Question

Show that $f\left(x\right)=2x+{\cot }^{-1}x+\log (\sqrt{1+x^2}-x)$ is increasing in $R$.

Answer 1

Consider the given function.

$f\left(x\right)=2x+{\cot }^{-1}x+\log (\sqrt{1-x^2}-x)$

On differentiating with respect to $x$, we get

$f^{\prime }\left(x\right)=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}\left(\frac{d(\sqrt{1+x^2}-x)}{dx}\right)$

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

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

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

$f^{\prime }\left(x\right)=2-\left[\frac{1+\sqrt{1+x^2}}{1+x^2}\right]>0$

Hence, $f\left(x\right)$ is increasing in $R$.