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 $$\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right)$$ is equal to

We are given $$\int_{\cos x}^{1} t^2 f(t)\, dt = \sin^3 x + \cos x where f is differentiable on (0, \frac{\pi}{2}), and we need to find \frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right).$$

Differentiating both sides with respect to $$x, the Leibniz rule gives \frac{d}{dx}\int_{\cos x}^{1} t^2 f(t)\,dt = -\frac{d(\cos x)}{dx} \cdot \cos^2 x \cdot f(\cos x) = \sin x \cdot \cos^2 x \cdot f(\cos x), while the right side yields \frac{d}{dx}(\sin^3 x + \cos x) = 3\sin^2 x \cos x - \sin x. Equating and dividing by the positive \sin x on (0, \frac{\pi}{2}) gives \cos^2 x \cdot f(\cos x) = 3\sin x \cos x - 1.$$

Substituting $$u = \cos x (so \sin x = \sqrt{1-u^2}) leads to u^2 \cdot f(u) = 3u\sqrt{1-u^2} - 1 and hence f(u) = \frac{3\sqrt{1-u^2}}{u} - \frac{1}{u^2}.$$

Differentiating $$f(u) with respect to u, we set g(u) = \frac{3\sqrt{1-u^2}}{u} = 3u^{-1}(1-u^2)^{1/2} and apply the product rule to get g'(u) = 3\Bigl[-u^{-2}(1-u^2)^{1/2} + u^{-1}\frac{-u}{(1-u^2)^{1/2}}\Bigr] = -\frac{3\sqrt{1-u^2}}{u^2} - \frac{3}{\sqrt{1-u^2}}, while for h(u) = -u^{-2} we have h'(u) = 2u^{-3} = \frac{2}{u^3}. Therefore f'(u) = -\frac{3\sqrt{1-u^2}}{u^2} - \frac{3}{\sqrt{1-u^2}} + \frac{2}{u^3}.$$

To evaluate at $$u = \frac{1}{\sqrt{3}}, note u^2 = \frac{1}{3},\quad u^3 = \frac{1}{3\sqrt{3}},\quad 1 - u^2 = \frac{2}{3},\quad \sqrt{1 - u^2} = \frac{\sqrt{2}}{\sqrt{3}}. Computing each term gives: Term 1: -\frac{3\cdot \frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{3}} = -\frac{9\sqrt{2}}{\sqrt{3}}, Term 2: -\frac{3}{\frac{\sqrt{2}}{\sqrt{3}}} = -\frac{3\sqrt{3}}{\sqrt{2}}, Term 3: \frac{2}{\frac{1}{3\sqrt{3}}} = 6\sqrt{3}, so f'\left(\frac{1}{\sqrt{3}}\right) = -\frac{9\sqrt{2}}{\sqrt{3}} - \frac{3\sqrt{3}}{\sqrt{2}} + 6\sqrt{3}.$$

Multiplying by $$\frac{1}{\sqrt{3}}} gives \frac{1}{\sqrt{3}}} f'\left(\frac{1}{\sqrt{3}}}\right) = -\frac{9\sqrt{2}}{3} - \frac{3}{\sqrt{2}} + 6 = -3\sqrt{2} - \frac{3}{\sqrt{2}} + 6, and since -3\sqrt{2} - \frac{3}{\sqrt{2}} = -\frac{9\sqrt{2}}{2} = -\frac{9}{\sqrt{2}}, it follows that \frac{1}{\sqrt{3}}} f'\left(\frac{1}{\sqrt{3}}}\right) = 6 - \frac{9}{\sqrt{2}}. The correct answer is Option C: 6 - \frac{9}{\sqrt{2}}.