
\begin{question}
What is the derivative of $f(x) = x^{7} e^{3.9x}$, evaluated at $x = 0.51$?
\end{question}

\begin{solution}
Using the product rule for $f(x) = g(x) \cdot h(x)$, where $g(x) := x^{7}$ and $h(x) := e^{3.9x}$, we obtain
\begin{eqnarray*}
f'(x) & = & [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\
      & = & 7 x^{7 - 1} \cdot e^{3.9x} + x^{7} \cdot e^{3.9x} \cdot 3.9 \\
      & = & e^{3.9x} \cdot(7 x^6 + 3.9 x^{7}) \\
      & = & e^{3.9x} \cdot x^6 \cdot (7 + 3.9x).
\end{eqnarray*}
Evaluated at $x = 0.51$, the answer is
\[ e^{3.9\cdot 0.51} \cdot 0.51^6 \cdot (7 + 3.9\cdot 0.51) = 1.155964. \]
Thus, rounded to two digits we have $f'(0.51) = 1.16$.
\end{solution}

%% \extype{num}
%% \exsolution{1.16}
%% \exname{derivative exp}
%% \extol{0.01}
