Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 32

Find the $y''$ of $\displaystyle y = \frac{3}{x^4} - \frac{1}{x}$
We have $y = 3x^{-4} - x^{-1}$, so

$
\begin{equation}
\begin{aligned}
y' &= 3 \cdot \frac{d}{dx} (x^{-4}) - \frac{d}{dx} (x^{-1})\\
\\
y' &= 3(-4) x^{-4-1} - (-1) x^{-1-1}\\
\\
y' &= -12x^{-5} + x^{-2}\\
\\
\text{or}\\
\\
&= \frac{-12}{x^5} + \frac{1}{x^2}
\end{aligned}
\end{equation}
$

Then,

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} \left[ -12x^{-5} + x^{-2} \right]\\
\\
y'' &= -12 \cdot \frac{d}{dx} (x^{-5}) + \frac{d}{dx} (x^{-2})\\
\\
y'' &= -12(-5) x^{-5 - 1} + (-2)x^{-2-1}\\
\\
y'' &= 60x^{-6} - 2x^{-3}\\
\\
\text{or}\\
\\
&= \frac{60}{x^6} - \frac{2}{x^3}
\end{aligned}
\end{equation}
$