Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 16

Take the derivative of $\displaystyle g(x) = \frac{3x^7 - x^3}{x}$: first, use the Quotient Rule;
then, by dividing the expression before differentiating. Compare your results as a check.
By using Quotient Rule,

$
\begin{equation}
\begin{aligned}
g'(x) &= \frac{x \cdot \frac{d}{dx} (3x^7 - x^3) - (3x^7 - x^3) \cdot \frac{d}{dx} (x)}{(x)^2}\\
\\
&= \frac{x(21x^6 - 3x^2) - (3x^7 - x^3)(1)}{x^2}\\
\\
&= \frac{21x^7 - 3x^3 - 3x^7 + x^3}{x^2}\\
\\
&= \frac{18x^7 - 2x^3}{x^2}\\
\\
&= 18x^{7 - 2} - 2x^{3 - 2}\\
\\
&= 18x^5 - 2x
\end{aligned}
\end{equation}
$


By dividing the expression first,

$
\begin{equation}
\begin{aligned}
g(x) &= \frac{3x^7 - x^3}{x} \\
\\
&= \frac{3x^7}{x} - \frac{x^3}{x} \\
\\
&= 3x^{7 - 1} - x^{3 - 1}\\
\\
&= 3x^6 - x^2\\
\\
g'(x) &= \frac{d}{dx} (3x^6 - x^2) \\
\\
&= 18x^5 - 2x
\end{aligned}
\end{equation}
$


Both results agree.