If $u={\cos }^{-1}\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)$ then$\frac{{\partial }^{2}u}{\partial x^2}-\frac{{\partial }^{2}u}{\partial y^2}=$

The correct option is

C

$\frac{2y(1+x^2{)}^2+2x(1+y^2{)}^2}{\left[\right(1+x^2\left)\right(1+y^2){]}^2}$

Given : $u={\cos }^{-1}\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)$
Apply the chain rule w.r.t 'x' potentially
$\frac{\partial u}{\partial x}=\frac{-1}{1-\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)}\frac{\partial }{\partial x}\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)$
$=\frac{-1}{1-\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)}\times \frac{1}{\sqrt{1+y^2}}\frac{\partial }{\partial x}\left(\frac{x-y}{\sqrt{1+x^2}}\right)$
$\frac{-xy+1}{\sqrt{{(xy+1)}^2}(1+x^2)}=\frac{-(xy+1)}{(xy+1)(1+x^2)}=\frac{-1}{1+x^2}$
Differentiate w.r.t 'x' partially again
$\frac{{\partial }^{2}y}{\partial x^2}=\frac{2x}{(1+x^2)}\to \left(1\right)$
Apply the chain rule differentiate w.r.t 'y' partially
$\frac{\partial u}{\partial y}=\frac{-1}{1-\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)}\times \frac{\partial }{\partial y}\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)$
$=\frac{-1}{1-\left(\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\right)}\times \frac{1}{\sqrt{1+x^2}}\frac{\partial }{\partial y}\left(\frac{x-y}{\sqrt{1+y^2}}\right)$
$=-\frac{-xy-1}{\sqrt{{(xy+1)}^2}\sqrt{1+y^2}}$
$=-\frac{-(xy+1)}{(xy+1)(1+y^2)}$
$\frac{\partial u}{\partial y}=\frac{+1}{(1+y^2)}$
Differentiate w.r.t 'y' partially again
$\frac{{\partial }^{2}u}{\partial y^2}=-\frac{2y}{{(1+y^2)}^2}$
$\frac{{\partial }^{2}u}{\partial x^2}-\frac{{\partial }^{2}u}{\partial y^2}=\frac{2x}{{(1+x^2)}^2}-\frac{2y}{{(1+y^2)}^2}$
$=\frac{2x}{{(1+x^2)}^2}+\frac{2y}{{(1+y^2)}^2}$
$\frac{{\partial }^{2}u}{\partial x^2}-\frac{{\partial }^{2}u}{\partial y^2}=\frac{2x{(1+y^2)}^2+2y{(1+x^2)}^2}{{(1+x^2)}^2{(1+y^2)}^2}$