Question

question determine q(7) to the nearest tenth when q(x)=8e^x ln(x). provide your answer below:

Explanation:

Step1: Apply product - rule.

If $q(x)=8e^{x}\ln(x)$, then $q'(x)=8(e^{x}\ln(x)+\frac{e^{x}}{x})$ by $(uv)' = u'v + uv'$ where $u = 8e^{x}$ and $v=\ln(x)$.

Step2: Substitute $x = 7$.

$q'(7)=8(e^{7}\ln(7)+\frac{e^{7}}{7})\approx8(1096.633\times1.946+\frac{1096.633}{7})\approx8(2134.04 + 156.66)=8\times2290.7=18325.6$

Answer:

$18325.6$