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

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
$y=\log(1+x)-\large\frac{2x}{2+x}$$,x > -1 Let f(x)=\log(1+x)-\large\frac{2x}{2+x} Differentiating w.r.t x we get, f'(x)=\large\frac{1}{1+x}-$$2\bigg[\large\frac{(x+2).1-x.1}{(x+2)^2}\bigg]$
Step 2:
On simplifying we get,
$\qquad=\large\frac{1}{(1+x)}-\frac{4}{(x+2)^2}$
$\qquad=\large\frac{(x+2)^2-4(1+x)}{(x+1)(x+2)^2}$
$\qquad=\large\frac{x^2}{(x+1)(x+2)^2}$
Step 3:
For $f(x)$ to be increasing.
$f'(x) > 0$
$\Rightarrow \large\frac{x^2}{(x+1)(x+2)^2}$$> 0 But \large\frac{x^2}{(x+2)^2} >$$ 0$
Hence $\large\frac{1}{(x+1)} $$>0 \Rightarrow x > -1 Hence y=\log x-$$\large\frac{2x}{x+2}$ is an increasing function of $x$ for all values of $x > -1$