y = \frac{ln x}{x}

$D = \{x \in ℝ | x > 0\}$
Nullstelle: $x = 1$
Polstelle: $x = 0$
Ableitungen:
$y^{'} = \frac{1}{x^{2}} - \frac{ln x}{x^{2}}$
$y^{″} = - \frac{2}{x^{3}} - \frac{1 - 2 ln x}{x^{3}} = \frac{- 3}{x^{3}} + \frac{2 ln x}{x^{3}}$
$y^{‴} = \frac{9}{x^{4}} + \frac{+ 2 \frac{x^{3}}{x} - 3 x^{2} \cdot 2 ln x}{)}$

$= \frac{9}{x^{4}} + \frac{2 - 3 \cdot 2 ln x}{x^{4}}$

$= \frac{11 - 6 ln x}{x^{4}}$

.
Extremwerte: $y^{'} = 0 \Rightarrow \frac{1}{x^{2}} = \frac{ln x}{x^{2}}$
$für x \neq 0 \Rightarrow 1 = l n x$
$e^{1} = e^{ln x} \Rightarrow e^{1} = x$
$y^{″} (e) = \frac{- 3}{e^{3}} + \frac{2 ln e}{e^{3}} = \frac{- 1}{e^{3}} < 0 \to Maximum bei x = e$
Wendepunkte: $y^{″} = \frac{- 3}{x^{3}} + \frac{2 ln x}{x^{3}} = 0$
$2 ln x = 3$
$ln x = \frac{3}{2} \to x = e^{\frac{3}{2}}$
$y^{‴} (e^{\frac{3}{2}}) = \frac{11}{e^{6}} - \frac{6 ln e^{\frac{3}{2}}}{e^{6}}$
$= \frac{11}{e^{6}} - \frac{6 \cdot \frac{3}{2} \cdot ln e}{e^{6}}$
$= \frac{11}{e^{6}} - \frac{9}{e^{6}} \neq 0 ✓$
Zeichnung
.

Abbildung 2: $y = \frac{ln x}{x}$