Let \( f \) be a differentiable function in \( \left(0, \frac{\pi}{2}\right) \). If \( \int_{\cos x}^{1} t^{2} f(t)\, dt = \sin^{3} x + \cos x \), then \( \frac{1}{\sqrt{3}} \, f^{\prime}\!\left(\frac{1}{\sqrt{3}}\right) \) is equal to
\( 6 - 9\sqrt{2} \)
\( 6 - \frac{9}{\sqrt{2}} \)
\( \frac{9}{2} - 6\sqrt{2} \)
\( \frac{9}{\sqrt{2}} - 6 \)
