Exam 1

1. Question

What is the derivative of $f\left(x\right)={x}^{8}{e}^{3.4x}$, evaluated at $x=0.7$?

Solution

Using the product rule for $f\left(x\right)=g\left(x\right)·h\left(x\right)$, where $g\left(x\right):={x}^{8}$ and $h\left(x\right):={e}^{3.4x}$, we obtain
 $\begin{array}{ccc}\multicolumn{1}{c}{f\text{'}\left(x\right)}& =\hfill & \left[g\left(x\right)·h\left(x\right)\right]\text{'}=g\text{'}\left(x\right)·h\left(x\right)+g\left(x\right)·h\text{'}\left(x\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & 8{x}^{8-1}·{e}^{3.4x}+{x}^{8}·{e}^{3.4x}·3.4\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{3.4x}·\left(8{x}^{7}+3.4{x}^{8}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & {e}^{3.4x}·{x}^{7}·\left(8+3.4x\right).\hfill \end{array}$

Evaluated at $x=0.7$, the answer is
 ${e}^{3.4·0.7}·0.{7}^{7}·\left(8+3.4·0.7\right)=9.236438.$

Thus, rounded to two digits we have $f\text{'}\left(0.7\right)=9.24$.