Exam 1

1. Question

   What is the derivative of $f(x)=x^7{e}^{3.9x}$, evaluated at $x=0.51$?
   

   Solution

   Using the product rule for $f(x)=g(x)·h(x)$, where $g(x):=x^7$ and $h(x):=e^{3.9x}$, we obtain
   

   ${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{f\text{'}(x)}& =\hfill & [g(x)·h(x)]\text{'}=g\text{'}(x)·h(x)+g(x)·h\text{'}(x)\hfill \\ \multicolumn{1}{c}{}& =\hfill & 7x^{7-1}·e^{3.9x}+x^7·e^{3.9x}·3.9\hfill \\ \multicolumn{1}{c}{}& =\hfill & e^{3.9x}·(7x^6+3.9x^7)\hfill \\ \multicolumn{1}{c}{}& =\hfill & e^{3.9x}·x^6·(7+3.9x).\hfill \end{array}}$

   
   Evaluated at $x=0.51$, the answer is
   

   ${\displaystyle e^{3.9·0.51}·0.{51}^6·(7+3.9·0.51)=1.155964.}$

   
   Thus, rounded to two digits we have $f\text{'}(0.51)=1.16$.