Show that


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


We have,

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

$\therefore \frac{d y}{d x}=\frac{1}{1+x}-\frac{(2+x)(2)-2 x(1)}{(2+x)^{2}}=\frac{1}{1+x}-\frac{4}{(2+x)^{2}}=\frac{x^{2}}{(1+x)(2+x)^{2}}$

Now, $\frac{d y}{d x}=0$

$\Rightarrow \frac{x^{2}}{(1+x)(2+x)^{2}}=0$

$\Rightarrow x^{2}=0 \quad[(2+x) \neq 0$ as $x>-1]$

$\Rightarrow x=0$

Since $x>-1$, point $x=0$ divides the domain $(-1, \infty)$ in two disjoint intervals i.e., $-10$.

When $-1

$x<0 \Rightarrow x^{2}>0$

$x>-1 \Rightarrow(2+x)>0 \Rightarrow\left(2+x^{2}\right)>0$

$\therefore y^{\prime}=\frac{x^{2}}{(1+x)(2+x)^{2}}>0$

Also, when $x>0$ :

$x>0 \Rightarrow x^{2}>0,(2+x)^{2}>0$

$\therefore y^{\prime}=\frac{x^{2}}{(1+x)(2+x)^{2}}>0$

Hence, function f is increasing throughout this domain.

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