Question

question 3. 1 point determine $r(x)$ when $r(x)=0.09e^{x}ln(x)$. select the correct answer below: $r(x)=0.09e^{x}ln(x)+\frac{0.09e^{x}}{x}$ $r(x)=\frac{0.09e^{x}}{ln(x)}$ $r(x)=\frac{0.09e^{x}}{x}$ $r(x)=0.09xe^{x - 1}ln(x)+\frac{0.09e^{x}}{x}$

Explanation:

Step1: Apply product - rule

$(uv)' = u'v+uv'$, where $u = 0.09e^{x}$ and $v=\ln(x)$.
$u'=0.09e^{x}$, $v'=\frac{1}{x}$.

Step2: Calculate $r'(x)$

$r'(x)=0.09e^{x}\ln(x)+\frac{0.09e^{x}}{x}$

Answer:

$r'(x)=0.09e^{x}\ln(x)+\frac{0.09e^{x}}{x}$