## Exam 1

1. #### Question

What is the derivative of $f\left(x\right)={x}^{7}{e}^{3.9x}$, evaluated at $x=0.61$?

1. $3.89$
2. $5.22$
3. $4.23$
4. $6.86$
5. $3.01$

#### Solution

Using the product rule for $f\left(x\right)=g\left(x\right)·h\left(x\right)$, where $g\left(x\right):={x}^{7}$ and $h\left(x\right):={e}^{3.9x}$, we obtain
 $\begin{array}{ccc}\multicolumn{1}{c}{f\text{'}\left(x\right)}& =\hfill & \left[g\left(x\right)·h\left(x\right)\right]\text{'}=g\text{'}\left(x\right)·h\left(x\right)+g\left(x\right)·h\text{'}\left(x\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & 7{x}^{7-1}·{e}^{3.9x}+{x}^{7}·{e}^{3.9x}·3.9\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{3.9x}·\left(7{x}^{6}+3.9{x}^{7}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{3.9x}·{x}^{6}·\left(7+3.9x\right).\hfill \end{array}$

Evaluated at $x=0.61$, the answer is
 ${e}^{3.9·0.61}·0.{61}^{6}·\left(7+3.9·0.61\right)=5.215814.$

Thus, rounded to two digits we have $f\text{'}\left(0.61\right)=5.22$.

1. False
2. True
3. False
4. False
5. False