Question

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

Answer 1

Given, $y=log(1+x)-\frac{2x}{(2+x)}$
On differentiating, we get $\frac{dy}{dx}=\frac{d}{dx}\left[log\right(1+x)-\frac{2x}{2+x}]=\frac{1}{1+x}-\frac{(2+x)\frac{d}{dx}\left(2x\right)-2x\frac{d}{dx}(2+x)}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4+2x-2x}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}=\frac{(2x+x{)}^2-4(1+x)}{(1+x)(2+x{)}^2}=\frac{4+x^2+4x-4-4x}{[1+x](2+x{)}^2}=\frac{x^2}{(1+x)(2+x{)}^2}$
When $xϵ(-1,\infty )$, then $\frac{x^2}{2+x^2}>0$ and $(1+x)>0$
$\therefore y>0$ when $x>-1$
Hence, y is an increasing function throughout $(x>-1)$ its domain.