x) + \alpha(\Delta x)]\Delta x \tag{$\ast$} \label{链式法则式1}
\end{equation*}
再规定$\alpha(0) = 0$，那么\eqref{链式法则式1}对$\Delta = 0$也成立。
那么
\begin{align*}
\Delta y & = f(\varphi(t + \Delta t)) - f(\varphi(t))\\
& = f(x + \Delta t) - f(x)\\
& = [\deriv{f}(x) + \alpha(\Delta x)]\Delta x\\
\end{align*}
\begin{equation*}
\frac{\Delta y}{\Delta t} = [\deriv{f}(x) + \alpha(\Delta x)]\frac{\Delta x}{\Delta t} \tag{$\ast \ast$} \label{链式法则式2}
\end{equation*}
令$\Delta t \to 0$，则$\alpha (\Delta x) \to 0 \quad (\Delta x \to 0)$，式\eqref{链式法则式2}可以化为
\begin{align*}
\frac{\Delta y}{\Delta x} & = \deriv{f}(x) - \deriv{\varphi}(b)\\
& = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper \qedhere
\end{align*}
\end{proof}
\subsection{反函数的导数}
\begin{theorem}
设$y = f(x)$在区间$I$上严格单调，$x_0 \in I$，$\deriv{f}(x_0) \neq 0$存在，则反函数$x = \invertfunc{f}(y)$在$y = \invertfunc{f}(x_0)$点可导，且
\[\deriv{(\invertfunc{f})}(y_0) = \frac{1}{\deriv{f}(x_0)}\]
\end{theorem}
\begin{remark}
上式用Leibniz记号表示为
\begin{equation*}
\frac{\dif x}{\dif y} = \frac{1}{\dfrac{\dif y}{\dif x}}
\quad
\begin{cases}
x = \invertfunc{f}(y)\\
y = f(x)
\end{cases}
\eqper
\end{equation*}
\end{remark}
\begin{proof}
记$y = f(x)$，$x = \invertfunc{f}(y)$。
\begin{equation*}
\frac{\invertfunc{f}(y) - \invertfunc{f}(y_0)}{y - y_0} = \frac{x - x_0}{f(x) - f(x_0)} = \frac{1}{\deriv{f}(x_0)} \quad (x \to x_0)
\end{equation*}
由函数的连续性可得$y \to y_0$时，$x = \invertfunc{f}(y) \to \invertfunc{f}(y_0) = x_0$。
\end{proof}
\begin{example}
指数函数求导：设$f(x) = e^x$，求$\deriv{f}(x)$。
\end{example}
\begin{proof}[解]
记$y = e^x$，则$x = \ln y$。那么
\[\frac{\dif x}{\dif y} = \frac{1}{y}\]
利用反函数求导公式
\[\frac{\dif y}{\dif x} = \frac{1}{\dfrac{\dif x}{\dif y}} = y\]
即$\deriv{(e^x)} = e^x$。
\end{proof}
\begin{corollary}
令$a > 0, a \neq 1$，则$a^x = e^{x\ln a}$，$\log_a x = \dfrac{\ln x}{\ln a}$。
因此
\[\deriv{(a^x)} = \deriv{(e^{x\ln a})} = (e^{x\ln a})\deriv{(x\ln a)} = a^x \ln a \eqco\]
\[\deriv{(\log_a x)} = \frac{\deriv{(\ln x)}}{\ln a} = \frac{1}{x\ln a}\]
\end{corollary}
\subsection{参数方程求导法}
\begin{theorem}
设函数$y = f(x)$由参数方程
\begin{equation*}
\begin{cases}
x = \varphi(t)\\
y = \psi(t)
\end{cases}
\end{equation*}
确定，设$\deriv{\varphi}(t)$，$\deriv{\psi}(t)$都存在，且$\deriv{\varphi}(t) \neq 0$，$x = \varphi(t)$存在可导的反函数$t = \invertfunc{\varphi}(x)$，则
\[\frac{\dif y}{\dif x} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)} \eqper\]
\end{theorem}
\begin{proof}
分析函数关系：
\[x = \varphi(t) \Rightarrow t= \invertfunc{\varphi}(x)\]
那么$y$通过$t$成为$x$的复合函数
\[y = \psi[\invertfunc{\varphi}(x)]\]
再利用复合函数和反函数微分法，得
\[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)} \eqper \qedhere\]
\end{proof}
\subsection{基本导数公式}
\begin{multicols}{2}
\begin{enumerate}
\item $\deriv{(C)} = 0$
\item $\deriv{(x^\alpha)} = \alpha x^{\alpha-1}$
\item $\deriv{(e^x)} = e^x$
\item $\deriv{(a^x)} = a^x \ln a$
\item $\deriv{(\ln x)} = \dfrac{1}{x}$
\item $\deriv{(\log_a x)} = \dfrac{1}{x \ln a}$
\item $\deriv{(\sin x)} = \cos x$
\item $\deriv{(\cos x)} = -\sin x$
\item $\deriv{(\tan x)} = \dfrac{1}{\cos^2 x} = \sec^2 x$
\item $\deriv{(\cot x)} = -\dfrac{1}{\sin^2 x} = -\csc^2 x$
\item $\deriv{(\sec x)} = \sec x \tan x$
\item $\deriv{(\csc x)} = -\csc x \cot x$
\item $\deriv{(\arcsin x)} = \dfrac{1}{\sqrt{1 - x^2}}$
\item $\deriv{(\arccos x)} = -\dfrac{1}{\sqrt{1 - x^2}}$
\item $\deriv{(\arctan x)} = \dfrac{1}{1 + x^2}$
\item $\deriv{(\arccot x)} = -\dfrac{1}{1 + x^2}$
\end{enumerate}
\end{multicols}
\section{高阶导数及其计算}
\subsection{高阶导数的计算}
回忆：初等函数函数的导数仍是初等函数。
\begin{definition}[高阶导数]
设$\deriv{f}: I \to \realnum$为$f$的导函数，定义
\[f^{\prime \prime}(x_0) = \tolim{x}{x_0}\frac{\deriv{f}(x) - \deriv{f}(x_0)}{x - x_0} \text{（若极限存在）}\]
称为$f$在$x_0$点的二阶导数，也即$f^{\prime \prime} = \deriv{(\deriv{f})}$。进一步可以定义三阶导数$f^{\prime \prime \prime} = (f^{\prime \prime})^\prime$……以至$n$阶导数，记为$f^{(n)} = \deriv{(f^{(n-1)})}, n = 1, 2, 3, \cdots$。记$f^{(0)} = f$。
\end{definition}
\begin{remark}
高阶导数的Leibniz记号：令$y = f(x)$，
\[\frac{\dif^2 y}{\dif x^2} = \frac{\dif}{\dif x}\left(\frac{\dif y}{\dif x}\right),\ \frac{\dif^n y}{\dif x^n} = \frac{\dif}{\dif x}\left(\frac{\dif^{n-1} y}{\dif x^{n-1}}\right), \ n = 1, 2, \cdots\]
\end{remark}
\subsection{高阶导数的计算}
\begin{theorem}
设函数$f$，$g$都在$I$上有$n$阶导数，则
\begin{enumerate}
\item $(f \pm g)^{(n)} = f^{(n)} \pm g^{(n)}$；
\item $(cf)^{(n)} = cf^{(n)}$，$c$为常数；
\item （Leibniz公式）$(fg)^{(n)} = \sum \limits_{k=0}^n \dbinom{k}{n}f^{(k)}g^{(n-k)}$。
\end{enumerate}
\end{theorem}
\begin{theorem}
对于由参数方程
\begin{equation*}
\begin{cases}
x = \varphi(t)\\
y = \psi(t)
\end{cases}
\end{equation*}
确定的函数$y = f(x)$，有
\begin{equation*}
\begin{cases}
x = \varphi(t)\\
\deriv{y} = \dfrac{\deriv{\psi(t)}}{\deriv{\varphi(t)}} = \omega(t)
\end{cases}
\end{equation*}
那么$\deriv{y} = g(x)$又是一个由上述参数方程确定的函数，因此有
\[y^{\prime \prime} = \frac{\dif }{\dif x} (\deriv{y}(x)) = \frac{\deriv{\omega}(t)}{\deriv{\varphi(t)}}\]
\end{theorem}
\section{微分/导数中值定理}
目标：研究区间内可导函数的导数性质。
\begin{definition}[极值]
设$f$在$x_0$附近有定义（包括$x_0$），若$\exists \delta > 0$，使得
\begin{enumerate}
\item $\forall \vert x - x_0 \vert < \delta$，$f(x) \leq f(x_0)$，称$f(x_0)$为$f$的极大值，$x_0$为极大值点；
\item $\forall \vert x - x_0 \vert < \delta$，$f(x) \geq f(x_0)$，称$f(x_0)$为$f$的极小值，$x_0$为极小值点；
\end{enumerate}
\end{definition}
\begin{remark}
极值不一定是最值。极值点必须在区间内部。
\end{remark}
\begin{theorem}[Fermat引理]
设函数$f$在其极值点$x_0$可导，则必有$\deriv{f}(x_0) = 0$。
\end{theorem}
\begin{proof}
不妨令$x_0$为极大值点，则$\exists \delta > 0$，使得
\[\forall \vert x - x_0 \vert < \delta,\ f(x) \leq f(x_0)\]
考察$x_0$点的左右导数，结合极限的保序性
\begin{align*}
\deriv{f}_- (x_0) = \tolim{x}{x_0^-} \frac{f(x) - f(x_0)}{x-x_0} \geq 0\\
\deriv{f}_+ (x_0) = \tolim{x}{x_0^+} \frac{f(x) - f(x_0)}{x - x_0} \leq 0
\end{align*}
而$f$在$x_0$可导，$\deriv{f}(x_0) = \deriv{f}_- (x_0) = \deriv{f}_+ (x_0) = 0$。
\end{proof}
\begin{remark}
Fermat引理的逆不成立。例如$f(x) = x^3$在$x = 0$导数为0，但$f(0) = 0$不是极值。
\end{remark}
\begin{definition}[临界点]
若$\deriv{f}(x_0) = 0$，则称$x_0$为$f$的临界点/驻点。
\end{definition}
\begin{theorem}[Rolle中值定理]
设函数$f \in C[a,b]$，且在$(a,b)$内可导。若$f(a) = f(b)$，则存在$c \in (a,b)$使得$\deriv{f}(c) = 0$。
\end{theorem}
\begin{proof}
若$f(x)$为常数，则$\forall x \in (a,b)$，$\deriv{f}(x) = 0$；
若$f(x)$非常数，不妨设$\exists x_0 \in (a,b)$，使得$f(x_0) > f(a) = f(b)$。由连续函数最值性质，
\[\exists \overline{x} \in (a,b),\ \forall x \in [a,b] f(x) \leq f(\overline{x})\]
那么$\overline{x} \in (a,b)$是极大值点，由Fermat引理，由$\deriv{\overline{x}} = 0$。
\end{proof}
如果将其推广，取消$f(a) = f(b)$的限制，我们就得到
\begin{theorem}[Lagrange中值定理]
设函数$f \in C[a,b]$，且在$(a,b)$内可导。则存在$c \in (a,b)$使得$\deriv{f}(c) = \dfrac{f(b) - f(a)}{b-a}$。
\end{theorem}
\begin{proof}
构造函数$F(x) = f(x) - \frac{f(b) - f(a)}{b-a}(x-a)$，$F(x) \in C[a,b]$。注意到$F(b) f(b) - [f(b) - f(a)] = f(a) = F(a)$，应用Rolle定理，$\exists c \in (a,b)$使得$\deriv{F}(c) = 0$，即
\[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper \qedhere\]
\end{proof}
\begin{remark}
Lagrange中值定理的公式的其它形式：记$x = a$，$x + \Delta x = b$，则
\[f(x + \Delta x) - f(x) = \deriv{f}(x + \theta \Delta x) \Delta x,\ \theta = \frac{c-a}{b-a}\in (0,1)\]
\end{remark}
\begin{corollary}
设函数$f$在$(a,b)$内可导，则$f(x)$为常数的充分必要条件是$\deriv{f}(x) = 0$，$\forall x \in (a, b)$。
\end{corollary}
\begin{proof}
必要性明显，计算导数即可。
为证充分性，令$\deriv{f}(x) \equiv 0$。任取$x_1, x_2 \in (a, b)$，$x_1 < x_2$，则$f \in C[x_1, x_2]$且在$(x_1, x_2)$内可导。应用Lagrange中值定理，$\exists c \in (x_1, x_2)$，使得
\[f(x_2) - f(x_1) = \deriv{f}(c)(x_2 - x_1) = 0\]
注意上式中$x_1, x_2$的任意性，因此$f(x)$为常数。
\end{proof}
\begin{corollary}
设函数$f, g$在$(a,b)$内可导，且$\deriv{f} \equiv \deriv{g}$，则$f - g$为常数，即$\exists C_0 \in \realnum$，使得$f(x) = g(x) + C_0$，$\forall x \in (a,b)$。
\end{corollary}
\begin{theorem}[Cauchy中值定理]
设函数$f,g \in C[a,b]$，且在$(a,b)$内可导，$\deriv{g}(x) \neq 0$，则存在$c \in (a,b)$使得
\[\frac{\deriv{f}(c)}{\deriv{g}(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\eqper\]
\end{theorem}
\begin{proof}
注意到，如果$g(x) = x$，即转化为Lagrange定理的情况。考虑Lagrange定理的证明方法，类似构造辅助函数：令
\[F(x) = f(x) - \frac{f(b) - f(a)}{g(b) - g(a)}[g(x) - g(a)] \in C[a,b]\]
那么
\[F(b) = f(b) - [f(b) - f(a)] = f(a) = F(a)\]
应用Rolle定理，$\exists c \in (a,b)$，使得$\deriv{F}(c) = 0$，即
\[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b) - f(a)}{g(b) - g(a)}\deriv{g}(c) = 0\]
即可导出结论。
\end{proof}
\section{中值定理与函数单调性}
\begin{theorem}[单调性的判别]
设函数$f \in C[a,b]$，且在$(a,b)$内可导，那么
\begin{enumerate}
\item $f$在$[a,b]$上单调增$\Leftrightarrow \forall x \in (a,b),\ \deriv{f}(x) \geq 0$；
\item $f$在$[a,b]$上单调减$\Leftrightarrow \forall x \in (a,b),\ \deriv{f}(x) \leq 0$。
\end{enumerate}
\end{theorem}
\begin{proof}
根据导数定义
\[\deriv{f}(x_0) = \tolim{x}{x_0}\frac{f(x) - f(x_0)}{x - x_0}\]
根据极限的保号性可得导数的符号，因此``$\Rightarrow$''成立。
反之，$\forall x_1, x_2 \in (a,b)$，$x_1 < x_2$，应用Lagrange中值公式，$\exists c \in (x_1, x_2)$使得$f(x_1) - f(x_2) = \deriv{f}(c)(x_1 - x_2)$。由此导出函数$f$的单调性，因此``$\Leftarrow$''成立。
\end{proof}
\begin{theorem}[严格单调性的判别]
设函数$f \in C[a,b]$，且在$(a,b)$内可导，那么
\begin{enumerate}
\item $f$在$[a,b]$上严格单调增，若$\forall x \in (a,b),\ \deriv{f}(x) > 0$；
\item $f$在$[a,b]$上严格单调减，若$\forall x \in (a,b),\ \deriv{f}(x) < 0$。
\end{enumerate}
\end{theorem}
证明与上类似。
\begin{remark}
函数严格单调不能保证导数严格不等于0，但个别点导数为0可以导出函数严格单调：
\begin{enumerate}
\item 若除了有限个点之外，在$(a,b)$内$\deriv{f}(x) > 0$，则$f$在$[a,b]$上严格单调增；
\item 若除了有限个点之外，在$(a,b)$内$\deriv{f}(x) < 0$，则$f$在$[a,b]$上严格单调减。
\end{enumerate}
\end{remark}
\section{利用导数研究函数}
\subsection{极值问题}
\begin{proposition}[极值的必要条件【Fermat引理】]
若$x=x_0$为$f(x)$的极值点，则$\deriv{f}(x_0) = 0$，除非$\deriv{f}(x_0)$不存在。
\end{proposition}
\begin{proposition}[充分条件一]
设$f$在$x_0$点连续，则$f(x_0)$为$f$的
\begin{enumerate}[label=(\alph{*})]
\item 极大值，若在$x_0$左侧附近$\deriv{f}(x) \geq 0$，在$x_0$右侧附近$\deriv{f}(x) \leq 0$
\item 极小值，若在$x_0$左侧附近$\deriv{f}(x) \leq 0$，在$x_0$右侧附近$\deriv{f}(x) \geq 0$
\end{enumerate}
\end{proposition}
\begin{proof}
只需证（a）：由题意$\exists > 0$，使
\begin{equation*}
\left.
\begin{aligned}
x \in (x_0 - \delta, x_0),\ \deriv{f}(x) \geq 0,\ \therefore f(x) \leq f(x_0)\\
x \in (x_0, x_0 + \delta),\ \deriv{f}(x) \leq 0,\ \therefore f(x) \leq f(x_0)
\end{aligned}
\right\}
\text{极大值}\qedhere
\end{equation*}
\end{proof}
\begin{proposition}[充分条件二]
设$f$在$x_0$点存在2阶导数，且$\deriv{f}(x_0) = 0$
\begin{enumerate}[label=(\alph{*})]
\item 若$f^{\prime \prime}(x_0) < 0$，则$f(x_0)$为严格极大值
\item 若$f^{\prime \prime}(x_0) > 0$，则$f(x_0)$为严格极小值
\end{enumerate}
\end{proposition}
\begin{proof}
只需证（a）：已知
\[f^{\prime \prime}(x_0) = \tolim{x}{x_0}\frac{\deriv{f}(x) - \deriv{f}(x_0)}{x-x_0} = \tolim{x}{x_0} \frac{\deriv{f}(x)}{x - x_0} < 0\]
根据极限的保号性，
\[\exists \delta > 0,\ \forall 0 < \vert x - x_0 \vert < \delta,\ \frac{\deriv{f}(x)}{x-x_0} < 0\]
这说明
\begin{equation*}
\left.
\begin{aligned}
x \in (x_0 - \delta, x_0),\ \deriv{f}(x) > 0,\ \therefore f(x) < f(x_0)\\
x \in (x_0, x_0 + \delta),\ \deriv{f}(x) < 0,\ \therefore f(x) < f(x_0)
\end{aligned}
\right\}
\text{严格极大值}\qedhere
\end{equation*}
\end{proof}
\subsection{函数的凸性}
\begin{definition}
设$f: I \to \realnum$，$I$为区间，如果$\forall x_1, x_2 \in I$，$\lambda \in (0,1)$，总有
\[f(\lambda x_1 + (1 - \lambda)x_2) \leq \lambda f(x_1) + (1 - \lambda)f(x_2)\]
则称，$f$为$I$中的凸（下凸）函数。
若反向不等式成立，即有
\[f(\lambda x_1 + (1 - \lambda)x_2) \leq \lambda f(x_1) + (1 - \lambda)f(x_2)\]
则称$f$为上凸。
\end{definition}
\begin{corollary}
$f$是下凸函数当且仅当$-f$是上凸函数。
\end{corollary}
\begin{proposition}[等价凸性]
设$f: I \to \realnum$，$f$在区间$I$中下凸等价于$\forall x_1, x_2, \cdots, x_n \in I$，$\lambda_1, \lambda_2, \cdots, \lambda_n \in (0,1)$满足$\sum \limits_{i=1}^n \lambda_i = 1$，总有
\[f(\lambda_1 x_1 + \cdots + \lambda_n x_n) \leq \lambda_1f(x_1) + \cdots + \lambda_nf(x_n)\eqper\]
\end{proposition}
\begin{proof}
$n = 2$时与定义相同。用递推即可得出$n = 3, 4, \cdots$的情况。以$n = 3$为例，已知$n = 2$时有上面不等式，现在任取$x_1, x_2, x_3 \in I$，$\lambda_1, \lambda_2, \lambda_3 \in (0,1)$满足$\lambda_1 + \lambda_2 + \lambda_3 = 1$，注意$\lambda_2 + \lambda_3 = (1 - \lambda_1)$，$\dfrac{\lambda_2}{\lambda_2 + \lambda_3} + \dfrac{\lambda_3}{\lambda_2 + \lambda_3} = 1$。
\begin{align*}
f(\lambda_1x_1 + \lambda_2x_2 + \lambda_3x_3) & \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)f\left(\frac{\lambda_2 x_2 + \lambda_3 x_3}{\lambda_2 + \lambda_3}\right)\\
& \leq \lambda_1f(x_1) + (\lambda_2 + \lambda_3)\left(\frac{\lambda_2}{\lambda_2 + \lambda_3}f(x_2) + \frac{\lambda_3}{\lambda_2 + \lambda_3}f(x_3)\right)\\
& = \lambda_1 f(x_1) + \lambda_2 f(x_2) + \lambda_3 f(x_3) \eqper \qedhere
\end{align*}
\end{proof}
\begin{corollary}[Jensen不等式]
设$f: I \to \realnum$为下凸函数，则$\forall x_1, \cdots x_n \in I$，$\forall \lambda_1, \cdots, \lambda_n > 0$，总有
\[f\left(\frac{\lambda_1 x_1 + \cdots + \lambda_n x_n}{\lambda_1 + \cdots + \lambda_n}\right) \leq \frac{\lambda_1 f(x_1) + \cdots + \lambda_n f(x_n)}{\lambda_1 + \cdots + \lambda_n}\eqper\]
\end{corollary}
\begin{proposition}[等价凸性]
设$f: I \to \realnum$，$I$为区间，则$f$在$I$中下凸等价于对任意的$x_1, x_2 \in I$，$x_1 < x_2$，$x \in (x_1, x_2)$都有
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper\]
\end{proposition}
\begin{figure}[H]
\centering
\begin{tikzpicture}
\draw[line width=1pt, ->] (-0.2,0)--(4.3,0) node[below] {$x$};
\draw[line width=1pt, ->] (0,-0.2)--(0,4) node[left] {$f(x)$};
\draw[red, line width=1pt] (0.4, {((0.4-1.5)^2)/3 + 1.2})--(2.1, {((2.1-1.5)^2)/3 + 1.2})--(3.7, {((3.7-1.5)^2)/3 + 1.2})--cycle;
\draw[domain=0.2:4, line width=1pt] plot(\x,{((\x-1.5)^2)/3 + 1.2});
\node[left] at (0,{((0.4-1.5)^2)/3 + 1.6}) {$f(x_1)$};
% \draw (0,{((0.4-1.5)^2)/3 + 1.6})--(0,{((0.4-1.5)^2)/3 + 1.2});
\node[left] at (0,{((2.1-1.5)^2)/3 + 1.2}) {$f(x)$};
\node[left] at (0,{((3.7-1.5)^2)/3 + 1.2}) {$f(x_2)$};
\node[below] at (0.4,0) {$x_1$};
\node[below] at (2.1,0) {$x$};
\node[below] at (3.7,0) {$x_2$};
\draw[dashed] (0.4,0)--(0.4, {((0.4-1.5)^2)/3 + 1.2})--(0,{((0.4-1.5)^2)/3 + 1.2});
\draw[dashed] (2.1,0)--(2.1, {((2.1-1.5)^2)/3 + 1.2})--(0,{((2.1-1.5)^2)/3 + 1.2});
\draw[dashed] (3.7,0)--(3.7, {((3.7-1.5)^2)/3 + 1.2})--(0,{((3.7-1.5)^2)/3 + 1.2});
\end{tikzpicture}
\end{figure}
\begin{proof}
注意$x_1 < x < x_2$，可以将$x$表示为
\[x = \lambda x_1 + (1 - \lambda) x_2 = \frac{x_2 - x}{x_2 - x_1}x_1 + \frac{x - x_1}{x_2 - x_2}x_2\]
利用下凸的性质
\[f(x) \leq \frac{x_2 - x}{x_2 - x_1}f(x_1) + \frac{x - x_1}{x_2 - x_2}f(x_2)\]
由此得
\begin{equation*}
\begin{cases}
f(x) - f(x_1) \leq \dfrac{x - x_1}{x_2 - x_1}[f(x_2) - f(x_1)]\\
f(x) - f(x_2) \leq \dfrac{x - x_2}{x_2 - x_2}[f(x_2) - f(x_1)]
\end{cases}
\end{equation*}
综合即得原式。过程中的推导全部为等价变换，因此得证。
\end{proof}
\begin{proposition}[用一阶导判别凸性]
设$f: I \to \realnum$可导，则$f$在$I$中下凸等价于$\deriv{f}$在$I$中单调增。
\end{proposition}
\begin{proof}
先证``$\Rightarrow$''：令$f$下凸，则由等价凸性$\forall x_1, x_2 \in I$，$x_1 < x_2$，$x \in (x_1, x_2)$，有
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x} \tag{$\ast$} \label{凸性判别式1}\]
在两个不等式中分别令$x \to x_1^+$与$x \to x_2^-$，得到
\[\deriv{f}(x_1) \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \deriv{f}(x_2)\eqper\]
为证``$\Leftarrow$''，令$\deriv{f}$单调增，要证明式\eqref{凸性判别式1}，应用Lagrange中值公式，$\forall x_1, x_2 \in I$，$x_1 < x_2$，$x \in (x_1, x_2)$，存在$c_1 \in (x_1, x)$，$c_2 \in (x, x_2)$使得
\[\deriv{f}(c_1) \frac{f(x) - f(x_1)}{x - x_1},\ \deriv{f}(c_2) = \frac{f(x_2) - f(x)}{x_2 - x}\]
那么利用$\deriv{f}$的单调性，$c_1 < x < c_2$，有
\[\frac{f(x) - f(x_1)}{x - x_1} = \deriv{f}(c_1) \leq \deriv{f}(c_2) = \frac{f(x_2) - f(x)}{x_2 - x}\]
再利用初等分式不等式
\[\frac{a_1}{b_1} \leq \frac{a_2}{b_2}\text{且}b_1, b_2 > 0,\ \text{则}\frac{a_1}{b_1} \leq \frac{a_1 + a_2}{b_1 + b_2} \leq \frac{a_2}{b_2}\]
由此即可得到
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper \qedhere\]
\end{proof}
\begin{corollary}[二阶导数判别凸性]
设$f:I \to \realnum$处处2次可导，则$f$在$I$中下凸等价于在$I$上$f^{\prime \prime} > 0$。
\end{corollary}
\section{L'Hospital法则}
\begin{theorem}[L'Hospital]
设$f,g$在$(a,b)$内可导，满足$\tolim{x}{a^+}f(x) = \tolim{x}{a^+}g(x) = 0$，且$g(x) \neq 0$，若
\[\tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\]
存在（或为$\infty$），那么
\[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
\end{theorem}
\begin{proof}
补充定义$f(a) = g(a) = 0$，从而使得$f,g$在$[a,b)$上连续。利用Cauchy中值定理，对$x \in (a,b)$，存在$a < \xi < x$使得
\[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)}\]
那么当$x \to a^+$时，有$\xi \to a^+$，因此
\[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere\]
\end{proof}
\begin{remark}
若$\tolim{x}{a^+}\dfrac{\deriv{f}(x)}{\deriv{g}(x)}$不存在，说明L'Hospital法则失效，不能说明原极限不存在。
\end{remark}
\begin{theorem}
设函数$f,g$在区间$(b,+\infty)$内可导，满足$\tolim{x}{+\infty}f(x) = \tolim{x}{+\infty}g(x) = 0$，且$g(x) \neq 0$对$x \in (a, +\infty)$成立，若
\[\tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\]
存在（或为$\infty$），那么
\[\tolim{x}{+\infty} \frac{f(x)}{g(x)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
\end{theorem}
\begin{proof}
作变换$x = \dfrac{1}{t}$。那么$x \to +\infty$相当于$t \to 0^+$。那么我们有
\[\tolim{t}{0^+} f\left(\frac{1}{t}\right) = \tolim{t}{0^+}g\left(\frac{1}{t}\right) = 0\]
那么
\begin{align*}
\tolim{x}{+\infty} & = \tolim{t}{0^+}\frac{f \left(\dfrac{1}{t}\right)}{g \left(\dfrac{1}{t}\right)} = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)\left(-\dfrac{1}{t^2}\right)}\\
& = \tolim{t}{0^+}\frac{\deriv{f}\left(\dfrac{1}{t}\right)}{\deriv{g}\left(\dfrac{1}{t}\right)} = \tolim{x}{+\infty}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper \qedhere
\end{align*}
\end{proof}
\begin{theorem}
设函数$f,g$在$(a,b)$内可导，满足$\tolim{x}{a^+} g(x) = \infty$，且$g(x) \neq 0$对$x \in (a,b)$成立，若
\[\tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\]
存在（或为$\infty$），那么
\[\tolim{x}{a^+}\frac{f(x)}{g(x)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
\end{theorem}
\begin{remark}
只有``$\dfrac{0}{0}$型''和``$\dfrac{\infty}{\infty}$型''才能使用L'Hospital法则，其它形式的未定式要化成这两种形式。
\end{remark}
\section{函数作图}
曲线作图步骤：对于方程$y = f(x)$，
\begin{enumerate}
\item 确定函数$f$的定义域
\item 检查$f$的对称性：奇偶性与周期性
\item 检查曲线的渐近线：垂直、水平斜渐近线
\item 研究导函数$\deriv{f}$：确定函数的增减区间、极值点、临界点
\item 研究二阶到函数$f^{\prime \prime}$：确定曲线的上下凸区间与拐点
\item 计算在若干关键点的$f$的极值点
\item 绘图
\end{enumerate}
\begin{definition}[曲线渐近线]
考虑曲线$y = f(x)$
\begin{enumerate}
\item 若$\tolim{x}{a^\pm}f(x) = \infty$，则$x = a$是垂直渐近线；
\item 若$\tolim{x}{\pm \infty}f(x) = b$，则$y = b$是水平渐近线；
\item 若$\tolim{x}{+\infty} [f(x) - (ax + b)] = 0$或$\tolim{x}{-\infty} [f(x) - (ax + b)] = 0$，则$y = ax + b$是斜渐近线。
\end{enumerate}
\end{definition}
注意到$\tolim{x}{+/- \infty}\dfrac{f(x) - (ax + b)}{x} = 0$，因此$\tolim{x}{+/- \infty}\dfrac{f(x)}{x} - a = 0$。因此
\begin{proposition}
直线$y = ax + b$是曲线$y = f(x)$的渐近线当且仅当$a = \tolim{x}{+/-\infty}\dfrac{f(x)}{x}$，$b = \tolim{x}{+/-\infty}[f(x) - ax]$两极限存在。
\end{proposition}
\begin{definition}[函数/曲线拐点]
函数/曲线上下凸性发生改变的点称为函数/曲线的拐点，即$\deriv{f}(x)$增减发生改变的点，也即$f^{\prime \prime} = 0$的点。
\end{definition}
Reference in New Issue View Git Blame Copy Permalink