Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 24

Differentiate $\displaystyle y = \frac{3x^4 + 2x}{x^3 - 1}$

By applying Long Division first before differentiating, we get



$\displaystyle y = \frac{3x^4 + 2x}{x^3 -1} = 3x + \frac{5x}{x^3 - 1}$


Then,

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left[ 3x + \frac{5x}{x^3 - 1} \right] = \frac{d}{dx} (3x) + \frac{d}{dx} \left( \frac{5x}{x^3 - 1} \right)\\
\\
&= 3 + \left[ \frac{(x^3 - 1) \cdot \frac{d}{dx} (5x) - (5x) \cdot \frac{d}{dx}(x^3 - 1) }{(x^3 - 1)^2} \right]\\
\\
&= 3 + \left[ \frac{(x^3 - 1)(5) - (5x) (3x^2) }{(x^3 - 1)^2} \right]\\
\\
&= 3 + \left[ \frac{5x^3 - 5 - 15x^3}{(x^3 - 1)^2} \right]\\
\\
&= 3 + \left[ \frac{-10x^3 - 5}{(x^3 - 1)^2} \right]\\
\\
&= \frac{3(x^3 - 1)^2 - 10x^3 - 5}{(x^3 - 1)^2}\\
\\
&= \frac{3\left[(x^3)^2 - 2(x^3)(1) + (-1)^2 \right] -10x^3 -5 }{(x^3 - 1)^2}\\
\\
&= \frac{3 \left[ x^6 - 2x^3 + 1 \right] - 10x^3 - 5}{(x^3 - 1)^2}\\
\\
&= \frac{3x^6 - 6x^3 + 3 - 10x^3 - 5}{(x^3 - 1)^2}\\
\\
&= \frac{3x^6 - 16x^3 - 2}{(x^3 - 1)^2}
\end{aligned}
\end{equation}
$