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

Differentiate $\displaystyle y = \frac{3x - 4}{x^3}$
We have $\displaystyle y = \frac{3x}{x^3} - \frac{4}{x^2} = 3x^{-2} - 4x^{-3}$, so

$
\begin{equation}
\begin{aligned}
\frac{dy}{dx} &= \frac{d}{dx} \left( 3x^{-2} - 4x^{-3} \right) \\
\\
&= \frac{d}{dx} \left( 3x^{-2} \right) - \frac{d}{dx} \left( 4x^{-3} \right)\\
\\
&= 3 \cdot \frac{d}{dx} \left( x^{-2} \right) - 4 \cdot \frac{d}{dx} (x ^{-3})\\
\\
&= 3(-2)\left(x^{-2-1}\right) - 4 (-3) \left( x^{-3-1} \right)\\
\\
&= -6x^{-3} + 12x^{-4} \text{ or } \frac{-6}{x^3} + \frac{12}{x^4}
\end{aligned}
\end{equation}
$