Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 28

Determine $y'$ and $y''$ of $\displaystyle y = \frac{\ln x}{x^2}$
Solving for $y'$

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


Solving for $y''$

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} \left( \frac{1-2 \ln x}{x^3} \right)\\
\\
y'' &= \frac{x^3 \frac{d}{dx}(1- 2 \ln x) - (1 -2 \ln x) \frac{d}{dx} (x^3)}{(x^3)^2}\\
\\
y'' &= \frac{(x^3)\left( \frac{-2}{x} \right)-(1-2\ln x)(3x^2) }{x^6}\\
\\
y'' &= \frac{-2x^2 - 3x^2 + 6x^2 \ln x}{x^6}\\
\\
y'' &= \frac{-5x^2 + 6x^2 \ln x}{x^6}\\
\\
y'' &= \frac{x^2(6 \ln x - 5)}{x^6}\\
\\
y'' &= \frac{6 \ln - 5}{x^4}
\end{aligned}
\end{equation}
$