Let $f\left(x\right)=\left({\sin }^{-1}\left(\cos x\right)\right)\left({\cos }^{-1}\left(\sin x\right)\right),\forall x\in [0,2\pi ].$ Then which of the following is NOT TRUE?

$f\left(x\right)={\sin }^{-1}\left(\cos x\right).{\cos }^{-1}\left(\sin x\right)=(\frac{\pi }{2}-{\cos }^{-1}\left(\cos x\right))(\frac{\pi }{2}-{\sin }^{-1}\left(\sin x\right))$

$=\{\begin{array}{c}(\frac{\pi }{2}-x)(\frac{\pi }{2}-x),0\le x\le \frac{\pi }{2}\\ (\frac{\pi }{2}-x)(\frac{\pi }{2}-(\pi -x)),\frac{\pi }{2}<x\le \pi \\ (\frac{\pi }{2}-(2\pi -x))(\frac{\pi }{2}-(\pi -x)),\pi <x\le \frac{3\pi }{3}\\ \left(\begin{array}{c}\pi \\ 2\end{array}\left(2\pi x\right)\right)\left(\begin{array}{c}\pi \\ 2\end{array}\left(x2\pi \right)\right),\begin{array}{c}3\pi \\ 2\end{array}<x\le 2\pi \end{array}=\{\begin{array}{c}{\left(\begin{array}{c}\pi \\ 2\end{array}x\right)}^2,0\le x\le \begin{array}{c}\pi \\ 2\end{array}\\ -{(\frac{\pi }{2}-x)}^2,\frac{\pi }{2}<x\le \pi \\ (x-\frac{3\pi }{2})(x-\frac{\pi }{2}),\pi <x\le \frac{3\pi }{2}\\ (x-\frac{3\pi }{2})(\frac{5\pi }{2}-x),\frac{3\pi }{2}<x\le 2\pi \end{array}$

$\therefore$ Clearly f (x) is non differentiable at x = $\pi$