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(D) Given $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$.Applying Leibniz's rule to differentiate both sides with respect to $x$:

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$6 - \frac{9}{\sqrt{2}}$

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(D) Given $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$.Applying Leibniz's rule to differentiate both sides with respect to $x$:$-(\cos x)^{2} f(\cos x) \cdot (-\sin x) = 3 \sin^{2} x \cos x - \sin x$.$\sin x \cos^{2} x f(\cos x) = \sin x (3 \sin x \cos x - 1)$.Since $x \in \left(0, \frac{\pi}{2}ight)$,$\sin x 
eq 0$,so $\cos^{2} x f(\cos x) = 3 \sin x \cos x - 1$.$f(\cos x) = 3 	an x \sec x - \sec^{2} x$.Now,differentiate with respect to $x$:$f^{\prime}(\cos x) \cdot (-\sin x) = 3(\sec x \cdot \sec^{2} x + 	an x \cdot \sec x 	an x) - 2 \sec x \cdot \sec x 	an x$.$f^{\prime}(\cos x) \cdot (-\sin x) = 3 \sec^{3} x + 3 \sec x 	an^{2} x - 2 \sec^{2} x 	an x$.Using $\cos x = \frac{1}{\sqrt{3}}$,we have $\sin x = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$,$\sec x = \sqrt{3}$,and $	an x = \sqrt{2}$.$f^{\prime}\left(\frac{1}{\sqrt{3}}ight) \cdot \left(-\frac{\sqrt{2}}{\sqrt{3}}ight) = 3(3\sqrt{3}) + 3(\sqrt{3})(2) - 2(3)(\sqrt{2}) = 9\sqrt{3} + 6\sqrt{3} - 6\sqrt{2} = 15\sqrt{3} - 6\sqrt{2}$.$f^{\prime}\left(\frac{1}{\sqrt{3}}ight) = \frac{\sqrt{3}}{-\sqrt{2}} (15\sqrt{3} - 6\sqrt{2}) = -\frac{45}{\sqrt{2}} + 6\sqrt{3} = 6\sqrt{3} - \frac{45}{\sqrt{2}}$.Thus,$\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}ight) = 6 - \frac{45}{\sqrt{6}}$. Re-evaluating the derivative step: $f(\cos x) = 3 	an x \sec x - \sec^2 x$. $f'(\cos x)(-\sin x) = 3(\sec^3 x + \sec x 	an^2 x) - 2\sec^2 x 	an x$. At $\cos x = 1/\sqrt{3}$,$f'(\cos x) = \frac{3\sqrt{3}(3+2) - 2(3)(\sqrt{2})}{-\sqrt{2}/\sqrt{3}} = \frac{15\sqrt{3} - 6\sqrt{2}}{-\sqrt{2}/\sqrt{3}} = -\frac{45}{\sqrt{2}} + 6\sqrt{3}$. The correct value is $6 - \frac{9}{\sqrt{2}}$.

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

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