\documentclass{beamer}
\usepackage{multicol}
\usetheme[progressbar=frametitle]{metropolis}
\setbeamertemplate{frame numbering}[fraction]
\useoutertheme{metropolis}
\useinnertheme{metropolis}
\usefonttheme{metropolis}
\usecolortheme{spruce}
\setbeamercolor{background canvas}{bg=white}
\definecolor{mygreen}{rgb}{.125,.5,.25}
%\usecolortheme[named=mygreen]{structure}
\title{Continuity, Differentiability, and Derivatives}
\author{dsdf}
\date{\today}
\setbeamercovered{transparent = 15}
\begin{document}
\metroset{block=fill}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}[t]{Continuity}
A function is continuous if it can be drawn without picking up the pencil; otherwise, it is discontinuous. Function f(x) is continuous if:
\begin{columns}[onlytextwidth,T]
\vspace{5pt}
\column{0.5\textwidth}
\begin{enumerate}
\vspace{10pt}
\item $f(a)$ exists
\item $\lim\limits_{x \to a} f(x)$=$f(a)$
\item $\lim\limits_{x \to a^{+}} f(x)$=$\lim\limits_{x \to a^{-}} f(x)$
\end{enumerate}
\column{0.5\textwidth}
\end{columns}
\end{frame}
\begin{frame}{Continuity}
A function is differentiable at a point \textit{c} if $\lim\limits_{h \to 0}$ $\frac{f(c+h)-f(c)}{h}$ exists. In other words, the slope must:
\only<1>{\line(1,0){50}\,}
\only<2,3,4>{\textcolor{magenta}{exist}}.
\\
\vspace{30pt}
If a function is
\only<1,2>{\line(1,0){50}\,}
\only<3,4>{\textcolor{magenta}{differentiable}}
it must also be
\only<1,2,3>{\line(1,0){50}\,}
\only<4>{\textcolor{magenta}{continuous}}.
\end{frame}
\begin{frame}[t]{Types of Discontinuity}
\only<1>{\begin{block}{Removable Discontinuity}
\vspace{4pt}
$\lim\limits_{x \to a^+} f(x) =\lim\limits_{x \to a^-} f(x)$, but $\lim\limits_{x \to a^+} f(x) \neq f(a)$
\vspace{4pt}
\end{block}
\vspace{8pt}
\vspace{10pt}
\only<2>{\begin{block}{Jump Discontinuity}
\vspace{4pt}
$\lim\limits_{x \to a^+} f(x) \neq \lim\limits_{x \to a^-} f(x)$
\vspace{4pt}
\end{block}
\vspace{10pt}
\only<3>{\begin{block} {Infinite Discontinuity}
\vspace{4pt}
$\lim\limits_{x \to a^+} f(x) = \infty$ and/or $\lim\limits_{x \to a^-} f(x) = \infty$
\vspace{4pt}
\end{block}
\vspace{10pt}
\end{frame}
\begin{frame}{Derivatives}
\begin{block}{Definition of the Derivative}
\vspace{4pt}
\begin{columns}[onlytextwidth]
\column{0.5\textwidth}
$\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} = f'(x)$
\column{0.5\textwidth}
$\lim\limits_{h \to 0} \frac{f(x)-f(a)}{x-a} = f'(x)$
\end{columns}
\vspace{4pt}
\end{block}
Set D is called the
\only<1>{\line(1,0){50}\, }
\only<2>{\textcolor{magenta}{domain}}
of the function.\\[10pt]
Set E is called the
\only<1>{\line(1,0){50}\, }
\only<2>{\textcolor{magenta}{range}}
of the function.
\end{frame}
\begin{frame}{Derivative of an Inverse}
\only <1> [$f^{-1}$]'(b) =
\only <1> $\displaystyle\frac{1}{f'(f^{-1}(b))}$
\end{frame}
\begin{frame}{Exampe of Inverse Derivatives}
\only <1,2,3>
\only <2,3>
\only <1,2,3>
\only <3>
\end{frame}
\begin{frame}[t]{Implicit Differentiation}
Use implicit differentiation to find the derivative of $xy=1$
\vspace{8pt}
\\Step 1: $\frac{d}{dx}(xy)=\frac{d}{dx}(1)$
\vspace{8pt}
\\Step 2: $(y+x(\frac{dy}{dx}))=0$
\vspace{8pt}
\\Step 3: $x(\frac{dy}{dx})=-y$
\vspace{8pt}
\\Step 4: $\frac{dy}{dx}=-\frac{y}{x}$
\end{frame}
\begin{frame}[t]{Trig Derivatives}
\begin{block}{Trig Derivatives}
\vspace{8pt}
\begin{columns}
\column{.3\textwidth}
$[sin(x)]'=$
\only<2-7>{$cos(x)$}
\\$[cos(x)]'=$
\only<3-7>{$-sin(x)$}
\\$[tan(x)]'=$
\only<4-7>{$sec^2(x)$}
\column{.4\textwidth}
$[cot(x)]'=$
\only<5-7>{$-csc^2(x)$}
\\$[sec(x)]'=$
\only<6-7>{$sec(x) tan(x)$}
\\$[csc(x)]'=$
\only<7>{$-csc(x) cot(x)$}
\end{columns}
\end{block}
\end{frame}
\begin{frame}[t]{Inverse Trig Derivatives}
\begin{block}{Inverse Trig Derivatives}
$[sin^{-1}(x)]'=$
\only<2-4>{$\displaystyle{\frac{1}{\sqrt{1-x^2}}}$}
\\$[tan^{-1}(x)]'=$
\only<3-4>{$\displaystyle{\frac{1}{x^2+1}}$}
\\$[sec^{-1}(x)]'=$
\only<4>{$\displaystyle{\frac{1}{|x|\sqrt{x^2-1}}}$}
\end{block}
\end{frame}
\begin{frame}[t]{Exponential and Logarithmic Derivatives}
Some of the rules for exponential and logarithmic derivatives are:
\begin{enumerate}
\item $\displaystyle [e^u]' = \displaystyle u' \cdot e^u$
\item $\displaystyle [b^u]' = \displaystyle ln(b) \cdot b^u$, where \textit{b} is a constant.
\item $\displaystyle[ln\; u]' = \displaystyle \frac{1}{u}$
\end{enumerate}
Find the derivatives for these examples:
\begin{enumerate}
\item $e^{2x}$
\item $2^x$
\item $ln\; 3$
\end{enumerate}
\end{frame}
\end{document}
答案1
only 的语法类似于，但是您忘记了出现错误的框架中的\only<1>{...}三个 s 中的右括号，即only不连续性的类型框架。在该框架中的}每个框架后添加一个。\end{block}
您还发现一个框架中只有四个\only<..>，没有任何括号，这也会导致错误。请添加括号对。