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Given the function $f(x) = \frac{\log x}{x}$, we need to find its derivative $f'(x)$ using the quotient rule. The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
Here, $u(x) = \log x$ and $v(x) = x$. Thus, $u'(x) = \frac{1}{x}$ and $v'(x) = 1$. Applying the quotient rule, we get: $$f'(x) = \frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$$
To find the intervals where $f(x)$ is increasing or decreasing, we need to find the critical points by setting $f'(x) = 0$. $$\frac{1 - \log x}{x^2} = 0$$ This implies $1 - \log x = 0$, so $\log x = 1$. Therefore, $x = e$ (since we are considering the natural logarithm).
We need to analyze the sign of $f'(x)$ in the intervals $(0, e)$ and $(e, \infty)$. Note that the domain of $f(x)$ is $x > 0$. - For $0 < x < e$, $\log x < 1$, so $1 - \log x > 0$. Since $x^2 > 0$, $f'(x) = \frac{1 - \log x}{x^2} > 0$. Thus, $f(x)$ is strictly increasing in the interval $(0, e)$. - For $x > e$, $\log x > 1$, so $1 - \log x < 0$. Since $x^2 > 0$, $f'(x) = \frac{1 - \log x}{x^2} < 0$. Thus, $f(x)$ is strictly decreasing in the interval $(e, \infty)$.
Final Answer: f(x) is strictly increasing in (0, e) and strictly decreasing in (e, ∞)