Question

Show that $f\left(x\right)=\frac{x}{\sqrt{1+x}}-\ln (1+x)$ is an increasing function for x > -1.

Answer 1

$X=\frac{x}{\sqrt{1+x}}\ln (1+x)$

$Y^1=\frac{1}{\sqrt{1+x}}+x\left(\frac{-1}{2}\right)(x+1{)}^{\frac{-3}{2}}-\frac{1}{(1+x)}$

$Y^1=0$

$\Rightarrow \frac{1}{\sqrt{1+x}}-\frac{x}{2(x+1{)}^{\frac{3}{2}}}-\frac{1}{(1+x)}=0$

$\Rightarrow 1-\frac{x}{(x+1{)}^{\frac{3}{2}}}-\frac{1}{\sqrt{1+x}}=0$

$\Rightarrow -1\frac{x}{x+1}-\frac{1}{\sqrt{1+x}}=0$

$\Rightarrow \frac{x+1-x}{x+1}=\frac{1}{\sqrt{x+1}}$

$\Rightarrow \sqrt{x+1}=x+1$

$\Rightarrow x+1=x^2+2x+1$

$\Rightarrow x^2+x=0$

$x(x+1)=0$

Hence, function is increasing for $x>-1$