If $y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to :

$\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x\end{aligned}$ <br/><br/>$\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}$