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

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