Let f be a differentiable function in (0, (π)/(2)) . If ∫ _ x ^ 1 t...

${∫}_{\cos x}^{1}{t}^{2}f\left(t\right)dt = \sin^{3}x + \cos x$

On differentiating, we get

$f\left(\cos x\right)\sin x \cdot \cos^{2}x = 3\sin^{2}xcosx - \sin x$

$f\left(\cos x\right) = 3\tan x - \sec^{2}x$

Again differentiating, we get

$- \sin x{f}^{'}\left(\cos x\right) = 3\sec^{2}x - 2\sec^{2}xtanx$

When $\cos x = \dfrac{1}{\sqrt{3}}$ then $\sec x = \sqrt{3}$, $\tan x = \sqrt{2} \text{ and } \sin x = \dfrac{\sqrt{2}}{\sqrt{3}}$

Then $- \dfrac{\sqrt{2}}{\sqrt{3}}{f}^{'}\left(\dfrac{1}{\sqrt{3}}\right) = 3 \times 3 - 2 \times 3\sqrt{2}$

$⇒ \dfrac{1}{\sqrt{3}}{f}^{'}\left(\dfrac{1}{\sqrt{3}}\right) = 6 - \dfrac{9}{\sqrt{2}}$