diff --git a/advanced-math/3-function-and-limit/function-and-limit.tex b/advanced-math/3-function-and-limit/function-and-limit.tex
index 05748ae..0adcab2 100644
--- a/advanced-math/3-function-and-limit/function-and-limit.tex
+++ b/advanced-math/3-function-and-limit/function-and-limit.tex
@@ -3,7 +3,6 @@
 % 使用颜色
 \definecolor{orange}{RGB}{255,127,0} 
 \definecolor{violet}{RGB}{192,0,255} 
-\definecolor{aqua}{RGB}{0,255,255} 
 \usepackage{geometry}
 \setcounter{tocdepth}{5}
 \setcounter{secnumdepth}{5}
@@ -28,6 +27,8 @@
 % 超链接
 \usepackage{tikz}
 % 绘图
+\usepackage{mathtools}
+% 有字的长箭头
 \author{Didnelpsun}
 \title{函数与极限}
 \begin{document}
@@ -542,7 +543,7 @@ $
     & \lim_{x\to-\infty}x(\sqrt{x^2+100}+x) \\
     & \lim_{x\to-\infty}x\dfrac{x^2+100-x^2}{\sqrt{x^100}-x} \\
     & = \lim_{x\to-\infty}\dfrac{100x}{\sqrt{x^2+100}-x} \\
-    & \Rightarrow^{\text{令}x=-t} \lim_{t\to+\infty}\dfrac{-100t}{\sqrt{t^2+100}+t} \\
+    & \xRightarrow{\text{令}x=-t}\lim_{t\to+\infty}\dfrac{-100t}{\sqrt{t^2+100}+t} \\
     & = \lim_{t\to+\infty}\dfrac{-100}{\sqrt{1+\dfrac{100}{t^2}}+1} \\
     & = -50
 \end{aligned}
@@ -558,7 +559,7 @@ $
 \begin{aligned}
     & \lim_{x\to 1^-}\ln x\ln(1-x) \\
     & = \lim_{x\to 1^-}(x-1)\ln(1-x) \\
-    & \Rightarrow^{令t=1-x} =-\lim_{t\to 0^+}t\ln t \\
+    & \xRightarrow{令t=1-x} =-\lim_{t\to 0^+}t\ln t \\
     & = -\lim_{t\to 0^+}\dfrac{\ln t}{\dfrac{1}{t}} \\
     & = -\lim_{t\to 0^+}\dfrac{\dfrac{1}{t}}{-\dfrac{1}{t^2}} \\
     & = \lim_{t\to 0^+}t \\
@@ -617,9 +618,9 @@ $\infty-\infty$型\textbf{例题11：}求极限$\lim_{x\to+\infty}[x^2(e^{\frac{
 $
 \begin{aligned}
     & \lim_{x\to+\infty}[x^2(e^{\frac{1}{x}}-1)-x] \\
-    & \Rightarrow^{\text{令}x=\frac{1}{t}}\lim_{t\to 0^+}\left(\dfrac{e^t-1}{x^2}-\dfrac{1}{t}\right) \\
+    & \xRightarrow{\text{令}x=\frac{1}{t}}\lim_{t\to 0^+}\left(\dfrac{e^t-1}{x^2}-\dfrac{1}{t}\right) \\
     & \lim_{t\to 0^+}\dfrac{e^t-1-t}{t^2} \\
-    & \Rightarrow^{\text{泰勒展开}e^x}\lim_{t\to 0^+}\dfrac{\dfrac{1}{2}x^2}{x^2} \\
+    & \xRightarrow{\text{泰勒展开}e^x}\lim_{t\to 0^+}\dfrac{\dfrac{1}{2}x^2}{x^2} \\
     & =\dfrac{1}{2}
 \end{aligned}
 $
diff --git a/advanced-math/4-single-variable-function-differential-calculus/single-variable-function-differential-calculus.tex b/advanced-math/4-single-variable-function-differential-calculus/single-variable-function-differential-calculus.tex
index ce24d29..84500b2 100644
--- a/advanced-math/4-single-variable-function-differential-calculus/single-variable-function-differential-calculus.tex
+++ b/advanced-math/4-single-variable-function-differential-calculus/single-variable-function-differential-calculus.tex
@@ -1,9 +1,10 @@
 \documentclass[UTF8]{ctexart}
 % UTF8编码，ctexart现实中文
-\usepackage{color}
+\usepackage{xcolor}
 % 使用颜色
 \definecolor{orange}{RGB}{255,127,0} 
 \definecolor{violet}{RGB}{192,0,255} 
+\definecolor{aqua}{RGB}{0,255,255} 
 \usepackage{geometry}
 \setcounter{tocdepth}{4}
 \setcounter{secnumdepth}{4}
@@ -22,6 +23,8 @@
 \usepackage{setspace}
 \renewcommand{\baselinestretch}{1.5}
 % 1.5倍行距
+\usepackage{tikz}
+% 绘图
 \author{Didnelpsun}
 \title{一元函数微分学}
 \begin{document}
@@ -151,11 +154,117 @@ $\therefore \rm{d}y\vert_{x=x_0}=A\Delta x=y'(x_0)\cdot\Delta x=y'(x_0)\cdot\rm{
 
 由此，可导必可微，可微必可导。
 
+\begin{tikzpicture}[scale=0.9]
+    \draw[-latex](-0.5,0) -- (4.5,0) node[below]{$x$};
+    \draw[-latex](0,-0.5) -- (0,4) node[above]{$y$};
+    \draw[black, thick, domain=1.5:3] plot (\x,{pow(\x-1,2)/2+1}) node[above]{$y(x)$};
+    \filldraw[black] (0,0) node[below]{$O$};
+    \draw[black, densely dashed](1.5,1.125) -- (1.5,0) node[below]{$x_0$};
+    \draw[black, densely dashed](1.5,1.125) -- (0,1.125) node[left]{$y_0$};
+    \draw[black, densely dashed](3,3) -- (3,0) node[below]{$x_0+\Delta x$};
+    \draw[black, densely dashed](3,3) -- (0,3) node[left]{$y_0+\Delta x$};
+    \draw[black, densely dashed](3,1.875) -- (0,0.375) node[left]{$\rm{d}yx+b$};
+    \draw[<->, black](1.5,1.125) -- (3,1.125);
+    \draw[<->, black](4,1.125) -- (4,3);
+    \draw[<->, black](3.25,1.125) -- (3.25,1.875);
+    \draw[<->, black](3.25,3) -- (3.25,1.875);
+    \draw[black](3,3) -- (4.5,3);
+    \draw[black](3,1.125) -- (4.5,1.125);
+    \draw[black](3,1.875) -- (3.75,1.875);
+    \filldraw[black] (2.25,0.75) node{$\Delta x$};
+    \filldraw[black] (4.3,2) node{$\Delta y$};
+    \filldraw[black] (3.5,1.5) node{\scriptsize{$\rm{d}y$}};
+    \filldraw[black] (3.5,2.5) node{\scriptsize{$o(\Delta x)$}};
+\end{tikzpicture}
+
+所以可微就是用简单线性取代复杂线性，如图用直线取替代曲线。
+
 \section{导数与微分计算}
 \subsection{四则运算}
+
+若函数可导：
+
+\begin{enumerate}
+    \item 和差的导数或微分：$[u(x)\pm v(x)]'=u'(x)\pm v'(x)$，$\rm{d}[u(x)\pm v(x)]=\rm{d}u(x)\pm\rm{d}v(x)$。
+    \item 积的导数或微分：$[u(x)v(x)]'=u'(x)v(x)+u(x)v'(x)$，$\rm{d}[u(x)v(x)]=u(x)\rm{d}v(x)+v(x)\rm{d}u(x)$，$[u(x)v(x)w(x)]'=u'(x)v(x)w(x)+u(x)v'(x)w(x)+u(x)v(x)+w'(x)$。
+    \item 商的导数：$\left[\dfrac{u(x)}{v(x)}\right]'=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}$，$\rm{d}\left[\dfrac{u(x)}{v(x)}\right]=\dfrac{v(x)\rm{d}u(x)-u(x)\rm{d}v(x)}{[v(x)]^2}$，$v(x)\neq 0$。
+    \item 复合函数的导数：链式求导法则$\dfrac{\rm{d}u}{\rm{d}x}=\dfrac{\rm{d}u}{\rm{d}y}\cdot\dfrac{\rm{d}y}{\rm{d}x}$。
+\end{enumerate}
+
 \subsection{分段函数的导数}
+
+设$f(x)=\left\{
+    \begin{array}{lcl}
+        f_1(x), & & x\geqslant x_0 \\
+        f_2(x), & & x<x_0 \\
+    \end{array}
+\right.$。
+
+在分段点用定义：判断$f'_+(x_0)=\lim_{x\to x_0^+}\dfrac{f_1(x)-f(x_0)}{x-x_0}\overset{?}{=}\lim_{x\to x_0^-}\dfrac{f_2(x)-f(x_0)}{x-x_0}$。
+
+非分段点使用导数公式求导：$x>x_0,f'(x)=f_1'(x),x<0，f'(x)=f_2'(x)$。
+
 \subsection{复合函数的导数与微分形式不变性}
+
+$u=g(x)$在$x$可导，$y=f(u)$在$u=g(x)$处可导，则$\{f[g(x)]\}'=f'[g(x)]g'(x)$，$\rm{d}\{f[g(x)]\}=f'[g(x)]g'(x)\rm{d}x$。
+
+一阶微分形式不变性指：$\rm{d}f(\varsigma)=f'(\varsigma)\rm{d}\varsigma$，无论$\varsigma$是什么。（类似导数的链式求导法则）
+
+\textbf{例题7：}设$f(x)=\Pi_{n=1}^{100}\left(\tan\dfrac{\pi x^a}{4}-n\right)$，则$f'(1)$为？
+
+原式=$\left(\tan\dfrac{\pi x}{4}-1\right)\left(\tan\dfrac{\pi x^2}{4}-2\right)\cdots\left(\tan\dfrac{\pi x^100}{4}-100\right)$。
+
+令$\left(\tan\dfrac{\pi x^2}{4}-2\right)\cdots\left(\tan\dfrac{\pi x^100}{4}-100\right)=g(x)$。
+
+$\therefore f(x)=\left(\tan\dfrac{\pi x}{4}-1\right)\cdot g(x)$。
+
+$\therefore f'(x)=\sec^2\dfrac{\pi x}{4}\cdot\dfrac{\pi}{4}\cdot g(x)+\left(\tan\dfrac{\pi x}{4}-1\right)\cdot g'(x)$。
+
+$\therefore$根据导数的四则运算，需要导数的乘积为每一项求导乘以其他不求导项的和，而$\tan\dfrac{\pi x}{4}-1$当$x=1$时为0，只要它不求导，其他的项都必然是0，所以原式的后面的结果都是0。
+
+$\therefore$
+
+$
+\begin{aligned}
+    f'(1) & =f'(x)\vert_{x=1} \\
+    & =\dfrac{\pi}{2}\cdot g(1)+0\cdot g'(x) \\
+    & =\dfrac{\pi}{2}\cdot g(1) \\
+    & =\dfrac{\pi}{2}(-1)(-2)\cdots(-99) \\
+    & =-\dfrac{\pi}{2}\cdot 99!
+\end{aligned}
+$
+
+\textbf{例题8：}设$y=e^{\sin(\ln x)}$，求$\rm{d}y$。
+
+$\because y=e^{\sin(\ln x)} \therefore$
+
+$
+\begin{aligned}
+    \rm{d}y &=\rm{d}e^{\sin(\ln x)} \\
+    & =e^{\sin(\ln x)}\cdot\rm{d}(\sin(\ln x)) \\
+    & =e^{\sin(\ln x)}\cdot\cos(\ln x)\cdot\rm{d}\ln x \\
+    & =e^{\sin(\ln x)}\cdot\cos(\ln x)\cdot\dfrac{1}{x}\rm{d}x
+\end{aligned}
+$
+
 \subsection{反函数导数}
+
+\textcolor{aqua}{\textbf{定理：}}$y=f(x)$可导，且$f'(x)\neq 0$，则存在反函数$x=\varphi(y)$，且$\dfrac{\rm{d}x}{\rm{d}y}=\dfrac{1}{\dfrac{\rm{d}y}{\rm{d}x}}$，即$\varphi'(x)=\dfrac{1}{f'(x)}$。
+
+$y=f(x)$可导，且$f'(x)\neq 0$就是指严格单调，而严格单调必有反函数。
+
+\textbf{例题9：}求$y=\arcsin x,x\in(-1,1)$与$y=\arctan x$的导数。
+
+首先反三角函数就是三角函数的反函数、
+
+求$y=\arcsin x$，即$x=\sin y$。
+
+$\therefore\dfrac{\rm{d}\arcsin x}{\rm{d}x}=\dfrac{1}{\dfrac{\rm{d}\sin y}{\rm{d}y}}=\dfrac{1}{\cos y}=\dfrac{1}{\sqrt{1-\sin^2y}}=\dfrac{1}{\sqrt{1-x^2}}$。
+
+求$y=\arctan x$，就$x=\tan y$。
+
+$\therefore\dfrac{\rm{d}\arctan x}{\rm{d}x}=\dfrac{1}{\dfrac{\rm{d}\tan y}{\rm{d}y}}=\dfrac{1}{\sec^2y}=\dfrac{1}{1+\tan^2y}=\dfrac{1}{1+x^2}$。
+
 \subsection{参数方程函数导数}
 \subsection{隐函数求导法}
 \subsection{对数求导法}
@@ -165,5 +274,10 @@ $\therefore \rm{d}y\vert_{x=x_0}=A\Delta x=y'(x_0)\cdot\Delta x=y'(x_0)\cdot\rm{
 \subsubsection{莱布尼茨公式}
 \subsubsection{泰勒公式}
 \subsection{变限积分求导公式}
+
+必然会考。
+
+已知更改区间限制的积分$s(x)=\int_{\varphi_1(x)}^{\varphi_2(x)}g(t)\rm{d}x$，$s'(x)=g[\varphi_2(x)]\cdot\varphi_2'(x)-g[\varphi_1(x)]\cdot\varphi_1'(x)$。
+
 \subsection{基本求导公式}
 \end{document}