Question

question
to the nearest tenth, what is $f(k)$ if $f(x)=0.09e^{x}ln(x)$ and $k = 8$?
sorry, thats incorrect. try again?

Explanation:

Step1: Apply product - rule for differentiation

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 0.09e^{x}$ and $v=\ln(x)$. The derivative of $u$ with respect to $x$ is $u'=0.09e^{x}$ (since the derivative of $e^{x}$ is $e^{x}$ and the constant multiple rule), and the derivative of $v$ with respect to $x$ is $v'=\frac{1}{x}$. Then $f'(x)=0.09e^{x}\cdot\frac{1}{x}+0.09e^{x}\ln(x)=0.09e^{x}(\frac{1}{x}+\ln(x))$.

Step2: Substitute $x = k = 8$

Substitute $x = 8$ into $f'(x)$. We have $f'(8)=0.09e^{8}(\frac{1}{8}+\ln(8))$. First, calculate $e^{8}\approx2980.957987$. Then $\frac{1}{8}+\ln(8)=\frac{1}{8}+2.079442\approx0.125 + 2.079442=2.204442$. Now, $f'(8)=0.09\times2980.957987\times2.204442$. $0.09\times2980.957987 = 268.286219$. Then $268.286219\times2.204442\approx591.5$.

Answer:

$591.5$