Calculus and Its Applications, Chapter 1, 1.4, Section 1.4, Problem 38

Determine the $f'(x)$ of the function $\displaystyle f(x) = \frac{2x }{x + 1}$

$
\begin{equation}
\begin{aligned}
\frac{f(x + h) - f(x)}{h} &= \frac{\left[ \frac{2(x + 1)}{(x + h) + 1} \right] - \left[ \frac{2x}{(x + 1)} \right]}{h}\\
\\
&= \frac{\frac{2(x + h)(x + 1) - 2x (x + h + 1)}{(x + 1)(x + h + 1)}}{h}
&& \text{Get the LCD}\\
\\
&= \frac{2x^2 + 2x + 2xh + 2h - 2x^2 - 2xh - 2x}{h(x + 1)(x + h + 1)}\\
\\
&= \frac{2h}{h(x + 1)(x + h + 1)}\\
\\
&= \frac{2}{x^2 + xh + x + x + h + 1}\\
\\
&= \frac{2}{x^2 + xh + 2x + h + 1}
\end{aligned}
\end{equation}
$

Thus,

$
\begin{equation}
\begin{aligned}
f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x) }{h} &= \frac{2}{x^2 + x(0) + 2x + 0 + 1}\\
\\
&= \frac{2}{x^2 + 2x + 1} \text{ or } \frac{2}{( x + 1)^2}
\end{aligned}
\end{equation}
$