From 4bb5f32a067452c178e95ba8359446542bac4a2e Mon Sep 17 00:00:00 2001 From: Lei Zhao Date: Wed, 28 Aug 2024 16:59:04 +0800 Subject: [PATCH] =?UTF-8?q?Finish=20chapter=204=20for=20=E5=BE=AE=E7=A7=AF?= =?UTF-8?q?=E5=88=86B?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- notes/calculus/preamble-all.tex | 1 + ...\345\276\256\347\247\257\345\210\206B.tex" | 491 +++++++++++++++++- 2 files changed, 484 insertions(+), 8 deletions(-) diff --git a/notes/calculus/preamble-all.tex b/notes/calculus/preamble-all.tex index 39636bf..da035d7 100644 --- a/notes/calculus/preamble-all.tex +++ b/notes/calculus/preamble-all.tex @@ -140,6 +140,7 @@ \newcommand*{\dv}{\diff v} \newcommand*{\dtheta}{\diff \theta} \newcommand*{\dd}[2][]{\frac{\diff#1}{\diff#2}} +\newcommand*{\ddn}[3][]{\frac{\diff^#3#1}{\diff#2^#3}} \newcommand*{\ddx}{\frac{\diff}{\dx}} \newcommand*{\ddt}{\frac{\diff}{\dt}} \newcommand*{\ddy}{\dd y} diff --git "a/notes/calculus/\345\276\256\347\247\257\345\210\206B.tex" "b/notes/calculus/\345\276\256\347\247\257\345\210\206B.tex" index 95837b4..70eccf6 100644 --- "a/notes/calculus/\345\276\256\347\247\257\345\210\206B.tex" +++ "b/notes/calculus/\345\276\256\347\247\257\345\210\206B.tex" @@ -5203,11 +5203,11 @@ \section{导数与微分的概念} \label{rem:derivgeom} \leavevmode \begin{enumerate} - \item 切线方程：曲线\(y = f(x)\)在\(\paren*{x_0, y_0}\)处的切线方程为 + \item 切线方程：曲线\(y = f(x)\)在点\(\paren{x_0, \,y_0}\)处的切线方程为 \begin{equation*} y = x\,f'(x_0) + y_0 - x_0\,f'(x_0). \end{equation*} - \item 法线方程：曲线\(y = f(x)\)在\(\paren*{x_0, y_0}\)处的法线方程为 + \item 法线方程：曲线\(y = f(x)\)在点\(\paren{x_0, \,y_0}\)处的法线方程为 \begin{equation*} y = - \frac{x}{f'(x_0)} + y_0 + \frac{x_0}{f'(x_0)}. \end{equation*} @@ -5825,7 +5825,8 @@ \section{导数与微分的运算} \end{proof} \end{theorem*} -\begin{theorem*} +\begin{theorem} + \label{thm:derivinv} 若函数\(f\)在点\(x_0\)处可导且\(f'(x_0) \ne 0\)，则其反函数\(\,f^{-1}\)在点\(f(x_0)\)处可导且 \begin{equation*} \dd[f^{-1}]y \Bigg\vert_{\,f(x_0)} \mkern-11mu = \frac1{\,f'(x_0)}. @@ -5844,7 +5845,7 @@ \section{导数与微分的运算} \end{split} \end{equation*} \end{proof} -\end{theorem*} +\end{theorem} \begin{example*} 求反正弦函数的导函数． @@ -5990,9 +5991,9 @@ \subsection*{练习} \item 函数\(f\)满足\(y = f\,\paren[\Big]{\frac{x+1}{x-1}} \)和\(\,f'(x) = \arctan\sqrt x\)．求\(\dydx\bigg\vert_{2}\)． \ifshowsol \begin{equation*} - \dydx\bigg\vert_{2} - = f'(3) \cdot \paren[\bigg]{\frac{x+1}{x-1}}'\bigg\vert_{2} - = \arctan\sqrt3 \cdot \paren[\bigg]{-\frac{2}{(x-1)^2}} \bigg\vert_{2} + \dydx\Bigg\vert_{2} + = f'(3) \cdot \paren[\bigg]{\frac{x+1}{x-1}}'\Bigg\vert_{2} + = \arctan\sqrt3 \cdot \paren[\bigg]{-\frac{2}{(x-1)^2}} \Bigg\vert_{2} = - \frac23 \pi. \end{equation*} \fi @@ -6040,21 +6041,495 @@ \subsection*{练习} \section{几种特殊函数的求导法、高阶导数} +\begin{theorem*} + 设有方程\(f(x,y) = 0\)．把\(x\)看成自变量，把\(y\)看成因变量，则由此方程确定的函数\(y\)在点\(\paren{x_0, \,y_0}\)处关于\(x\)的导数满足 + \begin{equation*} + \diff f(x,y) = \pdpdx\,f(x,y) \dx + \pdpdy\,f(x,y) \dy. + \end{equation*} +\end{theorem*} + +\begin{example*} + 设\(y\)是\(x\)由方程\(x = y + \varepsilon \sin y\)确定的一个函数，其中\(0 < \varepsilon < 1\)．求\(y'\)． + + \begin{remark} + 因为 + \begin{equation*} + \dx + = \diff(y + \varepsilon \sin y) + = \dy + \varepsilon \cos y \dy + = (1 + \varepsilon \cos y) \dy, + \end{equation*} + 所以 + \begin{equation*} + \dydx = \frac1{1 + \varepsilon \cos y}. + \end{equation*} + \end{remark} +\end{example*} + +\begin{example*} + 设\(y\)是\(x\)由\(e^{xy} + \tan xy = y\)确定的一个函数．求\(y'(0)\)． + + \begin{remark} + 当\(x = 0\)时，有 + \begin{math} + e^0 + \tan 0 = 1 = y. + \end{math} + 又因为 + \begin{gather*} + \begin{split} + \dy + &= \diff(e^{xy} + \tan xy) \\ + &= e^{xy} \paren{y \dx + x \dy} + \paren{y \dx + x \dy} \sec^2xy \\ + &= \paren{y \dx + x \dy} \paren{e^{xy} + \sec^2xy}, + \end{split} + \shortintertext{所以} + \dydx = \paren[\bigg]{y + x \dydx} \paren{e^{xy} + \sec^2xy}, \\ + \shortintertext{从而} + \dydx\Bigg\vert_{\substack{x = 0 \\ y = 1}} \mkern-11mu + = (1 + 0)(e^0 + \sec^2 0) + = 2. + \end{gather*} + \end{remark} +\end{example*} + +% https://mathcurve.com/courbes2d.gb/foliumdedescartes/foliumdedescartes.shtml +% https://mathworld.wolfram.com/FoliumofDescartes.html +% https://en.wikipedia.org/wiki/Folium_of_Descartes +\begin{example*} + 求曲线\(x^3 + y^3 = 9xy\)在点\(\paren{2,4}\)处的切线方程和法线方程． + + \begin{remark} + 因为 + \begin{equation*} + \diff(x^3+y^3) + = 3x^2 \dx + 3y^2 \dy + = 9(y \dx + x \dy) + = \diff(9xy), + \end{equation*} + 所以当\(x = 2,\ y = 4\)时就有 + \begin{gather*} + 12\dx + 48\dy + = 9(4\dx + 2\dy) + = 36\dx + 18\dy + \shortintertext{即} + \dydx\Bigg\vert_{\substack{x = 2 \\ y = 4}} \mkern-10mu + = \frac{36-12}{48-18} + = \frac{4}{5}. + \end{gather*} + 因此，这条曲线在此处的切线方程和法线方程分别是 + \begin{equation*} + y = \frac45 x + \frac{12}{5} + \txt{和} + y = - \frac54 x + \frac{13}{2}. + \end{equation*} + \end{remark} +\end{example*} + +\begin{theorem*}[参数方程的求导法] + 设参数方程 + \begin{equation*} + x = x(t), \quad + y = y(t). + \end{equation*} + 若函数\(\fdx\)在点\(t_0\)处连续且在此处附近异号，则 + \begin{equation*} + \dydx\Bigg\vert_{t=t_0} \mkern-15mu + = \lim_{t\to t_0} \frac{\fdy(t)}{\fdx(t)}. + \end{equation*} + 特别地，若函数\(x\)在点\(t_0\)处可导且\(x'(t_0) \ne 0\)，则\(y\)在点\(t = t_0\)处关于\(x\)的导数是 + \begin{equation*} + \dydx\Bigg\vert_{t=t_0} \mkern-15mu + = \frac{\dy/\!\dt}{\dx/\!\dt}\Bigg\vert_{t=t_0} \mkern-22mu. + \end{equation*} + % TODO: Add proof +\end{theorem*} + +\begin{example*} + 求曲线\(r = 1 + \cos\theta\)在\(\theta = \pi/4\)处的切线方程． + + \begin{remark} + 因为\(x = r \cos\theta = \cos\theta + \cos^2\theta\)和\(y = \sin\theta + \cos\theta \sin\theta\)，所以有 + \begin{equation*} + \dydx + = \frac{\dy/\!\dtheta}{\dx/\!\dtheta} + = \frac{\cos\theta + \cos2\theta}{-\sin\theta - \sin2\theta} + = - \frac{\cos\theta + \cos2\theta}{\sin\theta + \sin2\theta}. + \end{equation*} + 因此，当\(\theta = \pi/4\)时有 + \begin{equation*} + \dydx\Bigg\vert_{\theta=\frac\pi4} \mkern-11mu + = - \frac{1/{\scriptstyle\sqrt2}}{1/{\scriptstyle\sqrt2} + 1} + = - \frac{1}{1 + {\scriptstyle\sqrt2}} + = 1 - \sqrt2. + \end{equation*} + \end{remark} +\end{example*} + +\begin{example*} + 设参数方程\(x = 2t + \abs t,\ y = 5t^2 + 3t\abs t\)．求\(y\)在点\(t = 0\)处关于\(x\)的导数． + \begin{equation*} + \dydx\Bigg\vert_{t=0} \mkern-11mu + = \frac{\dy/\!\dt}{\dx/\!\dt} \Bigg\vert_{t=0} \mkern-11mu + = \lim_{t\to0} \frac{5t^2 + 3t\abs t}{2t + \abs t} + = 0. + \end{equation*} +\end{example*} + +\begin{theorem*}[对数求导法] + 函数\(f\)在非零点处满足 + \begin{equation*} + f'(x) = f(x) \,\ddx \ln\,f(x). + \end{equation*} +\end{theorem*} + +\begin{example*} + 求函数\(\,f(x) = x^{x^2}\)的导函数． + \begin{equation*} + f'(x) + = f(x) \,\ddx \ln\,f(x) + = x^{x^2} \paren{2x\ln x + x}. + \end{equation*} +\end{example*} + +\begin{example*} + 求函数 + \begin{equation*} + f(x) = \frac{\prod\limits_{k=1}^n g_k(x)}{\prod\limits_{k=1}^n h_k(x)} + \end{equation*} + 的导函数． + \begin{equation*} + f'(x) + = f(x) \,\ddx \ln\,f(x) + = \paren[\Bigg]{\prod_{k=1}^n \frac{\ g_k(x)}{h_k(x)}} + \paren[\Bigg]{\sum_{k=1}^n \paren[\bigg]{\frac{g_k'(x)}{g_k(x)} - \frac{h_k'(x)}{h_k(x)}}}. + \end{equation*} +\end{example*} + +\begin{example*} + 求函数\(f(x) = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}\)的导函数． + \begin{remark} + 当\(x \in \paren{-\infty,1} \cup \paren{2,3} \cup \paren{4,+\infty}\)时，有 + \begin{equation*} + f'(x) + = f(x) \,\ddx \ln\,f(x) + = \frac12 + \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} + \paren*{\frac1{x-1} + \frac1{x-2} - \frac1{x-3} - \frac1{x-4}}. + \end{equation*} + \end{remark} +\end{example*} + +\begin{definition*}[高阶导数] + \label{defn:derivhio} + 令\(\,f^{(0)} = \dd[^0f]{x^0} = f\)．当\(n > 1\)时，令 + \begin{equation*} + f^{(n)} = \paren[\big]{\,f^{(n-1)}}' + \txt{和} + \dd[^nf]{x^n} = \ddx \dd[^{n-1}f]{x^{n-1}\negthickspace}. + \end{equation*} + 我们把\(\,f^{(n)}\)称为\(n\)阶导数，把大于一阶的导数称为高阶导数． +\end{definition*} + +\begin{theorem*}[莱布尼茨公式] + \begin{equation*} + (\,fg)^{(n)} = \sum_{k=0}^n \binom nk \,f^{(n-k)} g^{(k)}. + \end{equation*} +\end{theorem*} + +\begin{example*} + 求函数\(\,f(x) = a^x\)的\(n\)阶导函数． + \begin{equation*} + f^{(n)}(x) = a^x \ln^n a. + \end{equation*} +\end{example*} + +\begin{example*} + 求函数\(\,f(x) = \frac1{1+x}\)的\(n\)阶导函数． + \begin{equation*} + f^{(n)}(x) = (-1)^n \frac{n!}{(1+x)^{n+1}}. + \end{equation*} +\end{example*} + +\begin{example*} + 求函数\(\,f(x) = \lnp{1+x}\)的\(n\)阶导函数． + \begin{equation*} + f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{(1+x)^n}. + \end{equation*} +\end{example*} + +\begin{example*} + 求正弦函数的\(n\)阶导函数． + \begin{equation*} + \sin^{(n)}x = \sinp{x + {n\pi}/{2}}. + \end{equation*} +\end{example*} + +\begin{example*} + 求余弦函数的\(n\)阶导函数． + \begin{equation*} + \cos^{(n)}x = \cosp{x + {n\pi}/{2}}. + \end{equation*} +\end{example*} + +\begin{example*} + 求函数\(y = \frac{x}{1+x^2}\)的\(n\)阶导函数． + + \begin{remark} + 因为\(x = (1+x^2)\,y\)，所以有 + \begin{gather*} + \begin{split} + \dd[^n]{x^n} (1+x^2)\,y + &= \sum_{k=0}^n \binom nk (1+x^2)^{(n-k)} y^{(k)} \\ + &= n(n-1)y^{(n-2)} + 2nxy^{(n-1)} + (1+x^2)\,y^{(n)} \\ + &= + \begin{cases} + 1, & n = 1, \\ + 0, & n > 1, + \end{cases} + \end{split} + \shortintertext{即} + y^{(n)} = + \begin{cases} + \frac{1 - 2xy}{1+x^2}, & n = 1, \\ + - \frac1{1+x^2}\paren[\big]{n(n-1)y^{(n-2)} + 2nxy^{(n-1)}}, & n > 1. + \end{cases} + \end{gather*} + \end{remark} +\end{example*} + +\begin{example*} + 求函数\(y = \frac1{(1+x)(2+3x)}\)的\(n\)阶导函数． + + \begin{remark} + 因为 + \begin{gather*} + y = \frac1{(1+x)(2+3x)} = \frac{3}{2+3x} - \frac1{1+x}, \\ + \shortintertext{所以} + y^{(n)} = (-1)^n \frac{3^{n+1}n!}{(2+3x)^{n+1}} - (-1)^n \frac{n!}{(1+x)^{n+1}}. + \end{gather*} + \end{remark} +\end{example*} + +\begin{example*} + 求函数\(y = x^2 \sin x\)的\(n\)阶导函数． + \begin{equation*} + \begin{split} + \dd[^n]{x^n} x^2 \sin x + &= \sum_{k=0}^n \binom nk (x^2)^{(n-k)} \sin^{(k)}x \\ + &= n(n-1) \sinp[\Big]{x+(n-2)\frac{\pi}{2}} + + 2nx \sinp[\Big]{x+(n-1)\frac{\pi}{2}} + + x^2 \sinp[\Big]{x+\frac{n\pi}{2}}. + \end{split} + \end{equation*} +\end{example*} + \subpdfbookmark{思考}{B1.4.3.P} \subsection*{思考} \begin{enumerate} -\item 把函数的二阶微分定义成．一阶微分具有形式不变性，二阶微分是否也有形式不变性？ +\item 把函数\(y = \,f(x)\)的二阶微分定义成\(\diff^2y = \,f^{(2)}(x) \dx^2\)．一阶微分具有形式不变性，二阶微分是否也有形式不变性？ + + \ifshowsolp + 没有．令\(y = \,f(u),\ u = g(x)\)，有 + \begin{gather*} + \diff_u^2 y + = f^{(2)}(u) \du^2 + = f^{(2)}\paren[\big]{g(x)} \paren[\big]{g'(x)}^2 \dx^2, \\ + \diff_x^2 y + = \paren[\Big]{\,f^{(2)}\paren[\big]{g(x)} \paren[\big]{g'(x)}^2 + f'\paren[\big]{g(x)}\,g^{(2)}(x)} \dx^2. + \end{gather*} + \fi + \item 由参数方程\(x = \varphi(t),\ y = \psi(t)\)确定的函数\(y = y(x)\)的求导公式\(\dy/\!\dx = \psi'(t)/\varphi'(t)\)是如何确定的？如何求二阶导数\(\diff^2y/\!\dx^2\)和三阶导数\(\diff^3y/\!\dx^3\)？ + + \ifshowsolp + 确定的方式多种多样，最简单的方式就是把\(t\)看成\(x\)的函数，然后根据定理~\ref{thm:derivcomp}和~\ref{thm:derivinv}就能得到．使用同样的方法，能得到 + \begin{gather*} + \dd[^2y]{x^2} + = \ddt \frac{\psi'}{\varphi'} \cdot \dd[t]x + = \frac{\psi^{(2)}}{\paren{\varphi'}^2} + - \frac{\psi'\varphi^{(2)}}{\paren{\varphi'}^3}, \\ + \siand + \begin{split} + \ddn[y]x3 + &= \ddt \paren[\Bigg]{\frac{\psi^{(2)}}{\paren{\varphi'}^2} + - \frac{\psi'\varphi^{(2)}}{\paren{\varphi'}^3}} \cdot \dd[t]x \\ + &= \frac{\psi^{(3)}}{(\varphi')^3} + - \frac{\psi^{(2)}\varphi^{(2)} + + \psi'\varphi^{(3)} + + 2\psi^{(2)}\varphi^{(2)}}{(\varphi')^4} + + \frac{3\psi'(\varphi^{(2)})^2}{(\varphi')^5}. + \end{split} + \end{gather*} + \fi \end{enumerate} \ifshowex \currentpdfbookmark{练习}{B1.4.3.E} \subsection*{练习} + +\begin{enumerate} +\item 设\(x\)的函数\(y\)由方程\(e^{x+y} = xy + 1\)确定．求\(\dy\)． + \ifshowsol + \begin{equation*} + e^{x+y} \dx + e^{x+y} \dy + = y \dx + x \dy + \implies + \dy = \frac{y-e^{x+y}}{e^{x+y}-x} \dx. + \end{equation*} + \fi + +\item 设函数\(f\)二阶可导且\(y = \,f(x^2)\)．求\(y^{(2)}\)． + \ifshowsol + \begin{equation*} + y^{(2)} + = \paren[\big]{\,f'(x^2) \cdot 2x}' + = 2\,f'(x^2) + 4x^2\,f^{(2)}(x^2). + \end{equation*} + \fi + +\item 设参数方程\(x = a \cos t,\ y = b \sin t\)．求\(\ddn[y]x2\)． + \ifshowsol + \begin{gather*} + \dydx = - \frac{b \cos t}{a \sin t}, \\[1ex] + \ddn[y]x2 + = - \ddt \frac{b \cos t}{a \sin t} \cdot \dd[t]x + = - \frac{b}{a^2 \sin t} - \frac{b \cos^2 t}{a^2 \sin^3 t} + = - \frac{b}{a^2} \csc^3 t. + \end{gather*} + \fi + +\item 设参数方程\(x = \sin t,\ y = t \sin t + \cos t\)．求\(\ddn[y]x2 \bigg\vert_{t=\frac\pi4}\mkern-24mu\)． + \ifshowsol + \begin{equation*} + \dydx = \frac{\sin t + t \cos t - \sin t}{\cos t} = t, + \quad + \ddn[y]x2 = \ddt t \cdot \dd[t]x = \sec t, + \quad + \ddn[y]x2 \Bigg\vert_{t=\frac\pi4} \mkern-11mu = \sqrt2. + \end{equation*} + \fi + +\item 设\(x\)的函数\(y\)由方程\(y - 2x = (x-y) \lnp{x-y}\)确定．求\(\ddn[y]x2\)． + \ifshowsol + \begin{gather*} + -2\dx + \dy + = \paren[\big]{\ln(x-y) + 1} \dx - \paren[\big]{\ln(x-y) + 1} \dy, \\[1ex] + \dydx = \frac{\ln(x-y)+3}{\ln(x-y)+2} = 1 + \frac{1}{\ln(x-y)+2}, \\[1ex] + \ddn[y]x2 + = - \frac{1-y'}{(x-y)\paren[\big]{\ln(x-y)+2}{}^2} + = \frac{1}{(x-y)\paren[\big]{\ln(x-y)+2}{}^3}. + \end{gather*} + \fi + +\item 设\(x\)的函数\(y\)由方程\(xy + e^y = x + 1\)确定．求\(\dy\Big\vert_{x=0}\)和\(\ddn[y]x2\bigg\vert_{x=0}\mkern-22mu\)． + \ifshowsol + \begingroup + \addtolength{\jot}{1ex} + \begin{gather*} + y \dx + x \dy + e^y \dy = \dx, \\ + \dydx = \frac{1-y}{x+e^y}, \\ + \dy\Big\vert_{x=0} = \dy\Big\vert_{\substack{x=0 \\ y =0}} = \dx, \\ + \ddn[y]x2 = \frac{-y'}{x+e^y} - \frac{(1-y)(1+e^yy')}{(x+e^y)^2}, \\ + \ddn[y]x2\Bigg\vert_{x=0} + = \ddn[y]x2\Bigg\vert_{\substack{x=0 \\ y =0}} + = -1 - 2 = -3. + \end{gather*} + \endgroup + \fi + +\item 求函数\(\,f(x) = x \sin 2x\)的\(5\)阶导函数． + \ifshowsol + \begin{equation*} + f^{(5)}(x) + = 5 \,\ddn x4 \sin 2x + x \,\ddn x5 \sin 2x + = 16 (5\sin 2x + 2x \cos 2x). + \end{equation*} + \fi + +\item 求函数\(\,f(x) = x^{\sin x}\)的导函数． + \ifshowsol + \begin{equation*} + f'(x) + = f(x) \,\ddx \ln\,f(x) + = x^{\sin x} \,\ddx \sin x \ln x + = x^{\sin x} \paren[\Big]{\cos x \ln x + \frac{\sin x}{x}}. + \end{equation*} + \fi + +\item 求函数\(\,f(x) = x^x\)的导函数． + \ifshowsol + \begin{equation*} + f'(x) + = f(x) \,\ddx \ln\,f(x) + = x^x \paren{\ln x + 1}. + \end{equation*} + \fi + +\item 设\(x\)的函数\(y\)由方程\(x^2 + x^2y^2 + y^2 = 3\)确定．求\(\dy/\!\dx\)． + \ifshowsol + \begin{gather*} + 2x \dx + 2xy^2 \dx + 2x^2y \dy + 2y \dy = 0, \\ + \dydx = - \frac{2x+2xy^2}{2x^2y+2y} = - \frac{x(1+y^2)}{(x^2+1)y}. + \end{gather*} + \fi +\end{enumerate} \fi \chapter{导数应用} +\section{微分中值定理} + + + +\subpdfbookmark{思考}{B1.5.1.P} +\subsection*{思考} + +\ifshowex +\currentpdfbookmark{练习}{B1.5.1.E} +\subsection*{练习} +\fi + +\section{L'Hôpital法则} + +\subpdfbookmark{思考}{B1.5.2.P} +\subsection*{思考} + +\ifshowex +\currentpdfbookmark{练习}{B1.5.2.E} +\subsection*{练习} +\fi + +\section{函数的单调性与极值} + +\subpdfbookmark{思考}{B1.5.3.P} +\subsection*{思考} + +\ifshowex +\currentpdfbookmark{练习}{B1.5.3.E} +\subsection*{练习} +\fi + +\section{函数的凸性与拐点} + +\subpdfbookmark{思考}{B1.5.4.P} +\subsection*{思考} + +\ifshowex +\currentpdfbookmark{练习}{B1.5.4.E} +\subsection*{练习} +\fi + +\section{Taylor公式} + +\subpdfbookmark{思考}{B1.5.5.P} +\subsection*{思考} + +\ifshowex +\currentpdfbookmark{练习}{B1.5.5.E} +\subsection*{练习} +\fi + \chapter{原函数与不定积分} \vskip-1.25em