Show that $y=\log (1+x)-\frac{2 x}{2+x}, x>-1,$ is an increasing function of $x$ throughout its domain.

(A) We have,$y=\log (1+x)-\frac{2 x}{2+x}$.
Differentiating with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$= \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2}$
$= \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$= \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$
$= \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2}$
$= \frac{x^2}{(1+x)(2+x)^2}$.
Since $x > -1$,we have $(1+x) > 0$ and $(2+x)^2 > 0$ for all $x$ in the domain.
Also,$x^2 \ge 0$ for all real $x$.
Thus,$\frac{d y}{d x} = \frac{x^2}{(1+x)(2+x)^2} \ge 0$ for all $x > -1$.
Since the derivative is non-negative and only zero at the isolated point $x=0$,the function $y$ is an increasing function throughout its domain $(-1, \infty)$.