Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 40

Determine the equations of the tangent lines to the curve $\displaystyle y = \frac{\ln x}{x}$ at the points $(1,0)$ and $\displaystyle \left( e, \frac{1}{e} \right)$. Illustrate by graphing the curve and its tangent lines.

Solving for slope at $(1,0)$

$\begin{equation} \begin{aligned} y' &= \frac{d}{dx}\left( \frac{\ln x}{x} \right)\\ \\ y' &= \frac{x \frac{d}{dx}(\ln x) - (\ln x) \frac{d}{dx}(x) }{x^2}\\ \\ y' &= \frac{\cancel{x}\left( \frac{1}{\cancel{x}} \right) - \ln x(1)}{x^2}\\ \\ y' &= \frac{1 - \ln x}{x^2}\\ \\ y' &= \frac{1 - \ln 1}{(1)^2}\\ \\ y' &= \frac{1- 0}{1}\\ \\ y' &= 1 \end{aligned} \end{equation}$


Solving for slope at $\displaystyle \left( e, \frac{1}{e} \right)$

$\begin{equation} \begin{aligned} y' &= \frac{1-\ln x}{x^2}\\ \\ y' &= \frac{1 - \ln e}{(e)^2}\\ \\ y' &= \frac{1-1}{e^2}\\ \\ y' &= \frac{0}{e^2}\\ \\ y' &= 0 \end{aligned} \end{equation}$


Using point slope form

$\begin{equation} \begin{aligned} y- y_1 &= m(x - x_1)\\ \\ y - \frac{1}{e} &= 0(x - e)\\ \\ y - \frac{1}{e} &= 0\\ \\ y &= \frac{1}{e} \end{aligned} \end{equation}$