is denoted by $f^{(k)}(x)$.    Thus
\begin{align*}
\tag{6}
f'(x)  &= na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \dotsb + 2a_{n-2} x + a_{n-1}, \\
\tag{7}
f''(x) &= n(n-1)a_0 x^{n-2} + (n-1)(n-2)a_1 x^{n-3} + \dotsb + 2a_{n-2},
\end{align*}
etc. Hence we have
\begin{align*} %[** F1: Not aligned w/previous display]
\tag{8}
f(x+h) &= f(x) + f'(x)h + f''(x)  \frac{h^2}{1\cdot 2}
                        + f'''(x) \frac{h^3}{1\cdot 2\cdot 3} \\
  & \qquad + \dotsb + f^{(r)}(x) \frac{h^r}{r!}
           + \dotsb + f^{(n)}(x) \frac{h^n}{n!},
\end{align*}
where $r!$ is the symbol, read $r$~\emph{factorial}, for the product $1\cdot 2\cdot 3\dotsm (r-1)r$.
Here $r$ is a positive integer, but we include the case $r = 0$ by the definition,
$0! = 1$.

This formula~$(8)$ is known as \emph{Taylor's theorem} for the present case of
a polynomial~$f(x)$ of degree~$n$. We call $f'(x)$ the (\emph{first}) \emph{derivative of}~$f(x)$,
and $f''(x)$ the \emph{second derivative} of~$f(x)$, etc. %[** F1: N.B. Inconsistent italicization]
Concerning the fact that
$f''(x)$ is equal to the first derivative of $f'(x)$ and that, in general, the $k$th
derivative $f^{(k)}(x)$ of $f(x)$ is equal to the first derivative of $f^{(k-1)}(x)$, see
Exs.~6--9 of the next set.

In view of~$(8)$, the limit of~$(4)$ as $h$ approaches zero is $f'(x)$. Hence
$f'(x)$ \emph{is the slope of the tangent to the graph of $y=f(x)$ at the point}~$(x, y)$.

In $(5)$ and~$(6)$, let every $a$ be zero except~$a_0$. Thus the derivative of
$a_0 x^n$ is $na_0 x^{n-1}$, and hence is obtained by multiplying the given term by
its exponent~$n$ and then diminishing its exponent by unity. For example,
the derivative of~$2x^3$ is~$6x^2$.

Moreover, the derivative of $f(x)$ is equal to the sum of the derivatives
of its separate terms.    Thus the derivative of $x^3 + 4x^2 - 11$ is $3x^2 + 8x$,
as found also in §~55.

\begin{center}
\textbf{EXERCISES}
\end{center}

% [** F1: \small?]
\begin{itemize}
\item[1.]  Show that the slope of the tangent to $y = 8x^3 - 22x^2 + 13x - 2$ at $(x, y)$ is
$24x^2 - 44x + 13$, and that the bend points are $(0.37, 0.203)$, $(1.46, -5.03)$, approximately.
Draw the graph.

\item[2.]  Prove that the bend points of $y = x^3 - 2x - 5$ are $(.82, -6.09)$, $(-.82, -3.91)$,
approximately.    Draw the graph and locate the real roots.

\item[3.]  Find the bend points of $y = x^3 - 6x^2 + 8x + 8$.    Locate the real roots.

\item[4.]  Locate the real roots of $f(x) = x^4 + x^3 - x - 2 = 0$.

Hints: The abscissas of the bend points are the roots of $f'(x) = 4x^3 + 3x^2 - 1 = 0$.
The bend points of $y = f'(x)$ are $(0, -1)$ and $(-\frac{1}{2}, -\frac{3}{4})$, so that $f'(x)= 0$ has a single real
root (it is just less than $\frac{1}{2}$). The single bend point of $y=f(x)$ is $(\frac{1}{2}, -\frac{37}{16})$, approximately.
\end{itemize}
% [** F1: End of \small]