From f705e8ee1f744ede26172bbfda2d4a4ec8b43b65 Mon Sep 17 00:00:00 2001 From: Didnelpsun <48906416+Didnelpsun@users.noreply.github.com> Date: Tue, 16 Feb 2021 15:57:34 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E7=9B=AE=E5=BD=95=E8=B6=85?= =?UTF-8?q?=E9=93=BE=E6=8E=A5?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../continuity-and-discontinuity.tex | 2 ++ .../derivative-and-differentiate.tex | 12 ++++++++++++ 2 files changed, 14 insertions(+) diff --git a/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex b/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex index 2b66829..d5259e5 100644 --- a/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex +++ b/advanced-math/exercise/2-continuity-and-discontinuity/continuity-and-discontinuity.tex @@ -20,6 +20,8 @@ % 因为所以 \usepackage{amsmath} % 数学公式 +\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref} +% 超链接 \author{Didnelpsun} \title{连续与间断} \date{} diff --git a/advanced-math/exercise/3-derivative-and-differentiate/derivative-and-differentiate.tex b/advanced-math/exercise/3-derivative-and-differentiate/derivative-and-differentiate.tex index 101a382..23d3417 100644 --- a/advanced-math/exercise/3-derivative-and-differentiate/derivative-and-differentiate.tex +++ b/advanced-math/exercise/3-derivative-and-differentiate/derivative-and-differentiate.tex @@ -20,6 +20,8 @@ % 因为所以 \usepackage{amsmath} % 数学公式 +\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref} +% 超链接 \author{Didnelpsun} \title{导数与微分} \date{} @@ -148,6 +150,10 @@ $\therefore\alpha-2>0$,从而$\alpha>2$。 题目会给出对应的导数以及相关条件,并要求求一个极限,这个极限式子并不是个随机的式子,而一个是与导数定义相关的极限式子,所需要的就是将极限式子转换为导数定义的相关式子。 +\subsubsection{导数定义式子} + +有时极限式子可以直接转换为导数定义式子,稍微变换就可以代入导数。 + \textbf{例题:}设$f(x)$是以3为周期的可导函数,且是偶函数,$f'(-2)=-1$,求$\lim\limits_{h\to 0}\dfrac{h}{f(5-2\sin h)-f(5)}$。\medskip 根据导数与函数的基本性质,原函数为偶函数,则其导函数为奇函数,所以$f'(5)=f'(2)=-f'(-2)=1$。 @@ -162,6 +168,12 @@ $=-2f'(5)=-2\times 1=-2$ $\therefore\lim\limits_{h\to 0}\dfrac{h}{f(5-2\sin h)-f(5)}=-\dfrac{1}{2}$。 +\subsubsection{定义近似式子} + +有时候极限式子不为导数定义的近似式子,这时候就需要先根据求极限的计算方式简化目标极限式子。 + + + \section{高阶导数} \subsection{导数存在性}