670 lines
30 KiB
TeX
670 lines
30 KiB
TeX
\chapter{函数的导数}
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\section{导数的概念}
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\begin{definition}[导数]
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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
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\[\deriv{f}(x_0) = \tolim{x}{x_0}\frac{f(x) - f(x_0)}{x - x_0}\]
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若极限存在，则称$f$在$x_0$点可导。
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\end{definition}
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\begin{remark}
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应用代换$\delx = x - x_0$，导数可以等价地表示为
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\[\deriv{f} = \tolim{\delx}{0} \frac{(x_0 + \delx) - f(x_0)}{\delx}\eqper\]
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\end{remark}
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Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
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\[\frac{\dif f}{\dif x} = \tolim{\delx}{0} \frac{\Delta f}{\delx}\]
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或记为
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\[\frac{\dif f}{\dif x}(x_0) = \frac{\dif f}{\dif x}\bigg|_{x=x_0} = \deriv{f}(x_0) \eqper\]
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\begin{definition}[单侧导数]
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左导数：$\deriv{f}_{-} (x_0) = \tolim{\delx}{0^-} \dfrac{(x_0 + \delx) - f(x_0)}{\delx}$
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右导数：$\deriv{f}_{+} (x_0) = \tolim{\delx}{0^+} \dfrac{(x_0 + \delx) - f(x_0)}{\delx}$
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\end{definition}
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\begin{corollary}
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$\deriv{f}$存在等价于$\deriv{f}_{-}(x_0)$和$\deriv{f}_{+}(x_0)$都存在且相等。
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\end{corollary}
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\begin{theorem}[函数可导与连续的关系]
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若$f$在$x_0$点可导，则$f$在$x_0$点连续。
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\end{theorem}
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\begin{proof}
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注意到有
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\begin{align*}
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f(x) & = f(x_0) + (f(x) - f(x_0))\\
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& = f(x_0) + \frac{f(x) - f(x_0)}{x-x_0}(x-x_0)
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\end{align*}
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对上面等式两边求极限，应用极限的四则运算性质：
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\begin{align*}
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\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)\\
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& = f(x_0) + \deriv{f}(x_0) \cdot 0\\
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& = f(x_0)
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\end{align*}
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因此$f$连续。
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\end{proof}
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\section{导数的计算}
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\subsection{利用定义直接计算}
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\begin{example}
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$f(x) = c$，$c$为常数，求$\deriv{f}(x)$。
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\end{example}
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\begin{proof}[解]
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对于任意的$x$，有
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\[\deriv{f}(x) = \tolim{\delx}{0}\frac{f(x + \delx) - f(x)}{\delx} = \tolim{\delx}{0} \frac{c - c}{\delx} = 0 \eqper\]
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\end{proof}
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\begin{example}
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$f(x) = x^n \quad (n \in \naturalnum)$，求$\deriv{f}(x)$。
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\end{example}
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\begin{proof}[解]
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应用二项式展开
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\begin{align*}
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\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}\\
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& = nx^{n-1} + \frac{n(n-1)}{2}x^{n-1}\delx + \cdots + \delx^{n-1} \to nx^{n-1}
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\end{align*}
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因此$\deriv{f}(x) = nx^{n-1}$，也即$\deriv{(x^n)} = nx^{n-1}$。
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\end{proof}
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\begin{example}
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$f(x) = \sin x$，求$\deriv{f}(x)$。
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\end{example}
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\begin{proof}[解]
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应用和差化积公式有
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\[\frac{\sin (x + \delx) - \sin x}{\delx} = \frac{1}{\delx}\cos \left(x + \frac{\Delta}{2}\right) \sin \left(\frac{\delx}{2}\right) \to \cos x\]
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因此$\deriv{(\sin x)} = \cos x$，类似有$\deriv{(\cos x)} = -\sin x$。
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\end{proof}
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\begin{example}
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$f(x) = \ln x,\ x > 0$，求$\deriv{f}(x)$。
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\end{example}
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\begin{proof}[解]
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利用对数的性质
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\[\frac{\ln(x + \delx) - \ln x}{\delx} = \frac{1}{\delx} \ln \frac{x + \delx}{x} = \frac{1}{x}\ln \left(1 + \frac{\delx}{x}\right)^\frac{x}{\delx}\]
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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
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\[\frac{\ln(x + \delx) - \ln x}{\delx} \to \frac{1}{x}\ln e = \frac{1}{x}\]
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因此$\deriv{(\ln x)} = \dfrac{1}{x}, \ x>0$。
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\end{proof}
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\subsection{导数的四则运算}
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\begin{theorem}[导数的四则运算]
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设函数$f$，$g$在$x$处可导，那么有
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\begin{enumerate}
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\item $\deriv{(f \pm g)}(x) = \deriv{f}(x) \pm \deriv{g}(x)$；
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\item $\deriv{(fg)} = \deriv{f}(x)g(x) + f(x)\deriv{g}(x)$；
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\item $\deriv{\left(\dfrac{f}{g}\right)} = \dfrac{\deriv{f}(x)g(x) - f(x)\deriv{g}(x)}{g^2(x)}$。
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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只证2，3。
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对于2，有
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\begin{align*}
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& \tolim{\delx}{0} \frac{f(x+\delx)g(x+\delx) - f(x)g(x)}{\delx}\\
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= & \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}\\
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= & \deriv{f}(x)g(x) + f(x)\deriv{g}(x)
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\end{align*}
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对于3，可先证$f = 1$的情况：
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\[\deriv{\left(\frac{f}{g}\right)}(x) = \frac{\deriv{f}(x)g(x) - f(x)\deriv{g}(x)}{g^2(x)}\text{简化为}\deriv{\left(\frac{1}{g}\right)}(x) = - \frac{\deriv{g}(x)}{g^2(x)}\]
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为此，考虑
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\[\tolim{\delx}{0}\frac{1}{\delx}\left[\frac{1}{g(x+\delx)}-\frac{1}{g(x)}\right] = \tolim{\delx}{0} -\frac{g(x+\delx)g(x)}{\delx} \cdot \frac{1}{g(x+\delx)g(x)} = - \frac{\deriv{g}(x)}{g^2(x)}\]
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对于一般的$f$可由乘法求导公式得到
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\[\deriv{\left(\frac{f}{g}\right)} = \deriv{\left(f \cdot \frac{1}{g}\right)} = \deriv{f}\frac{1}{g} + f \deriv{\left(\frac{1}{g}\right)} = \frac{\deriv{f}g-f\deriv{g}}{g^2}\eqper\]
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\end{proof}
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\begin{example}
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\begin{align*}
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\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}\\
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& = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x\\
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\deriv{(\cot x)} & = \deriv{\left(\frac{1}{\tan x}\right)} = -\frac{1}{\sin^2 x} = -\csc^2 x
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\end{align*}
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\end{example}
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\subsection{复合函数的求导}
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对于复合函数导数的计算，有链式法则：
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\begin{theorem}[链式法则]
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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
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\[\deriv{(f(\varphi(t)))} = \deriv{(f\circ \varphi)}(t) = \deriv{f}(\varphi(t)) \cdot \deriv{\varphi}(t)\eqper\]
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\end{theorem}
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\begin{remark}
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用Leibniz记号上式可化为
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\[\frac{\dif y}{\dif t} = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t}\]
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\end{remark}
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\begin{proof}
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$\deriv{f}(x) = \tolim{\delx}{0}\frac{\Delta y}{\Delta x}$。
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思路：
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\begin{align*}
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\tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} & = \tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}\\
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\intertext{注意：这里不能保证$\Delta x \neq 0$}
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& = \tolim{\Delta t}{0} \frac{\Delta x}{\Delta t} \cdot \tolim{\Delta x}{0} \frac{\Delta y}{\Delta x}\\
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& = \deriv{\varphi}(t) \cdot \deriv{f}(x)\\
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& = \deriv{\varphi}(t) \cdot \deriv{f}(\varphi((t)))
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\end{align*}
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因此证法为：$\Delta x \neq 0$时，引入记号
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\[\alpha (\Delta x) = \frac{f(x+\Delta) - f(x)}{\Delta x} - \deriv{f}(x) \to 0 \quad (\Delta x \to 0)\]
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则
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\begin{equation*}
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f(x+\Delta x) - f(x) = [\deriv{f}(x) + \alpha(\Delta x)]\Delta x \tag{$\ast$} \label{链式法则式1}
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\end{equation*}
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再规定$\alpha(0) = 0$，那么\eqref{链式法则式1}对$\Delta = 0$也成立。
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那么
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\begin{align*}
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\Delta y & = f(\varphi(t + \Delta t)) - f(\varphi(t))\\
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& = f(x + \Delta t) - f(x)\\
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& = [\deriv{f}(x) + \alpha(\Delta x)]\Delta x\\
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\end{align*}
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\begin{equation*}
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\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}
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\end{equation*}
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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
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\begin{align*}
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\frac{\Delta y}{\Delta x} & = \deriv{f}(x) - \deriv{\varphi}(b)\\
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& = \deriv{f}(\varphi(t)) \deriv{\varphi}(t)\eqper
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\end{align*}
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\end{proof}
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\subsection{反函数的导数}
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\begin{theorem}
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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
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\[\deriv{(\invertfunc{f})}(y_0) = \frac{1}{\deriv{f}(x_0)}\]
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\end{theorem}
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\begin{remark}
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上式用Leibniz记号表示为
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\begin{equation*}
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\frac{\dif x}{\dif y} = \frac{1}{\dfrac{\dif y}{\dif x}}
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\quad
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\begin{cases}
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x = \invertfunc{f}(y)\\
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y = f(x)
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\end{cases}
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\eqper
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\end{equation*}
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\end{remark}
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\begin{proof}
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记$y = f(x)$，$x = \invertfunc{f}(y)$。
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\begin{equation*}
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\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)
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\end{equation*}
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由函数的连续性可得$y \to y_0$时，$x = \invertfunc{f}(y) \to \invertfunc{f}(y_0) = x_0$。
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\end{proof}
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\begin{example}
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指数函数求导：设$f(x) = e^x$，求$\deriv{f}(x)$。
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\end{example}
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\begin{proof}[解]
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记$y = e^x$，则$x = \ln y$。那么
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\[\frac{\dif x}{\dif y} = \frac{1}{y}\]
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利用反函数求导公式
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\[\frac{\dif y}{\dif x} = \frac{1}{\dfrac{\dif x}{\dif y}} = y\]
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即$\deriv{(e^x)} = e^x$。
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\end{proof}
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\begin{corollary}
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令$a > 0, a \neq 1$，则$a^x = e^{x\ln a}$，$\log_a x = \dfrac{\ln x}{\ln a}$。
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因此
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\[\deriv{(a^x)} = \deriv{(e^{x\ln a})} = (e^{x\ln a})\deriv{(x\ln a)} = a^x \ln a \eqco\]
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\[\deriv{(\log_a x)} = \frac{\deriv{(\ln x)}}{\ln a} = \frac{1}{x\ln a}\]
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\end{corollary}
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\subsection{参数方程求导法}
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\begin{theorem}
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设函数$y = f(x)$由参数方程
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\begin{equation*}
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\begin{cases}
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x = \varphi(t)\\
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y = \psi(t)
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\end{cases}
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\end{equation*}
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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
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\[\frac{\dif y}{\dif x} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)} \eqper\]
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\end{theorem}
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\begin{proof}
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分析函数关系：
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\[x = \varphi(t) \Rightarrow t= \invertfunc{\varphi}(x)\]
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那么$y$通过$t$成为$x$的复合函数
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\[y = \psi[\invertfunc{\varphi}(x)]\]
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再利用复合函数和反函数微分法，得
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\[\frac{\dif y}{\dif x} = \frac{\dif y}{\dif t} \cdot \frac{\dif t}{\dif x} = \frac{\frac{\dif y}{\dif t}}{\frac{\dif x}{\dif t}} = \frac{\deriv{\psi}(t)}{\deriv{\varphi}(t)}\]
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\end{proof}
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\subsection{基本导数公式}
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\begin{multicols}{2}
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\begin{enumerate}
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\item $\deriv{(C)} = 0$
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\item $\deriv{(x^\alpha)} = \alpha x^{\alpha-1}$
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\item $\deriv{(e^x)} = e^x$
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\item $\deriv{(a^x)} = a^x \ln a$
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\item $\deriv{(\ln x)} = \dfrac{1}{x}$
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\item $\deriv{(\log_a x)} = \dfrac{1}{x \ln a}$
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\item $\deriv{(\sin x)} = \cos x$
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\item $\deriv{(\cos x)} = -\sin x$
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\item $\deriv{(\tan x)} = \dfrac{1}{\cos^2 x} = \sec^2 x$
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\item $\deriv{(\cot x)} = -\dfrac{1}{\sin^2 x} = -\csc^2 x$
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\item $\deriv{(\sec x)} = \sec x \tan x$
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\item $\deriv{(\csc x)} = -\csc x \cot x$
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\item $\deriv{(\arcsin x)} = \dfrac{1}{\sqrt{1 - x^2}}$
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\item $\deriv{(\arccos x)} = -\dfrac{1}{\sqrt{1 - x^2}}$
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\item $\deriv{(\arctan x)} = \dfrac{1}{1 + x^2}$
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\item $\deriv{(\arccot x)} = -\dfrac{1}{1 + x^2}$
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\end{enumerate}
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\end{multicols}
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\section{高阶导数及其计算}
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\subsection{高阶导数的计算}
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回忆：初等函数函数的导数仍是初等函数。
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\begin{definition}[高阶导数]
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设$\deriv{f}: I \to \realnum$为$f$的导函数，定义
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\[f^{\prime \prime}(x_0) = \tolim{x}{x_0}\frac{\deriv{f}(x) - \deriv{f}(x_0)}{x - x_0} \text{（若极限存在）}\]
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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$f^{(0)} = f$。
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\end{definition}
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\begin{remark}
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高阶导数的Leibniz记号：令$y = f(x)$，
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\[\frac{\dif^2 y}{\dif x^2} = \frac{\dif}{\dif x}\left(\frac{\dif y}{\dif x}\right),\ \frac{\dif^n y}{\dif x^n} = \frac{\dif}{\dif x}\left(\frac{\dif^{n-1} y}{\dif x^{n-1}}\right), \ n = 1, 2, \cdots\]
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\end{remark}
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\subsection{高阶导数的计算}
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\begin{theorem}
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设函数$f$，$g$都在$I$上有$n$阶导数，则
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\begin{enumerate}
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\item $(f \pm g)^{(n)} = f^{(n)} \pm g^{(n)}$；
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\item $(cf)^{(n)} = cf^{(n)}$，$c$为常数；
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\item （Leibniz公式）$(fg)^{(n)} = \sum \limits_{k=0}^n \dbinom{k}{n}f^{(k)}g^{(n-k)}$。
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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对于由参数方程
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\begin{equation*}
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\begin{cases}
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x = \varphi(t)\\
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y = \psi(t)
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\end{cases}
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\end{equation*}
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确定的函数$y = f(x)$，有
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\begin{equation*}
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\begin{cases}
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x = \varphi(t)\\
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\deriv{y} = \dfrac{\deriv{\psi(t)}}{\deriv{\varphi(t)}} = \omega(t)
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\end{cases}
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\end{equation*}
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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
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\[y^{\prime \prime} = \frac{\dif }{\dif x} (\deriv{y}(x)) = \frac{\deriv{\omega}(t)}{\deriv{\varphi(t)}}\]
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\end{theorem}
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\section{微分/导数中值定理}
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目标：研究区间内可导函数的导数性质。
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\begin{definition}[极值]
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设$f$在$x_0$附近有定义（包括$x_0$），若$\exists \delta > 0$，使得
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\begin{enumerate}
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\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；
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\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；
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\end{enumerate}
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\end{definition}
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\begin{remark}
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极值不一定是最值。极值点必须在区间内部。
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||
\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
|
||
\[\deriv{F}(c) = \deriv{f}(c) - \frac{f(b)-f(a)}{b-a} = 0\eqper\]
|
||
\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}
|
||
\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}
|
||
\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)
|
||
\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
|
||
\[\frac{f(x) - f(x_1)}{x - x_1} \leq \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_2) - f(x)}{x_2 - x}\eqper\]
|
||
\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
|
||
\[\tolim{x}{a^+} \frac{f(x)}{g(x)} = \tolim{x}{a^+} \frac{\deriv{f}(\xi)}{\deriv{g}(\xi)} = \tolim{x}{a^+}\frac{\deriv{f}(x)}{\deriv{g}(x)}\eqper\]
|
||
\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
|
||
\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$f$对对称性：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} |