{r data generation, echo = FALSE, results = "hide"} 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) }  Question ======== What is the derivative of $f(x) = x^{r a} e^{r b x}$, evaluated at $x = r c$? {r questionlist, echo = FALSE, results = "asis"} answerlist(sc$questions, markup = "markdown")  Solution ======== Using the product rule for$f(x) = g(x) \cdot h(x)$, where$g(x) := x^{r a}$and$h(x) := e^{r b x}, we obtain \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= r a x^{r a - 1} \cdot e^{r b x} + x^{r a} \cdot e^{r b x} \cdot r b \\ &= e^{r b x} \cdot(r a x^r a-1 + r b x^{r a}) \\ &= e^{r b x} \cdot x^r a-1 \cdot (r a + r b x). \end{aligned} Evaluated atx = r c$, the answer is $$e^{r b \cdot r c} \cdot r c^r a-1 \cdot (r a + r b \cdot r c) = r fmt(res, 6).$$ Thus, rounded to two digits we have$f'(r c) = r fmt(res)$. {r solutionlist, echo = FALSE, results = "asis"} answerlist(ifelse(sc$solutions, "True", "False"), markup = "markdown")  Meta-information ================ extype: schoice exsolution: r mchoice2string(sc\$solutions) exname: derivative exp