<>= sc <- NULL while(is.null(sc)) { ## parameters a <- sample(2:9, 1) b <- sample(seq(2, 4, 0.1), 1) c <- sample(seq(0.6, 0.9, 0.01), 1) ## solution res <- exp(b * c) * (a * c^(a-1) + b * c^a) ## schoice err <- c(a * c^(a-1) * exp(b * c), 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 = err, range = rg, delta = 0.1) } @ \begin{question} What is the derivative of $f(x) = x^{\Sexpr{a}} e^{\Sexpr{b}x}$, evaluated at $x = \Sexpr{c}$? <>= answerlist(sc$questions) @ \end{question} \begin{solution} Using the product rule for$f(x) = g(x) \cdot h(x)$, where$g(x) := x^{\Sexpr{a}}$and$h(x) := e^{\Sexpr{b}x}$, we obtain \begin{eqnarray*} f'(x) & = & [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ & = & \Sexpr{a} x^{\Sexpr{a} - 1} \cdot e^{\Sexpr{b}x} + x^{\Sexpr{a}} \cdot e^{\Sexpr{b}x} \cdot \Sexpr{b} \\ & = & e^{\Sexpr{b}x} \cdot(\Sexpr{a} x^\Sexpr{a-1} + \Sexpr{b} x^{\Sexpr{a}}) \\ & = & e^{\Sexpr{b}x} \cdot x^\Sexpr{a-1} \cdot (\Sexpr{a} + \Sexpr{b}x). \end{eqnarray*} Evaluated at$x = \Sexpr{c}$, the answer is $e^{\Sexpr{b}\cdot \Sexpr{c}} \cdot \Sexpr{c}^\Sexpr{a-1} \cdot (\Sexpr{a} + \Sexpr{b}\cdot \Sexpr{c}) = \Sexpr{fmt(res, 6)}.$ Thus, rounded to two digits we have$f'(\Sexpr{c}) = \Sexpr{fmt(res)}$. <>= answerlist(ifelse(sc$solutions, "True", "False")) @ \end{solution} %% \extype{schoice} %% \exsolution{\Sexpr{mchoice2string(sc\$solutions)}} %% \exname{derivative exp}