Question

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

Answer 1

Given:
$y=\log (1+x)-\frac{2x}{2+x}$

Differentiating w.r.t $x,$

$\frac{dy}{dx}=\frac{1}{1+x}-\frac{2(2+x)-2x\left(1\right)}{(2+x{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{4+2x-2x}{(2+x{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{(x+2{)}^2-4(x+1)}{(x+1)(x+2{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{x^2+4x+4-4x-4}{(x+1)(x+2{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{(x+1)(x+2{)}^2}$

For $x>-1$

$\Rightarrow \frac{dy}{dx}>0$

Hence, the function is an increasing function for $x>-1$.