Question

1. -/1 points if f(x) = ln(4x + ln(x)), find f(1). f(1) =

Explanation:

Step1: Apply chain - rule

Let $u = 4x+\ln(x)$. Then $f(x)=\ln(u)$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and the derivative of $u = 4x+\ln(x)$ with respect to $x$ is $4+\frac{1}{x}$. By the chain - rule $\frac{df}{dx}=\frac{1}{u}\cdot(4 + \frac{1}{x})=\frac{4+\frac{1}{x}}{4x+\ln(x)}$.

Step2: Substitute $x = 1$

When $x = 1$, substitute into $f'(x)=\frac{4+\frac{1}{x}}{4x+\ln(x)}$. We have $\ln(1)=0$, and $4+\frac{1}{1}=5$, $4\times1+\ln(1)=4$. So $f'(1)=\frac{4 + 1}{4+0}=\frac{5}{4}$.

Answer:

$\frac{5}{4}$