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

We have, $y = \log \left( {1 + x} \right) - \cfrac{{2x}}{{2 + x}},x > - 1$ …(i)

The domain of $y$ is $\left( { - 1,\infty } \right)$

Differentiating (i), w.r.t. $x$, we get
$\cfrac{{dy}}{{dx}} = \cfrac{1}{{1 + x}} - 2\cfrac{d}{{dx}}\left( {\cfrac{x}{{2 + x}}} \right) = \cfrac{1}{{1 + x}} - 2\left\{ {\cfrac{{\left( {2 + x} \right) - x\left( {0 + 1} \right)}}{{{{\left( {2 + x} \right)}^2}}}} \right\}$

$\cfrac{1}{{1 + x}} - \cfrac{4}{{{{\left( {2 + x} \right)}^2}}} = \cfrac{{{{\left( {2 + x} \right)}^2} - 4\left( {1 + x} \right)}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$

$= \cfrac{{{x^2}}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}\forall \,\,x > - 1$

Now ${x^2} \ge 0$, ${\left( {2 + x} \right)^2} \ge 0$ (Being perfect squares)
and $(1 + x) > 0\,\,\forall \,\,x > - 1$
$\Rightarrow \cfrac{{dy}}{{dx}} \ge 0$ for all $x > - 1$

Therefore we can say that $y$ is an increasing function of $x$ throughout its domain.