\section{Inverse Functions} \index{inverse function(s)} 
\begin{example}
  The inverse function of the function $f(x)=x^3$ is $g(x)=x^{\frac{1}{3}}$.
\end{example}

Intuitively $g(x)=x^{\frac{1}{3}}$ is the inverse of is $f(x)=x^3$ because $g$ undoes the action of $f$.  So if $f$ acts on the value $2$ so $f(2)=2^3=8$ and we act $g$ on the result, $g(8)=8^{\frac{1}{3}}=2$ we are returned to the original value.

One may wonder whether all functions have inverses.  The answer is no.  A necessary and sufficient condition for a function to have an inverse is that the function be \emph{one-to-one}.

\definition{A function $f$ with domain $A$ and range $B$ is said to be \textbf{one-to-one}\index{one-to-one function} if whenever $f(x_1)=f(x_2)$ (in $B$) one has that $x_1=x_2$ (in $A$)\index{inverse function(s)! definition of}.} 
A logically equivalent condition is that if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$.  In words, no two elements in the domain $A$ have the same image in the range $B$.

\subsection{Horizontal Line Test}

The \textbf{Horizontal Line Test}\index{Horizontal Line Test} says that a function $f(x)$ will be one-to-one if and only if every horizontal line intersects the graph of $y=f(x)$ at most once.

\begin{example}
  The horizontal line test shows that the function $y=x^2$ is not one-to-one while $y=x^3$ is one-to-one.

\begin{center}
\hfill\raisebox{13.7em}{\includegraphics{graphics/inverses_and_other_functions1}}\hfill
\includegraphics{graphics/inverses_and_other_functions2}\hfill~
\end{center}

\end{example}

The following theorem is intuitively true when one considers the Horizontal Line Test.
\theorem{Suppose a function $f$ has a domain $D$ consisting of an interval.  If the function is increasing everywhere or decreasing everywhere on $D$ then $f$ is one-to-one.}

\begin{example}\label{exam:onetoone}
  Determine whether the given functions are one-to-one:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle f(x)= 2x^3 + 5 $ 
    \item $\displaystyle f(x)= \frac{x}{x-4}$ 
    \item $\displaystyle f(x)= \frac{\sqrt{x+1}}{\sqrt{x}}$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
    \begin{enumerate}
    \item $\displaystyle f(x)= 2x^3 + 5 $
    \begin{itemize}
    \item Method 1: Using the definition:
    \begin{align*} 
    f(x_1)= f(x_2) & \implies 2x_1^3 +5 = 2x_2^3 + 5 \\
    & \implies 2x_1^3= 2x_2^3 \\
    & \implies x_1^3= x_2^3  \text{ (Take the cubic root of both sides.)} \\
    & \implies x_1= x_2
    \end{align*}
    \item Method 2: Using the derivative:
    \[f'(x) = 6x^2 > 0\ \text{for all\ } x\ \text{in}\ D=(-\infty,\infty)\]
    Therefore, $f(x)$ is always increasing on an interval which implies that $f(x)$ is one-to-one.
     \end{itemize}
   \item $\displaystyle f(x)= \frac{x}{x-4}$
     
	We use the definition to prove $f(x)$ is one-to-one: 
    \begin{align*}
     f(x_1)= f(x_2) & \implies\frac{x_1}{x_1-4} = \frac{x_2}{x_2-4} \\
    & \implies x_1(x_2-4)=x_2(x_1 -4) \\
    & \implies x_1x_2-4x_1 = x_2x_1 -4x_2 \\
    & \implies -4x_1 = -4x_2 \\
    & \implies x_1= x_2
    \end{align*}
    Note the derivative of $f(x)$ is negative everywhere it is defined:
   \[f'(x) = \frac{(1)(x-4)-x(1)}{(x-4)^2} = \frac{-4}{(x-4)^2}  < 0\]
    Therefore $f(x)$ is decreasing for all $x$ in the domain of $f$.  However the domain $D=\mathset{R}-\{4\}=(-\infty,4)\cup(4,\infty)$ is broken into two intervals so this is not sufficient to prove $f(x)$ is one-to-one.  Consider, as a counter-example, $f(x)=\tan x$ which increases everywhere on its domain but fails hopelessly to be one-to-one as is seen easily by the horizontal line test! Care must be taken to inspect the domain when using the derivative approach.  One would need to show that the range of the function on each interval it contained was disjoint (did not overlap) with the range of the function on the other intervals. This occurs in this example where $f$ maps $(-\infty,4)$ to $(-\infty,1)$ and maps $(4,\infty)$ to $(1,\infty)$.
  \item $\displaystyle f(x)= \frac{\sqrt{x+1}}{\sqrt{x}}$
    
	The square roots require $x\ge-1$ and $x\ge 0$.  Furthermore we cannot divide by zero. Therefore the domain of the function is composed of a single open interval, $D=(0, \infty)$.  We use the method of the derivative:
	\begin{align*}
	f'(x)&=\frac{\frac{1}{2}(x+1)^{-1/2}\sqrt{x}-\sqrt{x+1}\left(\frac{1}{2}x^{-1/2}\right)}{x} \\
	&=\frac{\frac{\sqrt{x}}{\sqrt{x+1}}-\frac{\sqrt{x+1}}{\sqrt{x}}}{2x}\cdot\frac{\sqrt{x}\sqrt{x+1}}{\sqrt{x}\sqrt{x+1}} = \frac{x-(x+1)}{2x^{3/2}\sqrt{x+1}} \\
	& = \frac{-1}{2x^{3/2}\sqrt{x+1}} < 0 \text{ for all $x > $ 0}
	\end{align*}
 Therefore, $f(x)$ is decreasing  on interval $(0, \infty)$ which implies that $f(x)$ is one-to-one.
      \end{enumerate}
      \end{solution*}

\begin{furtherquestions}
Determine whether the given functions are one-to-one:
\begin{enumerate}
\item $\displaystyle f(x)=2x^3+5$
\item $\displaystyle f(x)=\frac{3-x}{x+1}$
\end{enumerate}
\end{furtherquestions}
\end{example}

\definition{Suppose $f$ is a one-to-one function defined on domain $A$ with range $B$.  The \textbf{inverse function of $\bm f$} denoted by $\bm{f^{-1}}$ is defined on domain $B$ with range $A$ and satisfies
\[f^{-1}(y)=x\ \iff\ f(x)=y\]
for any $y$ in $B$.}

Here the symbol $\iff$ means ``if and only if''.  ``\emph{This} if and only if \emph{that}'' itself means that both the following hold
\begin{itemize}
\item ``If \emph{this} then \emph{that}.''
\item ``If \emph{that} then \emph{this}.''
\end{itemize}
Also observe that the notation $f^{-1}$ makes it clear that $f$ is the function for which this is the inverse.

Notes:
\begin{enumerate}
\item $\displaystyle f^{-1}\neq \frac{1}{f}$ We call $\frac{1}{f}$ the \emph{reciprocal} \index{reciprocal} of $f$.
  \textbook{\clearpage}
\item The definition says that if $f$ maps $x$ to $y$, then $f^{-1}$ maps $y$ back to $x$.
\begin{center}
\parbox[t]{.35\textwidth}{\vspace{0em}{\resizebox{.35\textwidth}{!}{\includegraphics{graphics/inverse}}}}
\end{center}
\item The domain of $f^{-1}$ is the range of $f$ while the range of $f^{-1}$ is the domain of $f$.
\item Reversing the roles of $x$ and $y$ gives
\[f^{-1}(x)=y\ \iff\ f(y)=x\]
or equivalently
\[f(y)=x\iff f^{-1}(x)=y\]
This implies that $f$ itself is the inverse function of $f^{-1}$.
\item The following hold (see last diagram)
\begin{eqnarray*}
f^{-1}\left(f(x)\right) & = & x\ \ \ \mbox{for every $x$ in $A$} \\
f\left(f^{-1}(y)\right) & = & y\ \ \ \mbox{for every $y$ in $B$} 
\end{eqnarray*}
The first relationship highlights the utility of the inverse function in \emph{solving equations} for if we have, say, $f(x)=3$ for some one-to-one function $f$ for which we know the inverse $f^{-1}(x)$, it follows, applying $f^{-1}$ to both sides that $f^{-1}\left(f(x)\right)=x=f^{-1}(3)$.  We are applying inverse functions all the time when we isolate variables in equations.

This explains why ``cube-rooting both sides'' of $x^3=64$ is a safe way to find the solution to this equation while ``square-rooting both sides'' of $x^2=64$ is not.  The latter finds only one of the two solutions.  (Applying a \emph{function} in this way can only produce one number.)

\end{enumerate}

\subsection{Finding Inverse Functions}
To find the inverse function of a one-to-one function $f$ proceed with the following steps:
\begin{enumerate}
\item Write $y=f(x)$
\item Solve the equation for $x$ in terms of $y$ (if possible).
\item Interchange the roles of $x$ and $y$.  The resulting equation is $y=f^{-1}(x)$.
\end{enumerate}
\begin{example}
  Find the inverse function of the given function:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle f(x)= 2x^3 + 5 $
    \item $\displaystyle f(x)= \frac{x}{x-4} $
    \item $\displaystyle f(x)= \frac{\sqrt{x+1}}{\sqrt{x}}$
    \end{enumerate}
  \end{multicols}
 \begin{solution}
 These are the same functions from Example~\ref{exam:onetoone}. As they were all shown to be one-to-one we know they are invertible.
 \begin{enumerate}
\item To find the inverse function of $\displaystyle f(x)= 2x^3 + 5 $, we solve $y=f(x)$ for $x$ in terms of $y$. \\
	\begin{align*}
	y= f(x) \implies & y= 2x^{3} +5 \implies y-5= 2x^3 \\
	 \implies & \frac{y-5}{2} = x^{3} \implies x= \sqrt[3]{\frac{y-5}{2}} \\
 &\text{Exchange $x$ and $y$}\implies y= \sqrt[3]{\frac{x-5}{2}} \\
	& \text{Therefore}\,\, f^{-1}(x)= \sqrt[3]{\frac{x-5}{2}}
	\end{align*}
\item To find the inverse function of $\displaystyle f(x)= \frac{x}{x-4}$, we solve $y=f(x)$ for $x$ in terms of $y$. \\
	\begin{align*}
	y=f(x) \implies y= \frac{x}{x-4} \implies& y(x-4)=x \implies yx-4y=x \\
	\implies& yx-x=4y \implies x(y-1)= 4y \\
	\implies& x= \frac{4y}{y-1} \\
        &\text{Exchange $x$ and $y$}\implies y= \frac{4x}{x-1} \\
	&\text{Therefore}\,\, f^{-1}(x)= \frac{4x}{x-1} \\
	\end{align*}


\item To find the inverse function of $\displaystyle f(x)= \frac{\sqrt{x+1}}{\sqrt{x}}$, we solve  $y=f(x)$ for $x$ in terms of $y$. \\
	\begin{align*}
	y=f(x) \implies& y= \frac{\sqrt{x+1}}{\sqrt{x}} = \sqrt{\frac{x+1}{x}} \\
	\implies& y^{2}= \frac{x+1}{x} \implies xy^{2}= x+1 \implies xy^{2}-x =1 \\
	\implies& x(y^{2}-1)= 1 \implies x= \frac{1}{y^{2}-1} \\
         &\text{Exchange $x$ and $y$}\implies y= \frac{1}{x^{2}-1} \\
	& \text{Therefore}\,\,  f^{-1}(x)= \frac{1}{x^{2}-1} \\
	\end{align*}
      \end{enumerate}
  \end{solution}

\begin{furtherquestions}
  Find the inverse function of the given function:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle f(x)=2x^3+5$
    \item $\displaystyle f(x)=\frac{3-x}{x+1}$
    \item $\displaystyle f(x)=x^2-9$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}
Note that if you are able to solve your expression uniquely for $x$ in terms of $y$ in the second step it follows that the function is one-to-one since, given any $y$ value in the range $B$ there can only be a single value $x$ in $A$ which maps to it, namely the value which results from evaluating your solved expression with $y$.

\subsection{Graphs of Inverse Functions} \index{inverse function(s)! graphs of}
The definition of the inverse function implies that if $(x,y)$ lies on
the graph of $y=f(x)$ then $(y,x)$ will lie on the graph of $f^{-1}$.
Geometrically this means that the graph of $f^{-1}$ may be obtained by
reflecting the graph of $f$ about the line $y=x$.

\begin{center}
\includegraphics[width=.6\textwidth]{graphics/inverses_and_other_functions3}%
\end{center}

Graphically a discontinuity in $f$ would imply a discontinuity in $f^{-1}$ and vice versa.  We have the following theorem.
\theorem{Suppose $f$ is a one-to-one continuous function defined on an interval then its inverse $f^{-1}$ is also continuous}.

\subsection{Derivative of an Inverse Function}\index{inverse function(s)! derivative of}

If we let $g(x)$ be the inverse of $f$ then our earlier relationship $x=f\left(f^{-1}(x)\right)=f(g(x))$.  Differentiating the left side with respect to $x$ just gives $1$.  Differentiating the right side of the equation with respect to $x$ can be done with the Chain Rule.  Solving for the derivative $g'(a)$ gives the following result.

\theorem{Suppose $f$ is a one-to-one differentiable function with inverse $g=f^{-1}$.  If $f'\left(g(a)\right)\neq 0$ then the inverse function is differentiable at $a$ with
\[g'(a)=\frac{1}{f'\left(g(a)\right)}\]
More generally the derivative of the inverse function is
\[g'(x)=\frac{1}{f'\left(g(x)\right)}\ .\]
}

\begin{example}
  For the function $\displaystyle f(x)=x^2+1,\ x\ge 0$:
  \begin{enumerate}
  \item Show that $f$ is one-to-one.
  \item\label{item:invexample}Calculate $g=f^{-1}$ and find its domain and range.
  \item Calculate $g'(5)$ using your result from part \ref{item:invexample}.
  \item Find $g'(5)$ from the formula $\displaystyle g'(x)=\frac{1}{f'(g(x))}$~.
  \end{enumerate}
  \begin{solution*}
    \begin{enumerate}
    \item Using the definition of one-to-one:
      \begin{align*} 
        f(x_1)= f(x_2) & \implies x_1^2 +1 = x_2^2 + 1 \\
                       & \implies x_1^2-x_2^2=0 \\
                       & \implies (x_1-x_2)(x_1+x_2)= 0 \\
                       & \implies x_1= x_2\ \text{or}\ x_1=-x_2
      \end{align*}
      Since the domain of $f$ is restricted to $x\ge 0$, $x_1=-x_2$ is not possible as one of the numbers would necessarily be negative, except in the case of $x_1=x_2=0$.  We conclude $f$ is one-to-one.
    \item Solve $y=f(x)$ for x:
      \begin{align*}
        y=f(x) \implies& y=x^2+1 \implies x^2= y-1 \\
        \implies& x= \pm\sqrt{y-1}\ \ \  (\leftarrow \text{Reject negative due to domain of }f.)\\
        \implies& x=\sqrt{y-1}\\
                       &\text{Exchange $x$ and $y$}\implies y= \sqrt{x-1} \\
                       & \text{Therefore}\,\,  g(x)=f^{-1}(x)=\sqrt{x-1}
      \end{align*}
      Domain of $g$ requires $x-1\ge0\implies x\ge 1$. So $D_g=[1,\infty)$. Range of $g$ equals the domain of $f$ and so $R_g=[0,\infty)$.
    \item Differentiating using the Chain Rule gives:
      \begin{align*}
        g'(x) & =\frac{1}{2}(x-1)^{-\frac{1}{2}}(1+0)=\frac{1}{2\sqrt{x-1}} \\
        \text{and so } g'(5)&=\frac{1}{2\sqrt{5-1}}=\frac{1}{4}
      \end{align*}
    \item Since $f'(x)=2x$ we have using the inverse derivative formula:
      \begin{align*}
        g'(5)&=\frac{1}{f'(g(5))}=\frac{1}{f'(\sqrt{5-1})}=\frac{1}{f'(2)}=\frac{1}{2(2)}=\frac{1}{4}
      \end{align*}
    \end{enumerate}
  \end{solution*}

\begin{furtherquestions}
For the function $\displaystyle f(x)=\frac{1}{x-1}$:
\begin{enumerate}
\item Show that $f$ is one-to-one.
\item\label{item:inv}Calculate $g=f^{-1}$ and find its domain and range.
\item Calculate $g'(2)$ using your result from part \ref{item:inv}.
\item Find $g'(2)$ from the formula $\displaystyle g'(x)=\frac{1}{f'(g(x))}$~.
\end{enumerate}
\end{furtherquestions}
\end{example}

\begin{example}
  Find $(f^{-1})'(a)$ for the given $a$:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle f(x)= x^3 +4x +3,\, a=-2 $
    \item $\displaystyle f(x)= \frac{x^3}{x^2+4},\, a=-\frac{1}{5}$
    \end{enumerate}
  \end{multicols}
  \begin{solution*}
    \begin{enumerate}
    \item $\displaystyle f'(x)= 3x^{2} + 4 > 0$ on $(-\infty, \infty) \implies f(x)$ has an inverse 
      \begin{align*} 
        (f^{-1})'(-2) &= \frac{1}{f'(f^{-1}(2))}
      \end{align*}
      We need to find $f^{-1}(-2)$.  To do so, call the unknown value $w$, use the properties of the inverse to get an equation for it, and then solve for $w$ as follows:
      \begin{align*}
        f^{-1}(-2) &= w \implies f(w)=-2 \implies w^3 + 4w +3 = -2 \\
                   &\implies w= -1 \,\, \text{(By inspection or direct solution.)}   \\
        f'(f^{-1}(-2)) &= f'(-1)= 3(-1)^{2} +4 =3 +4 = 7 \\
        \text{Therefore}\,\, (f^{-1})'(-2)&= \frac{1}{f'(f^{-1}(-2))}= \frac{1}{7}
      \end{align*}
    \item $\displaystyle f'(x)= \frac{3x^{2}(x^{2}+4)-x^{3}(2x)}{(x^{2}+4)^{2}} = \frac{x^{4}+12x^{2}}{(x^{2}+4)^{2}}$ 
      \begin{align*}
        \left(f^{-1}\right)'\left(-\frac{1}{5}\right) &= \frac{1}{f'(f^{-1}(-\frac{1}{5}))} \\
  	f^{-1}\left(-\frac{1}{5}\right) &= w \implies f(w)=-\frac{1}{5} 
        \implies \frac{w^{3}}{w^{2}+4} = -\frac{1}{5}\\
    &\implies 5w^3+w^2-4=0 \implies w = -1 \\
  	f'(-1) &= \frac{(-1)^{4}+ 12(-1)^{2}}{((-1)^{2} +4)^{2}} = \frac{1+ 12}{5^{2}} = \frac{13}{25} \\
  		\text{Therefore}\,\, (f^{-1})'\left(-\frac{1}{5}\right) &= \frac{1}{f'(f^{-1}(-\frac{1}{5}))}= \frac{1}{f'(-1)} = \frac{1}{\frac{13}{25}} = \frac{25}{13} \\
  \end{align*}
  \end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Find the following derivatives:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle \left(f^{-1}\right)'(1)$ if $f(x)=x^3+x+1$~.
    \item $g'(-1)$ if $\displaystyle f(x)=3x-\cos x$ and $g=f^{-1}$~.
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Creating Invertible Functions} \index{invertible functions}
So far one-to-one (and hence invertible) functions seem uncommon.  However this is only because we only considered functions defined on their natural domains, i.e. the set of numbers for which the function may be evaluated.  We can choose to define a function with a smaller domain and by suitable restriction we can create a function that is one-to-one and hence invertible.

\begin{example}
  Define the function $f(x)$ to have the value $f(x)=x^2$ but only be defined on the domain $A=[0,\infty)$.  Since $f$ is increasing everywhere on this interval it is one-to-one and hence has an inverse, $f^{-1}(x)=x^{\frac{1}{2}}=\sqrt{x}$. If we restricted the domain to be $A=(-\infty,0]$ the inverse would be $f^{-1}(x)=-\sqrt{|x|}$~!
\end{example}

\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/inverse-functions.tex}
\end{exercise}

\section{Exponential Functions}\index{exponential function(s)}

If we write the number $2^5$, then this, recall, means $2\times2\times2\times2\times2$.  We call $2$ the \emph{base} and $5$ the \emph{exponent}.  We have already seen that one way to create a function is to replace the base with a variable.  This produces \emph{power functions} \index{power function(s)} like

\hfill$f(x)=x^2$\hfill$y=x^{\frac{1}{2}}=\sqrt{x}$\hfill$y=x^{-1}=\frac{1}{x}$\hfill~

In general, a power function is of the form $y=x^r$ where $r$ is any real constant.

If, on the other hand, we let the exponent be a variable and the base a constant, like:
\[f(x)=2^x,\ y=\left(1/2\right)^x\]
we get \emph{exponential functions}.   
\definition{Let $a>0$.  The function
\[f(x)=a^x\]
is an exponential function.}

The graph of an exponential function has the following form depending on whether $a$ is greater than or less than 1.  Two typical values of $a$ are shown.
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/inverses_and_other_functions4}%
\end{center}

Notes: \index{exponential function(s)! properties of}
\begin{enumerate}
\item If $x=0$ then $a^x=a^0=1$.  Therefore all exponential functions go through the point $(0,1)$~.
\item If $x=n$, a positive integer then $a^x=a^n=\underbrace{a\cdot a\cdot\ldots\cdot a}_{\mbox{$n$ times}}$~.
\item If $x=-n$, $n$ a positive integer, then $a^x=a^{-n}=\frac{1}{a^n}$~.
\item If $x=\frac{1}{n}$, $n$ a positive integer, then $a^x=a^\frac{1}{n}=\sqrt[n]{a}$~.  (Hence $a<0$ is excluded.)
\item If $x$ is rational, $x=\frac{p}{q}$, then $a^x=a^\frac{p}{q}=\left(a^p\right)^{\frac{1}{q}}=\sqrt[q]{a^p}$~.
\item If $a\neq 1$ (and $a>0$) then $f(x)=a^x$ is a continuous function with domain \mathset{R} and range $(0,\infty)$.
\item If $0<a<1$ then $f(x)=a^x$ is a decreasing function.
\item If $a>1$ then $f(x)=a^x$ is an increasing function.
\item If $a,b>0$ and $x,y\in\mathset{R}$ then
\begin{inparaenum}
\item $a^xa^y=a^{x+y}$
\item $\frac{a^x}{a^y}=a^{x-y}$
\item $(a^x)^y=a^{xy}$
\item $(ab)^x=a^xb^x$~.
\end{inparaenum}
These relations are readily apparent when one considers $x$ and $y$ as positive integers.
\end{enumerate}

For two bases greater than one the base which is larger is the steeper curve while for two bases less than one the base which is smaller is steeper.
\begin{center}
\includegraphics[width=.5\textwidth]{graphics/inverses_and_other_functions5}%
\end{center}

As depicted in the previous graphs, we have the following limits:
\theorem{For exponential functions we have the following limits at infinity \index{exponential function(s)! limits at infinity of}:
If $a>1$, then $\displaystyle\mylim{x}{-\infty}a^x=0$ and $\displaystyle\mylim{x}{\infty}a^x=\infty$.\newline
If $0<a<1$, then $\displaystyle\mylim{x}{-\infty}a^x=\infty$ and $\displaystyle\mylim{x}{\infty}a^x=0$.\newline}

So the $x$-axis is a horizontal asymptote for $a^x$ provided $a>0$, $a\neq1$.

\subsection{The Natural Exponential Function}
Consider the derivative of $f(x)=a^x$:
\[f'(x)=\mylim{h}{0}\frac{a^{x+h}-a^{x}}{h}=\mylim{h}{0}\frac{a^{x}a^{h}-a^{x}}{h}=
\left(\mylim{h}{0}\frac{a^{h}-1}{h}\right)a^x\]
where here we are able to pull out the $a^x$ from the limit because $a^x$ does not involve the limit variable $h$.  The result shows the derivative is proportional to the function $f(x)=a^x$ itself with constant of proportionality $c$ given by the evaluation of the limit:
\[c=\mylim{h}{0}\frac{a^{h}-1}{h}\]
Due to the presence of the constant $a$ in the limit, one anticipates correctly that the constant $c$ depends on the choice of base $a$.  Interestingly, one can ask the question if there is some choice of base $a$ for which the constant is $c=1$. The answer is yes, the base is given by \emph{Euler's Number} \index{Euler's Number}:
\[e=2.71828\ldots\]
for which we have that $c=1$ in the above limit:
\[\mylim{h}{0}\frac{e^{h}-1}{h}=1\ .\]
More constructively, as opposed to $e$ being the solution of such a limit equation, it will be shown that $e$ may be written as the following limit:
\[e=\mylim{h}{0}\left(1+h\right)^{\frac{1}{h}}\ ,\]
or, setting $h=1/n$,
\[e=\mylim{n}{\infty}\left(1+\frac{1}{n}\right)^{n}\ .\]

\definition{If $a=e=2.71828\ldots$, then $f(x)=e^x$ is the \emph{natural exponential function} \index{natural exponential function}    .}

Since $e=2.71\ldots>1$ the natural exponential function shares all the aforementioned properties of $f(x)=a^x$ where $a>1$. (i.e. continuous, increasing function with domain \mathset{R}, range $(0,\infty)$, limits, etc.)
\begin{center}
\includegraphics{graphics/inverses_and_other_functions6}%
\end{center}

You should identify the natural exponential key $\boxed{e^x}$ on your calculator.
\textbook{\clearpage}
\subsection{Derivative of $e^x$}\index{natural exponential function!derivative of}
Furthermore from the preceding discussion we have the important result:
\theorem{The derivative of the natural exponential function is:
\[\displaystyle\frac{d}{dx}e^x=e^x\ .\]}

Proof is, as above,
\[\frac{d}{dx}e^x=\mylim{h}{0}\frac{e^{x+h}-e^{x}}{h}=\mylim{h}{0}\frac{e^{x}e^{h}-e^{x}}{h}=\left(\mylim{h}{0}\frac{e^{h}-1}{h}\right)e^x=(1)e^x=e^x\ .\]

A corollary of this theorem, applying the Chain Rule to the function $e^u$ with $u=g(x)$ is:
\theorem{\ \ \ \ $\displaystyle \frac{d}{dx}e^u=e^u\frac{du}{dx}$ \ \ \ or\ \ \ $\displaystyle\frac{d}{dx}\left[e^{g(x)}\right]=e^{g(x)}g'(x)$}
\begin{example}
Differentiate the following functions:
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle f(x)= e^{x^{2} +5}$ 
\item $\displaystyle y= \cos\left(e^{2x}\right) $ 
\item $\displaystyle f(t)= \left(t^{2}+5\right) e^{t^{3}}$ 
\item $ \displaystyle 2xy^{2} + xe^{y} = e^{3x} $ 
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle f(x)= e^{x^{2} +5}$\\
  Using the Chain Rule:
\begin{align*} 
  f'(x) &=e^{x^{2}+5}\left(x^{2}+5\right)'
  =e^{x^{2}+5}\left(2x+0\right)\\
&= 2x\,e^{x^{2}+5} 
\end{align*}
\item $\displaystyle y= \cos\left(e^{2x}\right) $  
\begin{align*}
  y'&= -\sin \left(e^{2x}\right)(e^{2x})'= -\sin \left(e^{2x}\right)e^{2x}(2)\\
    &= -2e^{2x}\sin\left(e^{2x}\right) 
\end{align*}
\item $\displaystyle f(t)= \left(t^{2}+5\right) e^{t^{3}}$\\
  Using Product and Chain Rules:
\begin{align*}
f'(t)&= 2te^{t^3}+ \left(t^{2}+5\right)e^{t^3}(3t^2) \\
&= 2te^{t^3}+ \left(3t^{4}+15t^2\right)e^{t^3}
\end{align*}
\item $ \displaystyle 2xy^{2} + xe^{y} = e^{3x}$ \\ 
Using implicit differentiation we have:
\begin{align*}
2y^{2} +4xyy' + e^{y} + xe^{y}\,y' &= 3e^{3x} \\
\implies \left(4xy + xe^{y}\right)y'&= 3e^{3x}-2y^{2}-e^{y} \\
\implies y'&= \frac{3e^{3x}-2y^{2}-e^{y}}{4xy+xe^{y}} 
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Find $\frac{dy}{dx}$ for the following:
\begin{multicols}{2}
\begin{enumerate}
\item$\displaystyle y=e^{2x}$
\item$\displaystyle y=e^{x^3+x}$
\item$\displaystyle y=e^{\sec x}+\sec\left(e^x\right)$
\item$\displaystyle y=e^{x^4}\sin\left(x^2+1\right)$
\item$\displaystyle xy+e^x=2xy^2$
\end{enumerate}
~%ensures ragged bottom of second column
\end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Integral of $e^x$} \index{exponential function(s)!integral of}
Since $\frac{d}{dx} e^x=e^x$, we have the following:

\theorem{The indefinite integral of $e^x$ is
\[\int e^x\,dx=e^x+C\]}

\begin{example}
Evaluate the following integrals:
 \begin{multicols}{2}
 \begin{enumerate}
 \item $\displaystyle\int e^{6x}\,dx $ \\
 \item $\displaystyle\int e^x\cos\left(e^x\right)\,dx $ \\
 \item $\displaystyle\int_0^1xe^{x^2+1} \,dx $ \\
 \item $\displaystyle \int\frac{e^{4x}}{\sqrt{2+e^{4x}}}\,dx$ \\
 \end{enumerate}
 \end{multicols}
 \begin{solution*}
 \begin{enumerate}
 \item  Using substitution $u=6x$ so $du= 6\,dx \implies\frac{1}{6}du=dx$ one has:
 \begin{align*}
 \int e^{6x}\, dx&= \int \frac{1}{6}e^u\,du= \frac{1}{6}e^u+C= \frac{1}{6}e^{6x} +C 
 \end{align*}
 \item Using substitution $u=e^x$ so $du=e^x\,dx$:
 \begin{align*} 
 \int e^x\cos (e^x)\, dx= \int \cos u \, du =\sin u + C 
 =\sin \left(e^x\right) + C 
 \end{align*}
 \item Let $u=x^{2}+1$ so $du=2x\,dx \implies\frac{1}{2}du=x\,dx$ and change limits:
 \begin{align*}
   &x=0 \implies u=1 \\
   &x=1 \implies u=2 \\
 \int_{0}^{1} xe^{x^{2}+1}\,dx &=\int_1^2e^{u}\frac{1}{2}\,du=\left.\frac{1}{2}e^u\right|_1^2 %\\ 
 =\frac{1}{2}\left(e^2-e^1\right)=\frac{1}{2}\left(e^2-e\right) 
 \end{align*}
 \item Let $u=2+e^{4x}$ so $du=4e^{4x}\,dx\implies \frac{1}{4}du=e^{4x}\,dx$:
 \begin{align*}
 \int\frac{e^{4x}}{2+e^{4x}}\,dx &= \frac{1}{4}\int\frac{1}{\sqrt{u}}\,du
 =\frac{1}{4}\int u^{-1/2}\,du
 =\frac{1}{4}\left(2\right)u^{1/2}+ C \\
 &=\frac{1}{2}\sqrt{2+e^{4x}} + C 
 \end{align*}
 \end{enumerate}
\end{solution*}

\begin{furtherquestions}
Evaluate the following integrals:
\begin{multicols}{2}
  \begin{enumerate}
\item $\displaystyle \int e^{2x}\,dx$
\item $\displaystyle \int x^3 e^{x^4+1}\,dx$
\item $\displaystyle \int_0^1 e^{-x}\,dx$
\item $\displaystyle \int \frac{2e^x}{\left(1+e^x\right)^2}\,dx$
\item $\displaystyle \int \frac{2+3e^x}{e^x}\,dx$
\item $\displaystyle \int \frac{1-4e^{3x}}{e}\,dx$
\end{enumerate}
\end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Simplifying Exponential Expressions} \index{exponential function(s)! simplifying}
Using the rules of exponents we are often able to consolidate expressions involving several exponents into an expression involving one exponent.
\begin{example}
The expression $\displaystyle \frac{e^2\sqrt{e^x}}{\left(2e^x\right)^3}$ may be simplified as follows:
\begin{eqnarray*}
\frac{e^2\sqrt{e^x}}{\left(2e^x\right)^3} & = & \frac{e^2\left(e^x\right)^{\frac{1}{2}}}{2^3\left(e^x\right)^3} \ \ \left(\ \mbox{since}\ \sqrt[n]{a}=a^{\frac{1}{n}},\ (ab)^x=a^xb^x\ \right) \\
 & = & \frac{e^2e^{\frac{1}{2}x}}{2^3e^{3x}} \ \ \left(\ \mbox{since}\ (a^x)^y=a^{xy}\ \right) \\
& = & \frac{e^{2+\frac{1}{2}x}}{8e^{3x}} \ \ \left(\ \mbox{since}\ a^xa^y=a^{x+y}\ \right) \\
& = & \frac{1}{8}e^{2+\frac{1}{2}x-3x} \ \ \left(\ \mbox{since}\ \frac{a^x}{a^y}=a^{x-y}\ \right) \\
& = & \frac{1}{8}e^{2-\frac{5}{2}x}
\end{eqnarray*}
\end{example}
The usefulness in consolidating exponents in this manner is clear when solving equations.
\begin{example}
Solving the equation
\[\frac{e^2\sqrt{e^x}}{\left(2e^x\right)^3}=\frac{1}{2}\]
Is equivalent, by using our previous result and multiplying both sides by 8, to
\[e^{2-\frac{5}{2}x}=4\]
Now if we could apply an \emph{inverse} to the natural exponential function on both sides we could solve for $x$.
\end{example}

\textbook{\clearpage}
\textbook{\addtoanswerkey{\clearpage}}
\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/exponential-functions.tex}
\end{exercise}

\section{Logarithmic Functions}\index{logarithmic function(s)}

We finished the last section by suggesting that inverses of exponential functions would be useful for, among other things, solving equations involving exponentials.  Since the exponential function $f(x)=a^x$ with constant $a>0$ and $a\neq 1$ is either everywhere decreasing ($0<a<1$) or increasing ($1<a$) on open interval $\mathset{R}=(-\infty,\infty$), the exponential function is one-to-one and hence has an inverse function~$f^{-1}$.

%Aside: The \log_a x command can't be put in bold using \bm{ } without
%generating an error?
\definition{Given constant $a>0$, $a\neq 0$, the \textbf{logarithmic function of base $\bm a$}, \index{logarithmic function(s)! of base $\emph a$}    written $\log_a x$ is defined by
\[\log_a x=y\iff a^y=x\]
That is, it is the inverse of the exponential function $f(x)=a^x$.}

In words, the logarithm of a value $x$ to a base $a$ is the \emph{exponent} to which you must take $a$ to get $x$.

For the case $\boxed{a>1}$, which, as you recall, is the case for $a=e=2.71\ldots$, a representative graph of $y=a^x$ and its inverse $y=\log_a x$ are as follows:
\begin{center}
\includegraphics{graphics/inverses_and_other_functions7}%
\end{center}

For base $a>1$ we saw that larger values of $a$ led to steeper $y=a^x$ curves, it follows that larger values of $a$ will make the logarithmic curves more horizontal in this case:
\begin{center}
\includegraphics{graphics/inverses_and_other_functions8}%
\end{center}

\subsection{Logarithmic Function Properties}\index{logarithmic function(s)!properties of}
Because of their relationship to exponentials as inverses the following are true for logarithmic functions:
\begin{enumerate}
\item $y=\log_a x$ has domain $(0,\infty)$ and range \mathset{R}.
\item $y=\log_a x$ is continuous on its domain. 
\item $y=f(x)=\log_a x$ is one-to-one with inverse function $f^{-1}(x)=a^x$.
\item $\log_a(1)=0$
\item The following limits hold (see graph for $a>1$ case):
\begin{itemize}
\item If $0<a<1$ then $f(x)=\log_a(x)$ is a decreasing function with\

\hfill $\displaystyle\lim_{x\rightarrow0^+}\log_ax=+\infty$\hfill $\displaystyle\lim_{x\rightarrow\infty}\log_a x=-\infty$ \hfill~ 
\item If $a>1$, then $f(x)=\log_a(x)$ is an increasing function with

\hfill $\displaystyle\lim_{x\rightarrow0^+}\log_ax=-\infty$\hfill $\displaystyle\lim_{x\rightarrow\infty}\log_a x=\infty$ \hfill~
\end{itemize} 
Note the $y$-axis is a vertical asymptote in either case.
\item The following inverse relations hold:
\begin{eqnarray*}
\log_a\left(a^x\right) & = & x\ \ \mbox{for any $x$ in \mathset{R}} \\
a^{\log_a x} & = & x\ \ \mbox{for any $x>0$}
\end{eqnarray*}
\end{enumerate}

The special multiplication, division, and power laws of exponents induce the following important logarithmic results.
\theorem{For $x>0$ and $y>0$ and any real number $r$ the following hold:
\begin{enumerate}
\item $\displaystyle\log_a(xy)=\log_ax+\log_ay$
\item $\displaystyle\log_a\left(\frac{x}{y}\right)=\log_ax-\log_ay$
\item $\displaystyle\log_a\left(x^r\right)=r\log_ax$
\end{enumerate}
}
To prove the theorem note that if $x>0$ and $y>0$ then $m=\log_ax$ and $n=\log_ay$ exist and, exponentiating both sides, it follows that $x=a^m$ and $y=a^n$.  Evaluating the first equation's left hand side we have:
\[\log_a(xy)=\log_a\left(a^ma^n\right)=\log_a\left(a^{m+n}\right)=m+n=\log_a x+\log_ay\]
The other conclusions are similarly proven.

These results can be used to simplify logarithmic expressions:\index{logarithmic function(s)!simplifying}
\begin{example}
  Simplify the following:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\log_{5}50+\log_{5}2-\log_{5}4 $
    \item $\displaystyle 5\log\left(10^{4}\right)-6\log\left(\sqrt{10}\right) +2\log\left(10^{3/2}\right) $
    \end{enumerate}
  \end{multicols}
  \textbook{\clearpage}
 \begin{solution*}
  \begin{enumerate}
 \item $\displaystyle\log_{5}50+\log_{5}2-\log_{5}4  =\log_{5}\left(\frac{(50)(2)}{4}\right) =\log_{5}\left(25\right)=\log_{5}\left(5^2\right)=2$
 \item Recalling that by convention $\log$ means $\log_{10}$ we have:
   \begin{align*}
 5\log\left(10^{4}\right)-6\log\left(\sqrt{10}\right) +2\log\left(10^{3/2}\right) &=(5)(4)\log (10)-6\frac{1}{2}\log 10+ 2\frac{3}{2}\log 10 \\
 &=20(1)-3(1)+3(1)= 20-3+3=20
 \end{align*}
 \end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Simplify the following:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle \log_2 4+\log_2 10 - \log_2 5$
    \item $\displaystyle \log_5 3 +\log_5 3^4 + \log_5 1$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{The Natural Logarithmic Function}\index{natural logarithmic function}
\definition{The logarithmic function with base $a$ equal to $e=2.71\ldots$ is called the \textbf{natural logarithmic function} and is denoted by $\bm{\ln x}$.  In symbols:
\[\ln x=\log_e x\]
}

All the properties for a logarithm with base $a>1$ apply to the natural logarithm.  In terms of the notation for natural logarithms and exponentials we have the definition:
\[\ln x=y \iff e^y=x\]
and the properties \index{natural logarithmic function! properties of} \index{natural exponential function! properties of}:
\begin{eqnarray*}
\ln e & = & 1 \\
\ln e^x & = & x\ \ \ (x\in\mathset{R}) \\
e^{\ln x} & = & x\ \ \ (x>0) \\
\ln(xy) & = & \ln x+\ln y \ \ \ (x,y>0) \\
\ln\left(\frac{x}{y}\right) & = & \ln x-\ln y \ \ \ (x,y>0)\\
\ln\left(x^r\right) & = & r\ln x \ \ \ (x>0, r\in\mathset{R})
\end{eqnarray*}

Note the following:
\[\log_a(x+y)\neq\log_a x+\log_a y\]
\[\log_a(x-y)\neq\log_a x-\log_a y\]
In the specific case of natural logarithms $(a=e)$:
\[\ln(x+y)\neq\ln x+\ln y\]
\[\ln(x-y)\neq\ln x-\ln y\]

You should identify the natural logarithm key $\boxed{\ln}$ on your calculator.  Note that the key $\boxed{\log}$ on the calculator means base 10 logarithm $\log_{10}x$.\footnote{However in other areas (some computer applications) the symbol $\log$ will often refer to a natural logarithm so one needs to be careful.}

\begin{example}
  Simplify the following:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\ln\left(\sqrt{e}\right)+2\ln(3e)-\ln\left(e^5\right)$
    \item $\displaystyle \ln\left(\frac{1}{e^{3/2}}\right)$
    \end{enumerate}
  \end{multicols}
 \begin{solution*}
  \begin{enumerate}
 \item $\displaystyle\ln\left(\sqrt{e}\right)+2\ln(3e)-\ln\left(e^5\right)=\ln(e^{\frac{1}{2}})+2(\ln 3+\ln e)-5=\frac{1}{2}+2(\ln 3+1)-5=-\frac{5}{2}+2\ln 3$
 \item $\displaystyle \ln\left(\frac{1}{e^{3/2}}\right)
   =\ln 1-\ln\left(e^{3/2}\right) =0-\frac{3}{2}\ln e =-\frac{3}{2}(1) =-\frac{3}{2} $
 \end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Simplify the following:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle \ln 5+2\ln 3+\ln 1$
    \item $\displaystyle \frac{1}{2}\ln(4t)-\ln(t^2+1)$
    \item $\displaystyle e^{\ln\left(x^2+1\right)}+3x^2-5$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Solving Exponential and Logarithmic Equations}
Solving equations involving logarithmic or exponential functions typically involves using properties of these functions to simplify those expressions involving the variable and then applying the appropriate inverse function to undo the exponential or logarithm.  Finally one may solve for the variable.\footnote{More complicated equations may allow themselves to be written as a product of factors equal to zero:
\[(\mbox{factor}_1)(\mbox{factor}_2)\cdot\ldots\cdot(\mbox{factor}_n)=0\]
where the factors themselves involve logarithms or exponentials.  Note
that a strictly exponential factor equalling zero will provide no
solution as $a^x\neq0$ for all $x$.  A strictly logarithmic factor
equalling zero will be equivalent to the argument of the logarithm
equalling 1.}

\begin{example}
We saw that the equation 
\[\frac{e^2\sqrt{e^x}}{\left(2e^x\right)^3}=\frac{1}{2}\]
could be written, using properties of exponentials, as
\[e^{2-\frac{5}{2}x}=4\ .\]
Applying $\ln$, the inverse of the exponential $e^x$, to both sides of the equation, gives
\[2-\frac{5}{2}x=\ln 4\ .\]
Solving for $x$ gives
\[x=\frac{2}{5}\left(2-\ln4\right)\ .\]
\end{example}

\begin{example}
  Solve the following equations for $x$.
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle e^{2x+5}=3$ 
    \item $\displaystyle \ln\left(\ln x\right)=2 $
    \item $\displaystyle \ln\left(x-2\right)= 2+\ln\left(x-3\right)$ 
    \end{enumerate}
  \end{multicols}
 \begin{solution*}
 \begin{enumerate}
 \item $\displaystyle e^{2x+5}=3$
   
 Take the natural logarithm of both sides.
 \begin{align*}
\ln\left(e^{2x+5}\right)&=\ln 3 \\
&\implies 2x+5=\ln 3 \\
&\implies 2x= \ln 3 -5 \\ 
&\implies x=\frac{1}{2}(\ln 3-5)
 \end{align*}
\item $\displaystyle \ln\left(\ln x\right)=2 $
  
 Exponentiate both sides.
 \begin{align*}
   &e^{\ln(\ln x)} =e^2 \implies \ln x=e^2
 \end{align*}
 Exponentiate both sides again.
 \begin{align*}
 &e^{\ln x} = e^{(e^2)} \implies x=e^{(e^2)} & 
 \end{align*}
 Note that by convention we can write $e^{(e^2)}=e^{e^2}$ (without parentheses).  This is because $(e^e)^2$ would just be written $e^{2e}$ using our rules of exponents.  In general, remember parentheses when you ladder exponents if you do not remember the convention.
\item $\displaystyle \ln\left(x-2\right)= 2+\ln\left(x-3\right)$
  
  First simplify the expression.
  \begin{align*}
    \ln(x-2)-\ln(x-3)&=2 \\ 
    \ln\left(\frac{x-2}{x-3}\right)&=2
  \end{align*}
  Next exponentiate both sides.
  \begin{align*}
    e^{\ln\left(\frac{x-2}{x-3}\right)} = e^2 & \implies\frac{x-2}{x-3}= e^2\implies x-2=e^{2}x-3e^{2} \\
    &\implies x-e^{2}x= 2-3e^{2} \implies x(1-e^2)= 2-3e^2 \\
    &\implies x=\frac{2-3e^2}{1-e^2} 
  \end{align*}
\end{enumerate}
\end{solution*}
\textbook{\clearpage}
\begin{furtherquestions}
Solve the following equations for $x$:
\begin{multicols}{2}
  \begin{enumerate}
  \item $\displaystyle 5e^{x-3}=4$
  \item $\displaystyle \ln\left(x^2-3\right)=0$
  \item $\displaystyle 4e^xe^{-2x}=6$
  \item $\displaystyle \ln(2\ln x-5)=0$
  \item $\displaystyle e^{x^2-5x+6}=1$
  \item $\displaystyle 3e^{2x-4}=10$
  \item $\displaystyle \ln\left(\frac{x-2}{x-1}\right)=1+\ln\left(\frac{x-3}{x-1}\right)$
  \end{enumerate}
  ~%Ensures ragged bottom of second column.
\end{multicols}
\end{furtherquestions}
\end{example}

The following relates logarithms in other bases to the natural logarithm \index{logarithmic function(s)! converting to the natural logarithm}.
\theorem{For $a>0$, $a\neq 1$ we have:
\[\log_ax=\frac{\ln x}{\ln a}\]}

Proof comes from observing that since $a^{\log_a x}=x$ we can take the natural logarithm of both sides and then use the power rule for the natural logarithm to get
\[(\log_a x)(\ln a)=\ln x\ .\]
Solving for $\log_a x$ gives our result.

This theorem is useful for evaluating an arbitrary base $a$ logarithm on a calculator.
\begin{example}
Write in terms of the natural logarithm ($\ln$):
\begin{multicols}{3}
  \begin{enumerate}
  \item $\displaystyle \log_{12}5$
  \item $\displaystyle \log_{2}(e^{5x})$
  \item $\displaystyle \log_{x}(x^2+2x+1)$
  \end{enumerate}
\end{multicols}
\begin{solution*}
  \begin{enumerate}
  \item $\displaystyle \log_{12}5=\frac{\ln 5}{\ln 12}$
  \item $\displaystyle \log_{2}(e^{5x})=\frac{\ln(e^{5x})}{\ln 2}=\frac{5x}{\ln 2}$
  \item $\displaystyle \log_{x}(x^2+2x+1)=\frac{\ln(x^2+2x+1)}{\ln x}=\frac{\ln\left((x+1)^2\right)}{\ln x}=\frac{2\ln(x+1)}{\ln x}$
  \end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Write in terms of the natural logarithm ($\ln$):
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle \log_5 7$
    \item $\displaystyle \log_{20}\left(x^2+1\right)$
    \item $\displaystyle \log_{10}\left(e^{2x}\right)$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Derivative of the Natural Logarithmic Function}\index{natural logarithmic function!derivative of}
\theorem{The derivative of the natural logarithmic function is
\[\frac{d}{dx}\left(\ln x\right)=\frac{1}{x}\]
}

To prove the theorem note that if $y=\ln x$ then by definition of the
logarithm as an inverse we have
\[e^y=x\ .\]
Differentiating this implicit equation with respect to $x$ on both sides gives
\[e^yy'=1\ ,\]
and so
\[y'=\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}\ .\]

A corollary of this result, applying the Chain Rule to the function $\ln u$ with $u=g(x)$ is
\theorem{\ \ \ \ $\displaystyle\frac{d}{dx}\ln u=\frac{1}{u}\frac{du}{dx}$\ \ \ \ or\ \ \ \  $\displaystyle\frac{d}{dx}\ln\left[g(x)\right]=\frac{g'(x)}{g(x)}$} 

\begin{example}
  Find the derivative of the given functions.
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle y=\ln(x^{3}+e^{x})$
    \item $\displaystyle f(x)=\ln(\tan x) $ 
    \item $\displaystyle f(t)=e^{2t+\ln t}$
    \end{enumerate}
  \end{multicols}
\begin{solution}
  In each of the following we use the Chain Rule.
\begin{enumerate}
\item $\displaystyle y=\ln(x^{3}+e^{x})$
\begin{align*}
  y'=\frac{1}{x^{3}+e^x}\left(x^{3}+e^x\right)'
  =\frac{3x^{2}+e^x}{x^{3}+e^x}
\end{align*}
\item $\displaystyle f(x)=\ln(\tan x) $ 
\begin{align*}
f'(x)=\frac{(\tan x)'}{\tan x}=\frac{\sec ^{2}x}{\tan x}
\end{align*}
\item $\displaystyle f(t)=e^{2t+\ln t}$
\begin{align*}
f'(t)=e^{2t+\ln t}\left(2t+\ln t\right)'=\left(2+\frac{1}{t}\right)e^{2t+\ln t}
\end{align*}
\end{enumerate}
\end{solution}

\begin{furtherquestions}
  Differentiate the following functions:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle y=\ln\left(x^2-3x+1\right)$
    \item $\displaystyle y=\ln\left(x+\ln x\right)$
    \item $\displaystyle y=\ln\left(\frac{x+1}{\sqrt{x+2}}\right)$
    \item $\displaystyle y=e^{\left(2+x\ln x\right)}$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Derivatives Using Arbitrary Bases}\index{logarithmic function(s)! derivatives using arbitrary bases}
\theorem{The derivative of the logarithm function to base $a>0$ ($a\neq 1$) is

\hfill$\displaystyle\frac{d}{dx}\left(\log_ax\right)=\frac{1}{x\ln a}$\hfill
$\displaystyle \frac{d}{dx}\left[\log_ag(x)\right]=\frac{g'(x)}{g(x)\ln a}$\hfill~}

Proof of the former derivative follows from the identity $\log_ax=\frac{\ln x}{\ln a}$:
\[\frac{d}{dx}\left(\log_a x\right)=\frac{d}{dx}\left(\frac{\ln x}{\ln a}\right)=\frac{d}{dx}\left(\frac{1}{\ln a}\cdot\ln x\right)=\frac{1}{\ln a}\cdot\frac{d}{dx}\left(\ln x\right)=\frac{1}{\ln a}\cdot\frac{1}{x}=\frac{1}{x\ln a}\]
Here note that we used that $\frac{1}{\ln a}$ is constant since $a$ is constant.  The latter derivative in the theorem follows from the Chain Rule applied to this former result.

\theorem{The derivative of an exponential function \index{exponential function(s)! derivative of} with base $a>0$, $a\neq 1$ is

\hfill$\displaystyle\frac{d}{dx}a^x=a^x\ln a$\hfill$\displaystyle\frac{d}{dx}\left[a^{g(x)}\right]=a^{g(x)}g'(x)\ln a$\hfill~} 

Proof of the former derivative follows by the observation that by our inverse identies the base $a$ may be written $a=e^{\ln a}$ and using the Chain Rule:
\[\frac{d}{dx}\left(a^x\right)=\frac{d}{dx}\left(e^{\ln a}\right)^x=\frac{d}{dx}e^{x\ln a}=e^{x\ln a}\frac{d}{dx}\left(x\ln a\right)=e^{x\ln a}\left(\ln a\right)=\left(e^{\ln a}\right)^x(\ln a)=a^x\ln a\]
Once again the latter derivative given in the theorem is just the result arising from using the Chain Rule with the former result.

\begin{example}
  Differentiate the following functions.
    \begin{enumerate}
    \item $\displaystyle y=5^{x^{2}-3x}$
    \item $\displaystyle f(\theta)= \log_3(\sin \theta) $ 
    \item $\displaystyle f(x)=\log_{5}(x^{2}+3x) + 10^{\ln x} $ 
    \end{enumerate}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle y=5^{x^{2}-3x}$
\begin{align*}
y'&=5^{x^{2}-3x}(\ln{5})(x^{2}-3x)' \\
&=(2x-3)\ln 5\cdot 5^{x^{2}-3x} 
\end{align*}
\item $\displaystyle f(\theta)= \log_3(\sin \theta) $ 
\begin{align*}
f'(\theta)= \frac{1}{\sin(\theta)\ln 3}\cdot(\sin\theta)'= \frac{\cos \theta}{(\ln 3)\sin \theta}
\end{align*}
\item $\displaystyle f(x)=\log_{5}(x^{2}+3x) + 10^{\ln x} $\\
  We differentiate term by term to get:
\begin{align*}
f'(x)&=\frac{1}{(x^{2}+3x)\ln 5}\cdot(x^2+3x)' + \cdot 10^{\ln x}\ln(10)(\ln x)'\\
&=\frac{2x+3}{x^{2}+3x}\cdot\frac{1}{\ln 5} + \frac{1}{x}\ln 10\cdot 10^{\ln x}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Differentiate the following functions:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle y=\log_{10}\left(3x^2+e^x\right)$
    \item $\displaystyle y=5^{2e^x+3x}$
    \item $\displaystyle y=a^{3x}\log_4x$
    \item $\displaystyle y=4^{\cos x}$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Logarithmic Differentiation}\index{logarithmic differentiation}
Using the properties of logarithms makes taking derivatives of logarithms of products, quotients, and powers easy.
\begin{example}
To differentiate $\displaystyle y=\ln[x(x^2+1)(x-3)]$ is easily done if we expand the logarithm first and then differentiate:
\begin{eqnarray*}
\frac{dy}{dx} & = & \frac{d}{dx}\ln[x(x^2+1)(x-3)] \\
& = & \frac{d}{dx}\left[\ln x+\ln(x^2+1)+\ln(x-3)\right] \\
& = & \frac{1}{x}+\frac{2x}{x^2+1}+\frac{1}{x-3}
\end{eqnarray*}
\end{example}
Wouldn't it be nice if when working with products, etc., we were always differentiating their logarithm?  In \emph{logarithmic differentiation} we take the \emph{logarithm of both sides} of an equation before differentiating. 
\begin{example}
To differentiate $y=x\left(2x^3+1\right)^3(x+5)^{\frac{1}{2}}\left(x^2+3x-1\right)^{\frac{1}{3}}$ one could use the (generalized) Product Rule.  Instead, try taking the logarithm of both sides of the equation to get:
\[\ln y=\ln x+3\ln\left(2x^3+1\right)+\frac{1}{2}\ln(x+5)+\frac{1}{3}\ln\left(x^2+3x-1\right)\]
Next differentiate both sides of the equation with respect to $x$ to get:
\[\frac{1}{y}y'=\frac{1}{x}+3\frac{6x^2}{2x^3+1}+\frac{1}{2}\frac{1}{x+5}+\frac{1}{3}\frac{2x+3}{x^2+3x-1}\]
Multiplying both sides by $y$ and substituting in its value gives our derivative:
\[y'=\left[\frac{1}{x}+\frac{18x^2}{2x^3+1}+\frac{1}{2(x+5)}+\frac{2x+3}{3\left(x^2+3x-1\right)}\right]x\left(2x^3+1\right)^3(x+5)^{\frac{1}{2}}\left(x^2+3x-1\right)^{\frac{1}{3}}\]
\end{example}
Note that implicit differentiation is used to differentiate the $\ln y$ that shows up on the left hand side of the equation with respect to $x$.  This gives the $\frac{1}{y}y'$ which is why we need to multiply both sides by $y$ (for which we have the function).

\subsection*{Steps in Logarithmic Differentiation}\index{logarithmic differentiation! steps }
\begin{enumerate}
\item Take logarithms of both sides of the equation $y=f(x)$~.
\item Differentiate with respect to $x$ on both sides, remembering to use implicit differentiation on $\ln y$ to get $\frac{1}{y}y'$.
\item Solve for $y'$ and substitute $f(x)$ for $y$.
\end{enumerate}

\begin{example}
Differentiate the following using logarithmic differentiation:
\begin{multicols}{3}
\begin{enumerate}
\item $\displaystyle y= (x+2)^{e^x}$
\item $\displaystyle y=\frac{(2x+5)^9\sqrt{x^2+3x}}{e^{\sin x}}$  
\item $\displaystyle f(x)= (\sin x +x)^{x}$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item First take the logarithm of both sides of $\displaystyle y= (x+2)^{e^x}$.  
\begin{align*}
\ln y & = \ln\left[(x+2)^{e^{x}}\right] \\
\implies \ln y & = e^{x}\ln (x+2) 
\end{align*}
Next differentiate both sides with respect to $x$.
\begin{align*}
\frac{1}{y}y'&=e^{x}\ln(x+2)+e^{x}\frac{1}{x+2}(1+0) \\
\implies y'(x) &=\left[e^{x}\ln(x+2)+\frac{e^{x}}{x+2}\right]\underbrace{(x+2)^{e^x}}_{=y} 
\end{align*}
\item Take the logarithm of $\displaystyle y=\frac{(2x+5)^9\sqrt{x^2+3x}}{e^{\sin x}}$  to get:
  \begin{align*}
    \ln y& =\ln\left[(2x+5)^9\right]+\ln\left[\left(x^2+3x\right)^{\frac{1}{2}}\right]-\ln\left[e^{\sin x}\right]\\
    \ln y& =9\ln\left(2x+5\right)+\frac{1}{2}\ln\left(x^2+3x\right)-\sin x
  \end{align*}
  Differentiate both sides with respect to $x$.
  \begin{align*}
    \frac{y'}{y}&=9\frac{2+0}{2x+5}+\frac{1}{2}\frac{2x+3}{x^2+3x}-\cos x\\
    y'&=\left[\frac{18}{2x+5}+\frac{x+3/2}{x^2+3x}-\cos x\right]\frac{(2x+5)^9\sqrt{x^2+3x}}{e^{\sin x}}
  \end{align*}
\item Taking the logarithms of both sides of $\displaystyle f(x)= (\sin x +x)^{x}$ we have: 
\begin{align*}
  \ln f(x)&= \ln\left[(\sin x +x)^{x}\right]= x\ln(\sin x +x)
\end{align*}
Differentiate both sides with respect to $x$.
\begin{align*}
\frac{f'(x)}{f(x)}&= (1)\ln (\sin x+x)+x\, \frac{\cos x +1} {\sin x +x} \\
\implies f'(x)&=\left[\ln(\sin x +x)+ x\, \frac{\cos x+1}{\sin x+ x}\right]f(x) \\
\implies f'(x)&= \left[\ln(\sin x + x)\,\frac{\cos x+1}{\sin x+x}\right](\sin x +x)^{x}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Differentiate the following  using logarithmic differentiation:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle y=\frac{\left(x^3+5\right)\left(x^2-3x\right)^4}{x-2}$
    \item $\displaystyle y=\left(x^2+3\right)^{x^3}$
    \item $\displaystyle f(x)=\left(e^x+1\right)^{\ln x}$
    \item $\displaystyle y=\left(2x+1\right)^{\sqrt{x}}$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

We note that logarithmic differentiation allows, as shown in the previous example, the ability to differentiate functions of the form $y=[f(x)]^{g(x)}$ which up until now we had no method to differentiate.  A function like $y=x^x$ for instance is neither a power function ($x^a$) nor an exponential function ($a^x$). Neither of the rules for differentiating those give the correct answer of $\frac{d}{dx}x^x=x^x\ln x+x^x$ obtained by logarithmic differentiation (show this!).  One approach, equivalent to logarithmic differentiation, is to rewrite the expression as $x^x=(e^{\ln x})^x=e^{x\ln x}$, so the base is now constant, and then use the rule for exponential derivatives along with the Chain Rule:
\[\frac{d}{dx}x^x=\frac{d}{dx}e^{x\ln x}=e^{x\ln x}\left[(1)\ln x+x\frac{1}{x}\right]=x^x(\ln x+1)\]
where we returned $e^{x\ln x}$ to $x^x$ in the last step. More generally one can write $[f(x)]^{g(x)}=e^{g(x)\ln[f(x)]}$ and differentiate that directly as well.  We prefer the logarithmic differentiation approach above as it avoids pushing the base into the exponent and returning again when you are done.

The approach of pushing the base into the exponent may be useful when the function one wishes to differentiate involves a sum of expressions only one of which requires logarithmic differentiation.  In that case one would need to evaluate the derivative of that single term on the side with logarithmic differentiation and insert the result in the final derivative. Pushing the base function into the exponent, on the other hand, can be done inline with the rest of the derivatives.  For example if \mbox{$f(x)=e^e+e^x+x^e+x^x$} then
  \begin{align*}
    f'(x)&=\frac{d}{dx}\left(e^e+e^x+x^e+x^x\right)=\frac{d}{dx}\left(e^e+e^x+x^e+e^{x\ln x}\right)\\
    &=0+e^x+(e-1)x^{e-1}+x^x(\ln x+1)
  \end{align*}
  where we differentiated the last term as above.  If we wanted to use logarithmic differentiation we would need to assign $y=x^x$, perform the logarithmic differentiation on the side to find the derivative $y'$, and then insert that answer into the derivative $f'(x)$.

\subsection{Integral of $\frac{1}{x}$ and $a^x$}
The Power Rule for integration is
\[\int x^n\,dx=\frac{1}{n+1}x^{n+1}\ \ \ (n\neq-1)\]
The answer for the indefinite integral clearly indicates that it cannot work for $n=-1$ as one would be dividing by zero.  Since $\frac{d}{dx}\ln x=\frac{1}{x}$
however, we now have an antiderivative for $x^{-1}=\frac{1}{x}$, namely $\ln
x$.  This will only work for values of $x>0$ since the domain
of $\ln x$ is only positive numbers.  However a second antiderivative
of $\frac{1}{x}$ that will work when $x<0$ is $\ln(-x)$,
since $\frac{d}{dx}\ln(-x)=\frac{1}{-x}\cdot(-1)=\frac{1}{x}$ by the
Chain Rule.  We can combine the results using absolute value bars in
the following theorem.
\theorem{The indefinite integral of $x^{-1}$ is
\[\int\frac{1}{x}\,dx=\ln|x|+C\]}

A useful corollary of this result is that one can now integrate the tangent function.  Using the substitution $u=\cos x$ (so $du=-\sin x\,dx$) one has
\[\int \tan x\,dx=\int\frac{\sin x}{\cos x}\,dx=-\int\frac{du}{u}=-\ln|u|+C=-\ln|\cos x|+C=\ln\left(|\cos x|^{-1}\right)+C=\ln|\sec x|+C\]

Since $\frac{d}{dx}a^x=a^x\ln a$ it follows, dividing both sides of the equation by the constant $\ln a$, that $\frac{d}{dx}\frac{a^x}{\ln a}=a^x$ and so we also have the result:
\theorem{\ \ $\displaystyle\int a^x\,dx=\frac{a^x}{\ln a}+C$\ \ }

\begin{example}
Evaluate the following integrals:
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle \int \frac{2}{x\ln x}\,dx$
\item $\displaystyle \int x^{2}10^{x^{3}+1}\,dx$
\item $\displaystyle \int \frac{4x-1}{4x^{2}-2x+10}\,dx $
\item $\displaystyle \int \frac{dx}{\left(3x-2\right)\ln \left(9x-6\right)} $
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item Let $\displaystyle u=\ln x$ so $\displaystyle du= \frac{1}{x}\,dx$. 
\begin{align*}
\int\frac{2}{x\ln x}\,dx = 2\int \frac{1}{u} \,du =2\ln|u| + C =2\ln|\ln x| + C 
\end{align*}
\item Let $\displaystyle u=x^{3}+1$, so $du=3x^{2}\,dx \implies \frac{1}{3}du= x^{2}\,dx$
\begin{align*}
\int x^{2}10^{x^{3}+1}\,dx = \frac{1}{3}\int 10^{u}\,du = \frac{1}{3}\cdot \frac{10^{u}}{\ln 10}+ C = \frac{10^{x^{3}+1}}{3\ln 10}+ C 
\end{align*}
\item Let $\displaystyle u=4x^{2}-2x+10$ so $du=(8x-2)dx \implies \frac{1}{2}du= (4x-1)dx$
\begin{align*}
\int \frac{4x-1}{4x^{2}-2x+10}\,dx = \frac{1}{2}\int\frac{1}{u}\,du =\frac{1}{2}\ln |u| + C  =\ln |4x^{2}-2x+10| + C
\end{align*}
\item Let $\displaystyle u= 9x-6$ so $du=9dx \implies \frac{1}{9}du= dx$. Also solve for the remaining factor in terms of $u$ to get $u=3(3x-2)\implies 3x-2= \frac{u}{3}$. Then
\begin{align*} 
\int \frac{dx}{(3x-2)\ln (9x-6)} = \frac{1}{9}\int \frac{1}{\frac{u}{3}\ln u}du =\frac{1}{3}\int \frac{1}{u\ln u}du 
\end{align*}
\\
Next let $w=\ln u$ so $dw=\frac{1}{u}du$. The integral becomes
\begin{align*}
=\frac{1}{3}\int \frac{1}{w}dw = \frac{1}{3}\ln |w| + C 
=\frac{1}{3}\ln |\ln u| + C
=\frac{1}{3}\ln |\ln(9x-6)| + C
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Evaluate the following integrals:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\int\frac{3}{2x}\,dx$
    \item $\displaystyle\int\frac{x^2}{x^3+5}\,dx$
    \item $\displaystyle\int\frac{\ln x}{x}\,dx$
    \item $\displaystyle\int_1^4\frac{10^{\sqrt{x}}}{\sqrt{x}}\,dx$
    \item $\displaystyle\int e^x5^{e^x}\,dx$
    \item $\displaystyle\int \frac{\sec x\tan x}{3+5\sec x}\,dx$
    \item $\displaystyle\int\cot x\,dx$
    \end{enumerate}
    ~%ensures ragged bottom of second column.
  \end{multicols}
\end{furtherquestions}
\end{example}

\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/logarithmic-functions.tex}
\end{exercise}

\newcommand{\expdepvar}{y}
\section{Exponential Growth and Decay}\index{exponential growth}
Many quantities, such as the number of cells being cultured in a lab dish or the number of radioactive nucleii of a particular isotope in a radioactive sample remaining undecayed, have a population $\expdepvar(t)$ that satisfies the differential equation
 \[\boxed{\frac{d\expdepvar}{dt}=k\expdepvar}\,.\]
 Here the constant $k$, called the \textbf{relative growth rate}\index{relative growth rate}, characterizes the population under consideration.  It will be positive ($k>0$) if the population $\expdepvar(t)$ is increasing in time and negative ($k<0$) if it is decreasing.  The preceding equation is called the \textbf{law of natural growth} \index{law of natural growth} or \textbf{law of natural decay}\index{law of natural decay} respectively. The constant $k$ is called \textbf{relative} since if we solve for it, $k=\frac{1}{y}\frac{d\expdepvar}{dt}$, we see the rate $d\expdepvar/dt$ is constant only relative to the population size $\expdepvar$ at an given time.

 The differential $d\expdepvar=\frac{d\expdepvar}{dt}dt$ satisfies
 \[d\expdepvar=k\expdepvar\,dt\,.\]
 Over a fixed time interval $\Delta t$ we have the analogous relation
 \[\Delta \expdepvar=k\expdepvar\Delta t\,,\]
 where $\expdepvar$, by the Mean Value Theorem, is evaluated at some time $t$ in the interval.  Assuming $\Delta t$ is small enough this can be effectively any time $t$ in the interval as $\expdepvar$ will be approximately constant over such an interval. The change $\Delta \expdepvar$ in the population $\expdepvar(t)$ over a fixed small time interval $\Delta t$ is therefore proportional to the population itself
 \[\Delta \expdepvar\propto \expdepvar\,.\]
 which is expected for a population whose growth (or loss) depends on the current size of the population.  Additionally the relation shows the change will also be approximately proportional to the length of the time interval $\Delta t$ considered, 
\[\Delta \expdepvar\propto \Delta t\,,\]
assuming again that $\Delta t$ is sufficiently small, a result that is also reasonable.

To understand how $\expdepvar$ changes in time we need to find the function $\expdepvar(t)$ that satisfies (solves) the differential equation
\[\frac{d\expdepvar}{dt}=k\expdepvar\,.\]
If the right hand side of the equation just involved $t$ explicitly, like $\frac{d\expdepvar}{dt}=t^2$, the answer would just be the antiderivative $\expdepvar(t)=\int t^2=\frac{1}{3}t^3+C$.  Our differential equation is not of this form, however, as it has the dependent variable $\expdepvar$ on the right hand side.  Solving such a differential equation such as ours can be done by the process of \textbf{separation of variables} \index{separation of variables}.  Inspired by the Leibniz notation, one formally proceeds by isolating, if possible, a function of the dependent variable $\expdepvar$ and its differential $d\expdepvar$ on one side of the equation and a function of the independent variable $t$ and its differential $dt$ on the other to get
\begin{align*}
\frac{d\expdepvar}{\expdepvar}&=k\,dt\,.
\end{align*}
One then integrates both sides:
\begin{align*}
\int\frac{d\expdepvar}{\expdepvar}&=\int k\,dt\\
\Rightarrow\ \ln \expdepvar&=kt+D\,,
\end{align*}
where we have combined the integration constants $C_1$ and $C_2$ arising from both sides of the integral into $D=C_2-C_1$.  Finally we can solve for $\expdepvar$ by taking the natural exponential of both sides:
\begin{align*}
e^{\ln \expdepvar}&=e^{kt+D}\\
\Rightarrow\ \expdepvar&=e^{kt}e^D
\end{align*}
Calling a new (positive) constant $C=e^D$ we have the final solution of the differential equation
\[\expdepvar(t)=Ce^{kt}\,.\]
Despite the lack of rigour in our separation of variable approach, one may readily confirm that $\expdepvar(t)=Ce^{kt}$ does satisfy the original differential equation as required.

The constant of integration, $C$ can be determined by providing an additional piece of information regarding the system.  If, for instance, one knows the initial size of the population is $\expdepvar(0)=\expdepvar_0$, then the solution of the resulting initial value problem gives
\[\expdepvar_0=\expdepvar(0)=Ce^{k(0)}=Ce^0=C(1)=C\,.\]
Placing this value for $C=\expdepvar_0$ back in $\expdepvar(t)$ gives
\[\boxed{\expdepvar(t)=\expdepvar_0e^{kt}}\,.\]
As such the population at arbitrary time, assuming it is undergoing exponential growth or decay, is characterized completely by the growth constant $k$ and its initial size $\expdepvar_0$\,.  The graphs of the cases where $k>0$ (growth) and $k<0$ (decay) are shown below.
\begin{center}
\includegraphics[width=0.4\textwidth]{graphics/exponential_growth}\hspace{.1\textwidth}\includegraphics[width=0.4\textwidth]{graphics/exponential_decay}
\end{center}
\textbook{\clearpage}
\begin{example}
A quantity $P(t)$ measuring the number of bacteria growing in a controlled laboratory setting increases exponentially.  Suppose it doubles in 8 hours.
\begin{enumerate}
\item Find the relative growth rate.
\item After how many hours has $P$ increased by 50$\%$?
\end{enumerate}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle P(8\ \text{hr})=2P_{0}$ where $P_{0}$ is the initial population.
\begin{align*}
\implies P_{0}e^{k(8\,\text{hr})} =2P_{0} \implies e^{(8\,\text{hr})k} &= 2 \\
 \ln\left(e^{(8\,\text{hr})k}\right)&= \ln 2 \\
(8\,\text{hr})k &=\ln 2 \\
k&=\frac{\ln 2}{8}\ (1/\text{hr})
\end{align*}
Therefore $\displaystyle P(t)=P_{0}e^{\frac{\ln 2}{8\, \text{hr}}t}$\,.
\item An increase of 50\% means that at the unknown time $t$ the population has grown to\\ $\displaystyle P(t)=P_{0}+(0.5)P_{0}=1.5P_{0}$\,.
\begin{align*}
\implies P_{0}e^{\frac{\ln 2}{8\,\text{hr}}t}= 1.5P_{0} 
\implies e^{\frac{\ln 2}{8\,\text{hr}}t} &=1.5 \\
\frac{\ln 2}{8\,\text{hr}}t &= \ln 1.5 \\
t=\frac{8\ln 1.5}{\ln 2}\ \text{hr} &\approx 4.68\,\text{hr}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}

%Fox squirrels breed twice a year with a litter of 2 to 5 young, typically 3.  So assuming 6 young to two parents a year (and an average of 10 year lifespan so we will neglect mortality) we can solve 8=2e^k to give k=ln4\approx 1.4/yr.  
Fox squirrels introduced into a city see their population increase from 50 to 12000 in 4 years.  Assuming the growth was exponential over this time period,
\begin{enumerate}
\item Find the relative growth rate $k$.
\item When will the squirrel population exceed 1 million?
\item Is the latter likely?  Explain.
\end{enumerate}
\end{furtherquestions}
\end{example}

When $k<0$ we have a decay formula and the amount $\expdepvar$ decreases over time.  Then the positive constant $\lambda=|k|=-k$, called the \textbf{decay constant} \index{decay constant}, may be introduced and our formula becomes
\[\expdepvar(t)=\expdepvar_0e^{-\lambda t}\,.\]
Rather than the decay constant, one often uses the \textbf{half-life}\index{half-life} constant $T$ for a radioactive sample.  It is defined to be the time required for half of the initial decaying substance to disappear (i.e. decay into a new form), and so $y(T)=\frac{1}{2}y_0$. This can be used to determine the decay constant $\lambda$\,.

\begin{example}
  The mass of a certain radioactive material is given by $m(t)=m_{0}e^{kt}$. If the half-life of this material is 1600 years, find the decay constant $\lambda=-k$\,.
  
\begin{solution*} 
\begin{align*}
m(t)&=m_{0} e^{kt} \implies m(1600\,\text{yr})=\frac{1}{2}m_{0} 
  \implies m_{0} e^{k(1600\,\text{yr})}=\frac{1}{2}m_{0} \implies e^{k(1600\,\text{yr})}=\frac{1}{2}\\
&\implies \ln \left(e^{k(1600\,\text{yr})}\right)=\ln \left(\frac{1}{2}\right)=\ln 1-\ln 2= 0-\ln 2 = -\ln 2 \\
    &\implies k(1600\,\text{yr})=-\ln 2
      \implies k=-\frac{\ln 2}{1600}\,(1/\text{yr})\\
  &\implies \lambda=\frac{\ln 2}{1600}\,(1/\text{yr})\approx 0.000433\,(1/\text{yr})
\end{align*}
\end{solution*}

\begin{furtherquestions}
Cobalt-60 is a radioactive isotope used in early radiotherapy and other applications.  Sixty is the \textbf{mass number} of the nucleus, the number of nucleons (protons and neutrons) it contains.  A sample of Cobalt-60 undergoes exponential decay with a half-life of 5.2714 years.
\begin{enumerate}
\item Find the decay constant $\lambda=-k$ of Cobalt-60.
\item How long would it take for a sample containing 40 grams of the isotope to decay to a sample containing only 10 grams of it?
\end{enumerate}
\end{furtherquestions}
\end{example}

Finally we note that there are many examples of quantities besides population counts which satisfy the differential equation $\frac{d\expdepvar}{dt}=k\expdepvar$ with solution $\expdepvar=\expdepvar_0e^{kt}$\,.  As an example, the voltage across a discharging capacitor in an electronic circuit containing only a resistor and a capacitor (an RC circuit) undergoes exponential decay from an initial voltage $V_0$\,.
% Note it would be good to make a problem here and introduce the definition
% of the time constant for the circuit, namely the time required for the voltage
% to fall to $V_0/e$.
\textbook{\clearpage}
\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/exponential-growth-and-decay.tex}
\end{exercise}

\section{Inverse Trigonometric Functions}\index{inverse trigonometric functions}
\subsection{Inverse Sine}\index{inverse sine function}
The sine function $y=\sin x$ on its natural domain $(-\infty,\infty)$ is not a one-to-one function.  It clearly fails the horizontal line test as the intersection with the line $y=\frac{1}{2}$ clearly shows:

\begin{center}
\includegraphics{graphics/inverses_and_other_functions9}%
\end{center}

However the function $y=\sin x$ on domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is a one-to-one function:
\begin{center}
\includegraphics{graphics/inverses_and_other_functions10}%
\end{center}

\definition{The inverse function of $y=\sin x$; $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
is called the \textbf{inverse sine function} \index{inverse sine function! definition of} or \textbf{arcsine function} and is denoted by $\bm{{y=\sin^{-1}x}}$ or $\bm{y=\arcsin x}$. It satisfies
\[y=\sin^{-1}x\iff x=\sin y\ \ \ \left(-\frac{\pi}{2}\le y\le \frac{\pi}{2}\right)\]}
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/inverses_and_other_functions11}%
\end{center}

The domain of inverse sine is $[-1,1]$ and range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.  The usual inverse identities apply:
\begin{eqnarray*}
\sin^{-1}(\sin x) & = & x\ \ \ \ \mbox{for}\ \ \ -\frac{\pi}{2}\le x\le\frac{\pi}{2} \\
\sin\left(\sin^{-1} x\right) & = & x\ \ \ \ \mbox{for}\ \ \  -1\le x\le 1
\end{eqnarray*}

\begin{example}
  Evaluate the following:
  \begin{multicols}{2}
  \begin{enumerate}
  \item $\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)$
  \item $\sec\left[\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)\right]$
  \end{enumerate}
  \end{multicols}
  \begin{solution*}
    \begin{enumerate}
    \item\label{itm:triginv1} Since inverse trigonometric functions return angles, call $\theta=\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)$. Then by definition of the inverse we have:
      \begin{align*}
        \sin\theta&=\frac{1}{\sqrt{2}}\ \text{with}\ -\frac{\pi}{2}\le\theta\le\frac{\pi}{2} \\
        \implies \theta & = \frac{\pi}{4}
      \end{align*}
      The latter result is by inspection of a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with side lengths $1$, $1$, $\sqrt{2}$ and recalling sine is the opposite over the hypotenuse.
    \item Evaluate the inverse trigonometric function as in part \ref{itm:triginv1} and then insert it to find
      \[\sec\left[\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)\right]=\sec\left(\frac{\pi}{2}\right)=\frac{1}{\cos\left(\frac{\pi}{2}\right)}=\frac{1}{1/\sqrt{2}}=\sqrt{2},\]
      where here we used the  $45^{\circ}-45^{\circ}-90^{\circ}$ triangle again to evaluate the cosine.
    \end{enumerate}
  \end{solution*}

\begin{furtherquestions}
  Evaluate the following:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$
    \item $\displaystyle \tan\left[\sin^{-1}\left(\frac{1}{2}\right)\right]$
    \item $\displaystyle \sin\left(2\sin^{-1}x\right)$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\theorem{The derivative of inverse sine is
\[\frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}\ ,\]
where $-1<x<1$~
\index{inverse sine function! derivative of}
.}

To prove the theorem note that if $y=\sin^{-1}x$ then by definition of the inverse:
\[\sin y=x\]
Implicit differentiation of both sides with respect to $x$ yields
\[\left(\cos y\right)y'=1\]
and so $\frac{dy}{dx}=\frac{1}{\cos y}$.  By the trigonometric identity $\cos^2y+\sin^2y=1$ it follows that
\[\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}\]
where here the positive solution was taken since $-\frac{\pi}{2}<y<\frac{\pi}{2}$ implies $\cos y>0$.  Inserting this in the formula for $\frac{dy}{dx}$ gives the result.

Note that the Chain Rule result is, as expected:
\[\frac{d}{dx}\left(\sin^{-1}g(x)\right)=\frac{g'(x)}{\sqrt{1-g^2(x)}}\ .\]


\begin{example}
  Differentiate the following function:
  \[\displaystyle y=\sin ^{-1} (e^{x}+2)\]
  \begin{solution}
    By the Chain Rule:\\
    $\displaystyle y'=\left(\sin ^{-1} (e^{x}+2)\right)'=\frac{1}{\sqrt{1-(e^x+2)^2}}(e^x+0)=\frac{e^x}{\sqrt{1-(e^x+2)^2}}$ \\
  \end{solution}
  
  \begin{furtherquestions}
    Differentiate the following functions:
    \begin{multicols}{2}
      \begin{enumerate}
      \item $\displaystyle y=\sin^{-1}\left(\ln x+3\right)$
      \item $\displaystyle y=e^{\sin^{-1}x}+\sin^{-1}\left(e^x\right)$
      \end{enumerate}
    \end{multicols}
  \end{furtherquestions}
\end{example}

\subsection{Inverse Cosine}
The function $y=\cos x$ on domain $\left[0,\pi\right]$ is a one-to-one function:
\begin{center}
\includegraphics{graphics/inverses_and_other_functions12}%
\end{center}

\definition{The inverse function of $y=\cos x$; $\left[0,\pi\right]$
is called the \textbf{inverse cosine function}\index{inverse cosine function} \index{inverse cosine function! definition of} or \textbf{arccosine function} and is denoted by $\bm{{y=\cos^{-1}x}}$ or $\bm{y=\arccos x}$. It satisfies
\[y=\cos^{-1}x\iff x=\cos y\ \ \ \left(0\le y\le \pi\right)\]}
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/inverses_and_other_functions13}%
\end{center}

The domain of inverse cosine is $[-1,1]$ and range is $\left[0,\pi\right]$.  The inverse identities are:
\begin{eqnarray*}
\cos^{-1}(\cos x) & = & x\ \ \ \ \mbox{for}\ \ \ 0\le x\le\pi \\
\cos\left(\cos^{-1} x\right) & = & x\ \ \ \ \mbox{for}\ \ \  -1\le x\le 1
\end{eqnarray*}

\theorem{The derivative of inverse cosine \index{inverse cosine function! derivative of} is
\[\frac{d}{dx}\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}}~,\]
where $-1<x<1$~ .}

\subsection{Inverse Tangent}
The function $y=\tan x$ on domain $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is a one-to-one function:
\begin{center}
\includegraphics{graphics/inverses_and_other_functions14}%
\end{center}
\definition{The inverse function of $y=\tan x$; $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
is called the \textbf{inverse tangent function}\index{inverse tangent function} \index{inverse tangent function! definition of}  and is denoted by $\bm{{y=\tan^{-1}x}}$ or $\bm{y=\arctan x}$. It satisfies
\[y=\tan^{-1}x\iff x=\tan y\ \ \ \left(-\frac{\pi}{2}\le y\le \frac{\pi}{2}\right)\]}
\begin{center}
\includegraphics{graphics/inverses_and_other_functions15}%
\end{center}

The domain of inverse tangent is $(-\infty,\infty)$ and range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.  The usual inverse identities apply:
\begin{eqnarray*}
\tan^{-1}(\tan x) & = & x\ \ \ \ \mbox{for}\ \ \ -\frac{\pi}{2}<x<\frac{\pi}{2} \\
\tan\left(\tan^{-1} x\right) & = & x\ \ \ \ \mbox{for}\ \ \  x\in\mathset{R}
\end{eqnarray*}
as well as the limits:

\hfill$\mylim{x}{-\infty}\tan^{-1}x=-\frac{\pi}{2}$\hfill$\mylim{x}{\infty}\tan^{-1}x=\frac{\pi}{2}$\hfill~

So $y=\pm\frac{\pi}{2}$ are horizontal asymptotes of the function.

\theorem{The derivative of inverse tangent is
\[\frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}~.\]
\index{inverse tangent function!derivative of}
}


\subsection{Other Trigonometric Inverses}
Similarly one defines $\displaystyle y=\csc^{-1}x$, $\displaystyle y=\sec^{-1}x$, and $y=\cot^{-1}x$.\footnote{Note that the convention for the inverse secant and inverse cosecant functions used here is that the domain of secant (cosecant), and hence the range of inverse secant (cosecant) is $[0,\pi/2)\cup[\pi,3\pi/2)$ ( $(0,\pi/2]\cup(\pi,3\pi/2]$ ).  One can also choose the more intuitive interval $[0,\pi/2)\cup(\pi/2,\pi]$ for secant and $[-\pi/2,0)\cup(0,\pi/2]$ for cosecant but then absolute value bars are required about the $x$ outside the radical in the derivative formulae.}

Notes:
\begin{itemize}
\item Since trigonometric functions are functions of angles, inverse trigonometric functions return angles.  All our angles above are in radians.  On your calculator you must have it set to radian mode to get these inverse trigonometric function results.  If you have your calculator set to degree mode you will get your answers in degrees.
\item The $-1$ in $\sin^{-1}x$ means inverse \emph{not} taking to the power of $-1$ (reciprocal) like the $2$ in $\sin^2x$ means.  If you mean take to the power of $-1$, i.e. $\frac{1}{\sin x}$ then you must write $(\sin x)^{-1}$ or simply use the reciprocal trigonometric function $\csc x$.
\item It is because none of the trig functions are one-to-one and hence not invertible on their natural domains that solving a trigonometric equation $\textrm{trig}(x)=\#$ requires more than just ``applying the inverse'' to both sides (unlike, say comparable logarithmic or exponential equations).  So to solve $\sin x=\frac{1}{2}$ the result $x=\sin^{-1}(1/2)=\pi/6$ is only one of many solutions.  (See the intersections between $y=\sin x$ and $y=1/2$ in our initial graph in this section.)
\end{itemize}  

A complete table of the inverse trigonometric derivatives is as follows \index{inverse trigonometric functions! table of}:

\hfill$\displaystyle\frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$ \hfill $\displaystyle\frac{d}{dx}\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}}$\hfill~

\hfill$\displaystyle\frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}$\hfill$\displaystyle\frac{d}{dx}\left(\cot^{-1}x\right)=-\frac{1}{1+x^2}$\hfill~

\hfill$\displaystyle\frac{d}{dx}\left(\sec^{-1}x\right)=\frac{1}{x\sqrt{x^2-1}}$\hfill$\displaystyle\frac{d}{dx}\left(\csc^{-1}x\right)=-\frac{1}{x\sqrt{x^2-1}}$\hfill~

Here the derivatives exist on the domains of the inverse trigonometric function except at those values where the expression is undefined.  For the Chain Rule formulae simply replace $x$ by $g(x)$ in each formula and multiply the result by $g'(x)$.
\textbook{\clearpage}
\begin{example}
Differentiate the following functions:
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle f(x)=\cos ^{-1}\left(\ln x\right)$
\item $\displaystyle y=\frac{\tan ^{-1}(x^{2}+3)}{x}$
\item $\displaystyle \tan^{-1}y+ \sin^{-1}x= e^y$
\item $y=\sec^{-1}(x^2)$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle f'(x)=\frac{d}{dx}\cos ^{-1}\left(\ln x\right)=\frac{-1}{\sqrt{1-(\ln x)^2}}\left(\frac{1}{x}\right)= -\frac{1}{x\sqrt{1-(\ln x)^2}}$
\item We can use the Product Rule:
  \begin{align*} y'&=\left(\tan ^{-1}(x^{2}+3)\cdot\left(\frac{1}{x}\right)\right)'=\frac{2x+0}{1+(x^{2}+3)^2}\left(\frac{1}{x}\right)+\tan^{-1}(x^{2}+3)\left(-\frac{1}{x^2}\right) \\
                   &=\frac{2}{1+(x^{2}+3)^2}-\frac{\tan^{-1}(x^{2}+3)}{x^2}
  \end{align*}
\item Using implicit differentiation differentiate both sides of $\displaystyle \tan^{-1}y+ \sin^{-1}x= e^y$ to get:
\begin{align*}
\implies \frac{1}{1+y^{2}}(y') + \frac{1}{\sqrt{1-x^{2}}} &= e^{y}y' \\
\implies y'\left(\frac{1}{1+y^{2}}-e^{y}\right) &= -\frac{1}{\sqrt{1-x^{2}}} \\
\implies y' &=\frac{-\frac{1}{\sqrt{1-x^2}}}{\frac{1}{1+y^2}-e^y}\ \  \left(\leftarrow\text{Multiply by }1=\frac{1+y^2}{1+y^2}\right)\\
\implies y'&=-\frac{1+y^2}{\sqrt{1-x^2}(1-e^y(1+y^2))}
\end{align*}
\item $\displaystyle y'=\frac{d}{dx}\sec^{-1}(x^2)=\frac{1}{x^{2}\sqrt{(x^{2})^{2}-1}}(2x)=\frac{2}{x\sqrt{x^{4}-1}}$
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Differentiate the following functions:
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle y=\sin^{-1}(2x-1)$
\item $\displaystyle y=\tan^{-1}\left(\frac{x}{3}\right)+\ln\sqrt{\frac{x-3}{x+3}}$
\item $\displaystyle y=x\cos^{-1}x-\sqrt{1-x^2}$
\item $\displaystyle y=\sin^{-1}\left(\tan^{-1}x\right)$
\item $\displaystyle y=\tan^{-1}\left(\ln x\right)e^{x^2+3}$
\item $\displaystyle y=\cos^{-1}\left(e^{2x}-5\right)$
\item $\displaystyle y=\tan^{-1}\left(x^2+3\right)-\tan\left(\cos^{-1}x+1\right)$
\item $\displaystyle f(t)=\sec^{-1}\left(e^t+\ln t\right)$
\item $\displaystyle \sin^{-1}y=x^2+y^2+e^y$
\end{enumerate}
\end{multicols}
\end{furtherquestions}
\end{example}
\textbook{\clearpage}
The derivatives of the inverse trigonometric functions give the following results:
\theorem{\label{thm:indefiniteint}We have the following indefinite integrals:
  
  \hfill$\displaystyle\int\frac{1}{\sqrt{1-x^2}}\,dx=\sin^{-1}x+C$,\hfill$\displaystyle\int\frac{1}{1+x^2}\,dx=\tan^{-1}x+C$,\hfill$\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,dx=\sec^{-1}\!x+C$\hfill~}

If one considers the more general integral
\[\int\frac{1}{\sqrt{a^2-x^2}}\,dx\ ,\]
where $a>0$ is constant this can be solved by first noting that
\[\sqrt{a^2-x^2}=\sqrt{a^2\left(1-\frac{x^2}{a^2}\right)}=\sqrt{a^2}\sqrt{1-\frac{x^2}{a^2}}=|a|\sqrt{1-\frac{x^2}{a^2}}=a\sqrt{1-\frac{x^2}{a^2}}\]
and then using the substitution $u=\frac{x}{a}$ (so $du=\frac{dx}{a}$) to get:
\[\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\int\frac{1}{a\sqrt{1-\frac{x^2}{a^2}}}\,dx=\int\frac{1}{\sqrt{1-u^2}}\,du=\sin^{-1}u+C=\sin^{-1}\frac{x}{a}+C\ .\]
Similar generalization can be done to the inverse tangent integral.  We thus have:

\theorem{For constant $a>0$,\label{thm:indefiniteintwitha}

\hfill$\displaystyle\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\frac{x}{a}+C$,\hfill$\displaystyle\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$,\hfill$\displaystyle\int\frac{1}{x\sqrt{x^2-a^2}}\,dx=\frac{1}{a}\sec^{-1}\frac{x}{a}+C$\hfill~ }

\subsubsection*{Note on Remembering Formulas}
Integral formulas appearing in Theorem~\ref{thm:indefiniteintwitha} with literal constants (i.e. $a$) are what one finds typically in integral tables.  The placement of the $a$'s need not be memorized as they can be inferred by \emph{dimensional analysis}.\index{dimensional analysis}  Suppose $x$ has units of length. Then from the form of the integrand $a$ would also have units of length since $a^2$ is added to or subtracted from $x^2$.  The resulting integrals involve inverse trigonometric functions whose arguments must be dimensionless since the result of any trigonometric function is a dimensionless ratio.  To achieve that we divide $x$ in our original integrals of Theorem~\ref{thm:indefiniteint} by $a$.  Next to remember whether we need a $\frac{1}{a}$ in front or not we consider the integrand.  For the first one we have $dx$ in the numerator which will have the same length dimension as $x$ since it is just an (infinitessimal) interval of $x$.  In the denominator we have the square root of a length-squared which is just a length.  As such for the first integral we are summing a dimensionless quantity since it is a length divided by a length and the result must be dimensionless.  The resulting arcsine function, which returns a dimensionless angle, therefore cannot have an $a$ out front as that would introduce a length dimension.  The latter two integrals, however, are sums of quantities of dimension $\text{length}/\text{length}^2=1/\text{length}$.  To achieve that dimension in our result we must multiply the dimensionless inverse trigonometric function by $\frac{1}{a}$.  Once again one notes how the differential $dx$ in our notation makes this work.   Thank-you Leibniz!
\textbook{\clearpage}
\begin{example}
  Evaluate the given integrals:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle \int \frac{1}{\sqrt{4-{t^2}}}\,dt $
    \item $\displaystyle \int_{0}^{2} \frac{dx}{4+x^{2}}\,dx$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item To get the leading $1$ we factor out $4$:
\begin{align*}
\int \frac{1}{\sqrt{4-{t^2}}}\,dt 
&=\int \frac{1}{\sqrt{4\left(1-\left(\frac{t}{2}\right)^{2}\right)}}\, dt
                                     =\frac{1}{2}\int \frac{1}{\sqrt{1-\left(\frac{t}{2}\right)^{2}}}\, dt
\end{align*}
Then let $\displaystyle u=\frac{t}{2}$ so $\displaystyle du=\frac{1}{2}dt \implies dt= 2du$\,.  The integral is then:
\begin{align*}
&=\frac{1}{2}\int \frac{1}{\sqrt{1-u^{2}}}\cdot 2\, du= \sin^{-1}u+ C \\
&=\sin^{-1}\left(\frac{t}{2}\right)+ C
\end{align*}
\item Again we factor out $4$ and to get the desired $1$:
\begin{align*}
\int_{0}^{2} \frac{dx}{4+x^{2}}\,dx
  &=\int_{0}^{2} \frac{1}{4}\cdot\frac{1}{1+(\frac{x}{2})^{2}}\, dx
\end{align*}
Substitute: $\displaystyle u=\frac{x}{2}$ so $\displaystyle du=\frac{1}{2}dx \implies 2du=dx$.\\
Change limits: $x=0 \implies u=0, \, x=2 \implies u=1$\\
The integral becomes:
\begin{align*}
&=\int_{0}^{1} \frac{1}{4}\cdot \frac{1}{1+u^{2}}\cdot 2\,du =\left.\frac{1}{2}\tan^{-1}\!u\right|_{u=0}^{u=1}\\
&= \frac{1}{2}\left[\tan^{-1}(1)-\tan^{-1}(0)\right]=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Evaluate the following integrals:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\int\frac{3}{\sqrt{4-2x^2}}\,dx$
    \item $\displaystyle\int\frac{\tan^{-1}x}{1+x^2}\,dx$
    \item $\displaystyle\int\frac{4}{t\left[9+\left(\ln t\right)^2\right]}\,dt$
    \item $\displaystyle\int\frac{e^x}{\sqrt{1-8e^{2x}}}\,dx$
    \item $\displaystyle\int\frac{3x+4}{2x^2+3}\,dx$
    \item $\displaystyle\int_0^{\frac{\pi}{2}}\frac{\cos x}{1+\sin^2x}\,dx$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\textbook{\clearpage}
\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/inverse-trigonometric-functions.tex}
\end{exercise}

\section{L'H\^{o}pital's Rule} \index{l'H\^{o}pital's Rule}
We have already evaluated limits that are indeterminate forms of the type $\frac{0}{0}$ and $\frac{\infty}{\infty}$.
\begin{example}
  Evaluate the following limits:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\lim_{x\to2}\frac{x^2+3x-10}{2x^2-8}$
    \item $\displaystyle\lim_{x\to\infty}\frac{7x^2+3x+1}{9x^3+4}$
    \end{enumerate}
  \end{multicols}
  \begin{solution*}
    \begin{enumerate}
      \item Substitution of $2$ in the rational function confirms a $\frac{0}{0}$ indeterminate form.  The polynomials evaluating to zero at $x=2$ suggest the presence of a factor of $(x-2)$ is the cause.  Factoring confirms this and allows evaluation of the limit:
      \begin{align*}
        \lim_{x\to2}\frac{x^2+3x-10}{2x^2-8}=\lim_{x\to2}\frac{(x-2)(x+5)}{2(x+2)(x-2)}=\lim_{x\to2}\frac{x+5}{2(x+2)}=\frac{7}{2(4)}=\frac{7}{8}
      \end{align*}
    \item As $x\to\infty$ the polynomials in the rational function both get large and we have an $\frac{\infty}{\infty}$ indeterminate form.  Factoring out the dominant term in each of the numerator and the denominator, in this case the highest power of $x$, allows the limit to be evaluated:
      \begin{align*}
        \lim_{x\to\infty}\frac{7x^2+3x+1}{9x^3+4}=\lim_{x\to\infty}\frac{x^2}{x^3}\cdot\frac{7+\frac{3}{x}+\frac{1}{x^2}}{9+\frac{4}{x^3}}=\lim_{x\to\infty}\frac{1}{x}\cdot\frac{7+\frac{3}{x}+\frac{1}{x^2}}{9+\frac{4}{x^3}}=0\cdot\frac{7+0+0}{9+0}=0
      \end{align*}
      \end{enumerate}
  \end{solution*}
  \begin{furtherquestions}
    Evaluate the following limits:
    \begin{multicols}{2}
      \begin{enumerate}
      \item $\displaystyle\mylim{x}{1}\frac{x^2-1}{x^2-3x+2}$
      \item $\displaystyle\mylim{x}{\infty}\frac{3x^2-5x+2}{4x^2+3x-10}$
      \end{enumerate}
    \end{multicols}
  \end{furtherquestions}
\end{example}

In general:
\begin{itemize}
\item If $\displaystyle\mylim{x}{a}f(x)=0$ and $\displaystyle\mylim{x}{a}g(x)=0$, then $\displaystyle\mylim{x}{a}\frac{f(x)}{g(x)}$ is called the \textbf{indeterminate form of type}~$\bm{\frac{0}{0}}$.
\item If $\displaystyle\mylim{x}{a}f(x)=\pm\infty$ and $\displaystyle\mylim{x}{a}g(x)=\pm\infty$, then $\displaystyle\mylim{x}{a}\frac{f(x)}{g(x)}$ is called the \textbf{indeterminate form of type}~$\bm{\frac{\infty}{\infty}}$ \index{indeterminate forms of limits} \index{indeterminate forms of limits! of type ${\frac{0}{0}}$ }
\index{indeterminate forms of limits! of type ${\frac{\infty}{\infty}}$ }
. 
\end{itemize}

Our techniques used above will not work for evaluating all limits of this type:
\begin{example}
  The limit $\displaystyle \mylim{x}{0}\frac{2^x-1}{x}$ is an indeterminate form of type $\frac{0}{0}$ while $\displaystyle\mylim{x}{\infty}\frac{\ln x}{x}$ is of type $\frac{\infty}{\infty}$.  Neither limit may be resolved using the methods of the previous example.
\end{example} 

\textbook{\clearpage}
\theorem{If $f$ and $g$ are differentiable functions with $g'(x)\neq0$ on an open interval containing the value $a$ and $\displaystyle\mylim{x}{a}\frac{f(x)}{g(x)}$ is an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, (i.e. $\displaystyle\mylim{x}{a}f(x)=0$ and $\displaystyle\mylim{x}{a}g(x)=0$
or
$\displaystyle\mylim{x}{a}f(x)=\pm\infty$ and $\displaystyle\mylim{x}{a}g(x)=\pm\infty$)
then
\[\mylim{x}{a}\frac{f(x)}{g(x)}=\mylim{x}{a}\frac{f'(x)}{g'(x)}\ ,\]
provided the limit on the right hand side either exists or is $\pm\infty$.  This is \textbf{L'H\^{o}pital's Rule}.}

\begin{example}
Evaluate the previous limits using L'H\^{o}pital's Rule:
\begin{enumerate}
\item $\displaystyle\mylim{x}{0}\frac{2^x-1}{x}=\mylim{x}{0}\frac{2^x\ln2-0}{1}=2^0\ln2=(1)(\ln 2)=\ln 2$
\item $\displaystyle\mylim{x}{\infty}\frac{\ln x}{x}=\mylim{x}{\infty}\frac{\frac{1}{x}}{1}=\mylim{x}{\infty}\frac{1}{x}=0$
\end{enumerate}
\end{example}
Note that when applying L'H\^{o}pital's Rule one is \textbf{not} using the Quotient Rule!  The derivatives in the numerator and denominator are taken separately.

\begin{example}
  Evaluate the given limits:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle \lim_{x\to 2} \frac{x^{3}-8}{x-2} $ 
    \item $\displaystyle \lim_{x\to 1} \frac{e^{2x}-e^{2}}{\ln x}$
    \item $\displaystyle \lim_{x \to\infty} \frac{x^{2}+1}{e^x}$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item The functions $f(x)=x^{3}-8$ and $g(x)=x-2$ are differentiable and
  \begin{align*}
    f(2)& =2^3-8=8-8=0\\
    g(2)& =2-2=0\,.
  \end{align*}
  Thus the quotient is of $\frac{0}{0}$ indeterminate form. Therefore we can apply L'H\^{o}pital's Rule:
\[\displaystyle \lim_{x\to 2} \frac{x^{3}-8}{x-2} \underset{\lh}=\lim_{x\to 2} \frac{3x^2}{1}=3(2)^2=12\]
\item The function $f(x)=e^{2x}-e^{2}$ and $g(x)=\ln x$ are differentiable and
  \begin{align*}
    f(1)& =e^{2}-e^{2}=0 \\
    g(1)& =\ln 1=0\,.
  \end{align*}
  Thus the quotient is of $\frac{0}{0}$ indeterminate form. Therefore, we can apply L'H\^{o}pital's Rule:
\[\displaystyle \lim_{x\to 1} \frac{e^{2x}-e^{2}}{\ln x} \underset{\lh}
=\lim_{x \to 1} \frac{2e^{2x}}{\frac{1}{x}}=\lim_{x \to 1} 2xe^{2x}=2(1)e^{2(1)}=2e^2\]
\item The function $f(x)=x^2+1$ and $g(x)=e^x$ are differentiable and
  \begin{align*}
    \lim_{x\to\infty}f(x)=\lim_{x\to\infty}\left(x^2+1\right)=\infty\\
    \lim_{x\to\infty}g(x)=\lim_{x\to\infty}\left(e^x\right)=\infty
  \end{align*}
  Thus the quotient is of $\frac{\infty}{\infty}$ indeterminate form. Therefore we can apply L'H\^{o}pital's Rule:
\[\lim_{x \to\infty} \frac{x^{2}+1}{e^x} \underset{\lh}= \lim_{x \to \infty} \frac{2x}{e^x} \underset{\lh}= \lim_{x \to \infty} \frac{2}{e^x}= \frac{2}{\infty}=0\]
Here we applied L'H\^{o}pital's Rule a second time because $\displaystyle\frac{2x}{e^x}$ was also of $\frac{\infty}{\infty}$ determinate form.
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Evaluate the following limits:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\mylim{x}{0}\frac{e^x-1}{x}$
    \item $\displaystyle\mylim{x}{0}\frac{\sin x}{x}$
    \item $\displaystyle\mylim{x}{0}\frac{\cos x+2x-1}{3x}$
    \item $\displaystyle\mylim{x}{2}\frac{x^2+3x+5}{x^2-4}$
    \item $\displaystyle\mylim{x}{\infty}\frac{\ln x}{\sqrt{x}}$
    \item $\displaystyle\mylim{x}{0}\frac{x-\sin x}{x^3}$
    \item $\displaystyle\mylim{x}{\infty}\frac{e^x}{\ln x}$
    \item $\displaystyle\mylim{x}{0}\frac{\cos x}{x^2-1}$
    \item $\displaystyle\mylim{x}{5}\frac{\sqrt{x-1}-2}{x^2-25}$
    \item $\displaystyle\mylim{x}{0}\frac{\sin x}{x-\tan x}$
    \item $\displaystyle\mylim{x}{0}\frac{3^x-1}{x}$
    \item $\displaystyle\mylim{x}{0}\frac{4e^{2x}-4}{e^x-1}$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}

\subsection{Indeterminate Forms of type $0\cdot\infty$ and $\infty-\infty$}
Consider the following indeterminate forms:
\begin{itemize}
\item If $\displaystyle\mylim{x}{a}f(x)=0$ and $\mylim{x}{a}g(x)=\pm\infty$ then $\mylim{x}{a}f(x)g(x)$ is called the \textbf{indeterminate form of type} $\bm{0\cdot\infty}$ \index{indeterminate forms of limits! of type 
${0\cdot\infty}$}
. 
\item If $\displaystyle\mylim{x}{a}f(x)=\infty$ and $\mylim{x}{a}g(x)=\infty$ then $\mylim{x}{a}[f(x)-g(x)]$ is called the \textbf{indeterminate form of type} $\bm{\infty-\infty}$ \index{indeterminate forms of limits! of type
${\infty-\infty}$}
. 
\end{itemize}
To solve an indeterminate form of type $0\cdot\infty$, write the product $f\cdot g$ as either

\hfill$\displaystyle f\cdot g=\frac{f}{1/g}$\ \ \ \ or\ \ \ \ $\displaystyle f\cdot g=\frac{g}{1/f}$\hfill~

This will convert the indeterminate form into a form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$ which can then potentially evaluated using L'H\^{o}pital's Rule.

For indeterminate forms of type $\infty-\infty$ try to convert the difference into a quotient (by using a common denominator or factoring out common terms or rationalization) to once again reduce the limit to type~$\frac{0}{0}$ or $\frac{\infty}{\infty}$.

\begin{example}
  Evaluate the given limits:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle \lim_{x\to \frac{\pi}{2}} \sec x \sin(2x) $ 
    \item $\displaystyle \lim_{x\to 2} \left[\frac{4}{x^2-4}-\frac{1}{x-2}\right]$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item As $\displaystyle x \to \frac{\pi}{2}$ we have $\sec x \to\infty$, and $\sin(2x) \to 0$. Thus the product is of $0 \cdot \infty $ indeterminate form. We write it as a $\displaystyle \frac{0}{0}$ form noting that cosine is the reciprocal of secant. 
\begin{align*}
\lim_{x \to \frac{\pi}{2}} \sec x \sin(2x) &=\lim_{x \to \frac{\pi}{2}} \frac{\sin(2x)}{\cos x}
\underset \lh =\lim_{x \to \frac{\pi}{2}} \frac{2\cos(2x)}{-\sin x}\\
&=-\frac{2\cos\left(2\frac{\pi}{2}\right)}{\sin \frac{\pi}{2}}
=-\frac{2\cos \pi}{(1)}
=-2(-1)=2
\end{align*}
\item As $\displaystyle x \to 2^+$ one has $\frac{4}{x^2-4} \to \infty $ and $\frac{1}{x-2} \to \infty$. As $\displaystyle x \to 2^-$ one has $\frac{4}{x^2-4} \to -\infty $ and $\frac{1}{x-2} \to -\infty$. Thus, the difference is of $\infty-\infty$ (or $-\infty+\infty$) indeterminate form. Convert the difference into a quotient using a common denominator. 
\begin{align*}
\lim_{x \to 2} \left[\frac{4}{x^2-4}-\frac{1}{x-2}\right]
=\lim_{x \to 2} \left[\frac{4}{(x+2)(x-2)}-\frac{1}{x-2}\right]
=\lim_{x \to 2} \frac{4-(x+2)}{x^2-4} \\
=\lim_{x \to 2} \frac{2-x}{x^2-4}\,\,\, \left(\frac{0}{0}\,\, \text{form}\right)
\underset\lh=\lim_{x \to 2} \frac{-1}{2x}=\frac{-1}{2(2)}=-\frac{1}{4}
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
  Evaluate the following limits:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\mylim{x}{0^+}x^2\ln x$
    \item $\displaystyle\mylim{x}{\frac{\pi}{2}}(2x-\pi)\sec x$
    \item $\displaystyle\mylim{x}{0}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)$
    \item $\displaystyle\mylim{x}{1}\left(\frac{1}{x^2-1}-\frac{1}{x-1}\right)$
    \item $\displaystyle\mylim{x}{\infty}\left(\sqrt{x^2+x}-x\right)$
    \item $\displaystyle\mylim{x}{\infty}\left(x-\ln x\right)$
    \end{enumerate}
  \end{multicols}
\end{furtherquestions}
\end{example}
\textbook{\clearpage}
\subsection{Exponential Indeterminate Forms}
Several indeterminate forms arise from $\displaystyle \mylim{x}{a}\left[f(x)\right]^{g(x)}$.
\begin{itemize}
\item If $\displaystyle\mylim{x}{a}f(x)=0$ and $\mylim{x}{a}g(x)=0$ then \textbf{indeterminate form of type} $\bm{0^0}$ \index{indeterminate forms of limits! of type ${0^0}$}. 
\item If $\displaystyle\mylim{x}{a}f(x)=\infty$ and $\mylim{x}{a}g(x)=0$ then \textbf{indeterminate form of type} $\bm{\infty^0}$ \index{indeterminate forms of limits! of type ${\infty^0}$}. 
\item If $\displaystyle\mylim{x}{a}f(x)=1$ and $\mylim{x}{a}g(x)=\pm\infty$ then \textbf{indeterminate form of type} $\bm{1^\infty}$ \index{indeterminate forms of limits! of type ${1^\infty}$}. 
\end{itemize}
Each of these can be evaluated either by taking the natural logarithm:
\[y=[f(x)]^{g(x)}\ \Rightarrow\ \ln y=g(x)\ln[f(x)]\ ,\]
or by writing the function as an exponential:
\[ [f(x)]^{g(x)}=e^{g(x)\ln[f(x)]}\,. \]
In either case an indeterminate form of type $0\cdot\infty$ will result. This resulting indeterminacy is the reason these three exponential forms are themselves indeterminate.  We will use the former approach, as we did with logarithmic differentiation, to evaluate these limits.

\begin{example}
As a practical example prove our limit formula for $e$ by evaluating $\displaystyle\mylim{x}{0}(1+x)^{\frac{1}{x}}$, a limit of indeterminate form $1^\infty$.

Let $\displaystyle y=(1+x)^{\frac{1}{x}}$.  Taking the natural logarithm of both sides results in
\[\ln y=\frac{1}{x}\ln(1+x)\]
Taking the limit as $x\rightarrow 0$ of the righthand side gives an indeterminate form of type $\frac{0}{0}$ readily evaluated using L'H\^{o}pital's Rule:
\[\mylim{x}{0}\ln y=\mylim{x}{0}\frac{\ln(1+x)}{x}=\mylim{x}{0}\frac{\frac{1}{1+x}}{1}=\mylim{x}{0}\frac{1}{1+x}=\frac{1}{1+0}=1\]
So \[\mylim{x}{0}(1+x)^{\frac{1}{x}}=\mylim{x}{0}y=\mylim{x}{0}e^{\ln y}=e^{\left(\mylim{x}{0}\ln y\right)}=e^1=e\]
This was the limit stated for $e$ given before.
\end{example}

\begin{example}
  Evaluate the following limits:
  \begin{multicols}{3}
    \begin{enumerate}
    \item $\displaystyle \lim_{x\to 0}(\sin x)^{x} $ 
    \item $\displaystyle \lim_{x\to \infty} (\ln x)^{\frac{1}{x}}$
    \item $\displaystyle \lim_{x \to 1} (2x-1)^{\frac{1}{\ln x}}$
    \end{enumerate}
  \end{multicols}
  \textbook{\newpage}
\begin{solution*}
\begin{enumerate}
\item As $x \to 0, (\sin x)^{x} \to 0^{0}$ indeterminate form. \\Let $y=(\sin x)^{x} \implies \ln y= \ln\left[(\sin x)^{x}\right]= x\ln (\sin x)$
\begin{align*}
\text{Then}\ \lim_{x \to 0} \ln y &= \lim_{x \to 0}x\ln (\sin x)\ \ \  (0\cdot \infty \ \text{form}) \\
&=\lim_{x \to 0} \frac{\ln (\sin x)}{x^{-1}}\ \ \ \left(\frac{\infty}{\infty}\ \text{form}\right) \\
&\underset{\lh}{=}\lim_{x \to 0} \frac{\frac{\cos x}{\sin x}}{-\frac{1}{x^{2}}}=\lim_{x \to 0} \frac{-x^{2}\cos x}{\sin x}\ \ \   \left(\frac{0}{0}\ \text{form}\right) \\
                                  &\underset{\lh}=\lim_{x \to 0} \frac{-2x\cos x+x^{2}\sin x}{\cos x}= \frac{0}{1}=0
\end{align*}
Finally exponentiate the result $\displaystyle \lim_{x \to 0} \ln y=0$ to get the desired limit:
\[\lim_{x\to0} (\sin x)^{x}=\lim_{x\to0}y=\lim_{x\to0}e^{\ln y}=e^{\lim_{x\to0}\ln y}=e^0=1\]
\item As $\displaystyle x \to \infty, (\ln x)^{\frac{1}{x}} \to \infty^{0}$ indeterminate form. \\
Let $\displaystyle y=(\ln x)^{\frac{1}{x}} \implies \ln y=\ln\left[ (\ln x)^{\frac{1}{x}}\right]= \frac{1}{x}\ln (\ln x)= \frac{\ln (\ln x)}{x}$
\begin{align*}
\text{Therefore}\  \lim_{x \to \infty} \ln y &= \lim_{x \to \infty}=\frac{\ln (\ln x)}{x} \ \ \ \left(\frac{\infty}{\infty}\ \text{form}\right) \\
&\underset{\lh}=\lim_{x \to \infty} \frac{\frac{1}{\ln x}\cdot \frac{1}{x}}{1}= \lim_{x \to \infty}\frac{1}{x\ln x}=\frac{1}{\infty}=0
\end{align*}
Exponentiating we have that $\displaystyle\lim_{x \to \infty} \ln y=0$ implies:
\[\lim_{x\to\infty}(\ln x)^{\frac{1}{x}} =\lim_{x \to \infty} y=\lim_{x\to\infty}e^{\ln y}=e^{\lim_{x\to\infty}\ln y}=e^0=1\]

\item As $\displaystyle x \to 1, (2x-1)^{\frac{1}{\ln x}}\to 1^{\infty}$ indeterminate form. \\
Let $\displaystyle y=(2x-1)^{\frac{1}{\ln x}} \implies \ln y= \ln\left[ (2x-1)^{\frac{1}{\ln x}}\right]= \frac{\ln (2x-1)}{\ln x}$
\begin{align*}
\text{Therefore}\  \lim_{x \to 1} \ln y &= \lim_{x \to 1} \frac{\ln (2x-1)}{\ln x}  \ \ \ \left(\frac{0}{0}\ \text{form}\right) \\ 
&\underset{\lh}= \lim_{x \to 1} \frac{\frac{2}{2x-1}}{\frac{1}{x}}= \lim_{x \to 1} \frac{2x}{2x-1} \\
                                        &= \frac{2(1)}{2(1)-1}=\frac{2}{2-1}=\frac{2}{1}=2
\end{align*}
Then $\displaystyle\lim_{x \to 1}\ln y= 2$ implies:
\[\lim_{x \to 1} (2x-1)^{\frac{1}{\ln x}}=\lim_{x\to1}y=\lim_{x\to1}e^{\ln y}=e^{\lim_{x\to1}\ln y}=e^2\]
\end{enumerate}
\end{solution*}
\textbook{\clearpage}
\begin{furtherquestions}
  Evaluate the following limits:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\mylim{x}{\infty}\left(1+e^x\right)^{e^{-x}}$
    \item $\displaystyle\mylim{x}{0^+}\left(e^x-1\right)^x$
    \item $\displaystyle\mylim{x}{\infty}\left(1+\frac{3}{x}+\frac{5}{x^2}\right)^x$
    \item $\displaystyle\mylim{x}{1^-}\left(1-x\right)^{\ln x}$
    \item $\displaystyle\mylim{x}{\frac{\pi}{2}^-}(\tan x)^{\cos x}$
    \end{enumerate}
    ~% ensures ragged bottom of second column
  \end{multicols}
\end{furtherquestions}
\end{example}

\begin{exercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/lhopitals-rule.tex}
\end{exercise}

\begin{reviewexercise}
\input{INVERSES_AND_OTHER_FUNCTIONS/inverse-and-other-functions-review.tex}
\end{reviewexercise}

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