Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 30

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



$
\begin{equation}
\begin{aligned}

y' =& \frac{(x^3 + x - 2) \displaystyle \frac{d}{dx} (x + 1) - \left[ (x + 1) \frac{d}{dx} (x^3 + x - 2) \right]}{(x^3 + x - 2)^2}
&& \text{Apply Quotient Rule}
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\\
y' =& \frac{(x^3 + x - 2)(1) - (x + 1) (3x^2 + 1)}{(x^3 + x - 2)^2}
&& \text{Expand the equation}
\\
\\
y' =& \frac{x^3 - \cancel{x} - 2 - 3x^3 - \cancel{x} - 3x^2 - 1}{(x^3 + x - 2)^2}
&& \text{Combine like terms}
\\
\\
y' =& \frac{-2x^3 - 3x^2 - 3}{(x^3 + x - 2)^2}
&&
\\
\\

\end{aligned}
\end{equation}
$