In flaviobarros/exams: Automatic Generation of Exams in R

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 $x = r c$?

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 at $x = r c$, 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 $f'(r c) = r fmt(res)$.

answerlist(ifelse(sc$solutions, "True", "False"), markup = "markdown")

Meta-information

extype: schoice exsolution: r mchoice2string(sc$solutions) exname: derivative exp

flaviobarros/exams documentation built on May 28, 2019, 8:39 p.m.