If the function $f\left(x\right)={\cos }^{-1}\left(x^{\frac{3}{2}}-\sqrt{1-x-x^2+x^3}\right)$ (where, $\forall 0<x<1$), then the value of $\left|\sqrt{3}{f}^{\text{'}}\left(\frac{1}{2}\right)\right|$ is equal to (take $\sqrt{3}=1.73$)

$f\left(x\right)={\cos }^{-1}\left\{x\sqrt{x}-\sqrt{\left(1-x\right)\left(1-x^2\right)}\right\}$
$={\cos }^{-1}\left\{x\sqrt{x}-\sqrt{1-{\left(\sqrt{x}\right)}^2}\sqrt{1-x^2}\right\}$
$={\cos }^{-1}x+{\cos }^{-1}\sqrt{x}$
Now, $f^{\prime }\left(x\right)=\frac{-1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}$
$f^{\text{'}}\left(\frac{1}{2}\right)=\frac{-1}{\sqrt{1-\frac{1}{4}}}-\frac{1}{\sqrt{\frac{1}{2}}}\times \frac{1}{2\sqrt{\frac{1}{2}}}=-\frac{2}{\sqrt{3}}-1$

$\left|\sqrt{3}{f}^{\text{'}}\left(\frac{1}{2}\right)\right|=\left|-\sqrt{3}\left(-\frac{2}{\sqrt{3}}-1\right)\right|=3.73$