first use the appropriate properties of logarithms to rewrite f(x), and then find f(x). f(x)=20x + ln…

Explanation:

Step1: Apply logarithm property

Recall that $\ln(ab)=\ln a+\ln b$. So for $\ln(20x)$, we have $\ln(20x)=\ln20+\ln x$. Then $f(x)=20x+\ln20+\ln x$.

Step2: Differentiate term - by - term

The derivative of $20x$ with respect to $x$ is $20$ (using the power rule $\frac{d}{dx}(ax)=a$ where $a = 20$). The derivative of a constant $\ln20$ with respect to $x$ is $0$. The derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$ (using the formula $\frac{d}{dx}(\ln x)=\frac{1}{x}$). So $f^{\prime}(x)=20 + 0+\frac{1}{x}=20+\frac{1}{x}$.

Answer:

Rewrite $f(x)$: $20x+\ln20+\ln x$ $f^{\prime}(x)$: $20+\frac{1}{x}$