Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 42

Find $f''(e)$ if $\displaystyle f(x) = \frac{\ln x}{x}$


$\begin{equation} \begin{aligned} \text{if } f(x) =& \frac{\ln x}{x}, \text{ then by using Quotient Rule} \\ \\ f'(x) =& \frac{\displaystyle x \cdot \frac{d}{dx} (\ln x) - (\ln x) \cdot \frac{d}{dx} (x) }{x^2 } \\ \\ f'(x) =& \frac{\displaystyle x \left( \frac{1}{x} \right) - \ln x (1)}{x^2} \\ \\ f'(x) =& \frac{1 - \ln x}{x^2} \end{aligned} \end{equation}$


Again, by using Quotient Rule


$\begin{equation} \begin{aligned} f''(x) =& \frac{\displaystyle x^2 \cdot \frac{d}{dx} (1 - \ln x) - (1 - \ln x) \cdot \frac{d}{dx} (x^2)}{(x^2)^2 } \\ \\ f''(x) =& \frac{\displaystyle x^2 \left( - \frac{1}{x} \right) - (1 - \ln x)(2x)}{x^4} \\ \\ f''(x) =& \frac{x(-1 - 2 + 2 \ln x)}{x^4} \\ \\ f''(x) =& \frac{-3 + 2 \ln x}{x^3} \end{aligned} \end{equation}$


Thus,


$\begin{equation} \begin{aligned} f''(e) =& \frac{-3 + 2 \ln (e)}{e^3} \\ \\ f''(e) =& \frac{-3 + 2 (1)}{e^3} \\ \\ f''(e) =& \frac{-1}{e^3} \end{aligned} \end{equation}$