diff --git a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf
index 2f06f1d..c78c4a6 100644
Binary files a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf and b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.pdf differ
diff --git a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex
index d7fd111..971b9ab 100644
--- a/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex
+++ b/advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex
@@ -1045,6 +1045,8 @@ $\therefore I_n=n!$。
 
 不用计算积分值，直接通过幂函数的极限来代替积分。
 
+\subsubsection{等级无穷小}
+
 \subsubsection{无穷区间}
 
 对于无限反常积分$\int_a^{+\infty}f(x)\,\textrm{d}x$，$f(x)$在$[a,+\infty)$上连续非负，对于常数$\rho$：
diff --git a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf
index 73c8175..77901e0 100644
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diff --git a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
index 4589d36..4d54718 100644
--- a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
+++ b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
@@ -166,6 +166,20 @@ $\therefore z=\dfrac{1}{2}x^2y+\dfrac{1}{2}xy^2+x+y^2$。
 
 通过微分定义和极限即可证明。
 
+\subsubsection{隐函数存在定理}
+
+当给出一个隐函数方程$f(x,y)$或$f(x,y,z)$时，各自对变量求偏导得到偏导函数，然后将判断的点的值代入偏导函数，若得到的值不为0则在该点的某个邻域内能确定关于这个变量的隐函数，否则不能。
+
+\textbf{例题：}已知方程$xy-z\ln y+e^{xz}=1$，若存在点$(0,1,1)$的一个邻域，求该邻域内该方程能确定哪些隐函数。
+
+解：对于$F=xy-z\ln y+e^{xz}-1$，则$F(0,1,1)=0$。
+
+分别求偏导：$F_x'=y+ze^{xz}$，$F_y'=x-\dfrac{z}{y}$，$F_z'=-\ln y+xe^{xz}$。
+
+代入点：$F_x'=1+1=2\neq0$，$F_y'=-1\neq0$，$F_z'=0$。
+
+所以根据隐函数存在定理$x$、$y$的隐函数存在。
+
 % \subsection{极限}
 
 \subsection{微分}