Function  ${\text{f:R}}^{\text{+}}\to \text{R}$  is differentiable and satisfies the condition $\text{f}\left(\text{x}\right)-\text{f}\left(\text{y}\right)\ge \text{ In}\left(\frac{\text{x}}{\text{y}}\right)\text{+ x}-\text{y,}$  for all x,  $\text{y}\in {\text{R}}^{\text{+}}\text{.}$If g denotes the derivative of f, then the value of the  $g\left(\frac{1}{2022}\right)$

$\text{f}\left(\text{x}\right)-\text{f}\left(\text{y}\right)-\text{ In y + x }-\text{ y}$       …..(1)
$x\to y\text{\hspace{0.17em}and y}\to \text{x}$
$\text{f}\left(\text{y}\right)-\text{f}\left(\text{x}\right)-\text{In\hspace{0.17em}x + y\hspace{0.17em}}-\text{x}$
$\Rightarrow \text{f}\left(\text{x}\right)-\text{f}\left(\text{y}\right)-\text{ In y + x}-\text{y}$   ……(2)
From equation (1) and (2) , f(x) – f (y) = ( In x+ x) – (In y + y)
$\Rightarrow \text{f}\left(\text{x}\right)\text{ = In\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}}\Rightarrow \text{g}\left(\text{x}\right)\text{=1+}\frac{\text{1}}{\text{x}}\Rightarrow \text{g }\left(\frac{\text{1}}{\text{2022}}\right)\text{= 2023}$