<<echo=FALSE, results=hide>>=
sc <- NULL
while(is.null(sc)) {
## parameters
a <- sample(2:9, 1)
b <- sample(2:4, 1)/10
c <- sample(6:9, 1)/10
## solution
res <- exp(b * c) * (a * c^(a-1) + b * c^a)
## schoice
err1 <- a * c^(a-1) * exp(b * c)
err2 <- a * c^(a-1) * exp(b * c) + c^a * exp(b * c)
rg <- if(res < 4) c(0.5, 5.5) else res * c(0.5, 1.5)
sc <- num_to_schoice(res, wrong = c(err1, err2), range = rg, delta = 0.1)
}
@

\begin{question}
What is the derivative of $f(x) = x^{\Sexpr{a}} e^{\Sexpr{b} x}$, evaluated at $x = \Sexpr{c}$?

<<echo=FALSE, results=tex>>=
answerlist(sc$questions)
@
\end{question}

\begin{solution}
Using the product rule we obtain
\[ f'(x) = e^{\Sexpr{b} x} \cdot (\Sexpr{a} \cdot x^\Sexpr{a-1} + \Sexpr{b} \cdot x^\Sexpr{a}). \]
Evaluated at $x = \Sexpr{c}$ and rounded to two digits the answer is
$f'(\Sexpr{c}) = \Sexpr{fmt(res, 6)} \approx \Sexpr{fmt(res, 2)}$.

<<echo=FALSE, results=tex>>=
answerlist(ifelse(sc$solutions, "True", "False"))
@
\end{solution}

%% \extype{schoice}
%% \exsolution{\Sexpr{mchoice2string(sc$solutions)}}
%% \exname{exp derivative}