Question

Prove the following identity:
$f^{\prime }\left(x\right)+f\left(x\right)+f\left(\frac{1}{x}\right)-\frac{1}{x}=0$ if $f\left(x\right)=\ln x$

Answer 1

$f\left(x\right)=lnx$

$f^{\prime }\left(x\right)=\frac{1}{x}$

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

$=\frac{1}{x}+lnx+ln{x}^{-1}-\frac{1}{x}$

$=\frac{1}{x}+lnx-lnx-\frac{1}{x}[ln{m}^n=n|nm]$

$=0$ (Hence proved)

So, $f^{\prime }\left(x\right)+f\left(x\right)+f\left(\frac{1}{x}\right)-\frac{1}{x}=0$.