Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 112

Differentiate $\displaystyle f(x) = \frac{7 - \frac{3}{2x}}{\frac{4}{x^2} + 5}$

If we simplify the function first before we take the derivative, we get

$
\begin{equation}
\begin{aligned}
f(x) &= \frac{\frac{14x - 3}{2x}}{\frac{4 + 5x^2}{x^2}}\\
\\
&= \frac{14x - 3}{2x} \left( \frac{x^2}{4 + 5x^2} \right)\\
\\
&= \frac{x(14x - 3)}{2(4 + 5x^2)}\\
\\
&= \frac{14x^2 - 3x}{8 + 10x^2}
\end{aligned}
\end{equation}
$


Now, by using Quotient Rule, we get

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{(8 + 10x^2) \cdot \frac{d}{dx} (14x^2 - 3x) - (14x^2 - 3x) \cdot \frac{d}{dx} (8 + 10x^2) }{(8 + 10x^2)^2}\\
\\
&= \frac{(8 +10x^2) (28x -3) - (14x^2 - 3x)(20x)}{(8 + 10x^2)^2}\\
\\
&= \frac{224x - 24 + 280x^3 - 30x^2 - 280x^3 + 60x^2}{(8 + 10x^2)^2}\\
\\
&= \frac{30x^2 + 224x - 24}{(8 + 10x^2)^2}
\end{aligned}
\end{equation}
$