fixed typos corbin pointed out

public/textbook/basicsOfDerivatives.tex

@@ -548,7 +548,7 @@ \section{Slopes of Tangent Lines via Limits}
           \addplot[color=penColor,fill=background,only marks,mark=*] coordinates{(4.5,0)};  %% open hole
         \end{axis}
 \end{tikzpicture}
-Answer 2.1.6: (c) a sketch of $f'(x)$.
+Answer 3.1.6: (c) a sketch of $f'(x)$.
 \end{marginfigure}
 \end{answer}
 \end{exercise}
@@ -1142,7 +1142,7 @@ \subsection*{The Derivative of $\textit{e}^\textit{x}$}
           \node at (axis cs:-.9,7) [anchor=west] {\color{penColor2!50!background}$(cf(x))'$};
         \end{axis}
 \end{tikzpicture}
-Answer 2.2.21.
+Answer 3.2.25.
 \end{marginfigure}
 \end{answer}
 \end{exercise}

public/textbook/chainRule.tex

@@ -157,14 +157,14 @@ \section{The Chain Rule}
 $g=g(x+h)$ and $g_0=g(x)$
 \[
 \frac{f(g(x+h))-f(g(x))}{h} = \left(f'(g(x)) +
-E(g(x))\right)\frac{g(x+h)-g(x)}{h}.
+E(g(x+h))\right)\frac{g(x+h)-g(x)}{h}.
 \]
 Taking the limit as $h$ approaches $0$, we see 
 \begin{align*}
 \lim_{h\to 0}\frac{f(g(x+h))-f(g(x))}{h} &= \lim_{h\to 0}\left(f'(g(x))
-+ E(g(x))\right)\frac{g(x+h)-g(x)}{h}\\
++ E(g(x+h))\right)\frac{g(x+h)-g(x)}{h}\\
 &= \lim_{h\to 0}\left(f'(g(x))
-+ E(g(x))\right)\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\
++ E(g(x+h))\right)\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\
 &= f'(g(x))g'(x).
 \end{align*}
 Hence, $\ddx f(g(x))= f'(g(x))g'(x)$.

public/textbook/curveSketching.tex

@@ -310,8 +310,8 @@ \section{The First Derivative Test}
   $a$, then $f(a)$ is a local maximum.
 \item If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of
   $a$, then $f(a)$ is a local minimum.
-\item If $f'(x)$ has the same sign to the left and right of $f'(a)$,
-  then $f'(a)$ is not a local extremum.
+\item If $f'(x)$ has the same sign to the left and right of $a$,
+  then $f(a)$ is not a local extremum.
 \end{itemize}
 \end{mainTheorem}
 

public/textbook/functions.tex

@@ -733,7 +733,7 @@ \subsection{A Word on Notation}
 \begin{answer}
   $h^{-1}(20)=7$.  This means that a height of 20 meters is achieved
   at 7 seconds in the restricted interval.  In fact, it turns out that
-  $h^{-1}(t) = 7 \cdot (\pi - \arcsin((t-20)/9))/\pi$ when $h$ is
+  $h^{-1}(t) = 7 \cdot (\pi - \arcsin((t-20)/18))/\pi$ when $h$ is
   restricted to the given interval.
 \end{answer}
 \end{exercise}