Key Concepts and Formulas

• Leibniz Integral Rule: If $F(x) = \int_{a(x)}^{b(x)} g(t) dt$, then $F'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$. This is a generalization of the Fundamental Theorem of Calculus for integrals with variable limits.
• Chain Rule: If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. This is crucial for differentiating composite functions, such as $f(\cos x)$.
• Trigonometric Identities: Knowledge of basic trigonometric identities like $\sin^2 x + \cos^2 x = 1$, $\sec x = 1/\cos x$, and $\tan x = \sin x / \cos x$ is essential for simplification.

Step-by-Step Solution

We are given the equation $\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x}$ and need to find $\frac{1}{\sqrt{3}}f'\left(\frac{1}{\sqrt{3}}\right)$.

Step 1: Differentiate both sides of the given equation with respect to $x$. To differentiate the left-hand side (LHS), we use the Leibniz Integral Rule. Let $F(x) = \int_{\cos x}^1 {t^2 f(t) dt}$. Here, $g(t) = t^2 f(t)$, the upper limit is $b(x) = 1$, and the lower limit is $a(x) = \cos x$. The derivative of the upper limit is $b'(x) = \frac{d}{dx}(1) = 0$. The derivative of the lower limit is $a'(x) = \frac{d}{dx}(\cos x) = -\sin x$. Applying the Leibniz Rule: $\frac{d}{dx} F(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$ $\frac{d}{dx} \left( \int_{\cos x}^1 {t^2 f(t) dt} \right) = (1^2 f(1)) \cdot (0) - ((\cos x)^2 f(\cos x)) \cdot (-\sin x)$ $= 0 - (-\sin x \cos^2 x f(\cos x)) = \sin x \cos^2 x f(\cos x)$ Now, we differentiate the right-hand side (RHS) with respect to $x$: $\frac{d}{dx} (\sin^3 x + \cos x) = \frac{d}{dx}(\sin^3 x) + \frac{d}{dx}(\cos x)$ Using the chain rule for $\sin^3 x$, we get $3\sin^2 x \cdot \cos x$. The derivative of $\cos x$ is $-\sin x$. So, the derivative of the RHS is: $3\sin^2 x \cos x - \sin x$ Equating the derivatives of both sides: $\sin x \cos^2 x f(\cos x) = 3\sin^2 x \cos x - \sin x$

Step 2: Solve for $f(\cos x)$. We can factor out $\sin x$ from the RHS: $\sin x \cos^2 x f(\cos x) = \sin x (3\sin x \cos x - 1)$ Since $x \in (0, \pi/2)$, $\sin x \neq 0$. We can divide both sides by $\sin x$: $\cos^2 x f(\cos x) = 3\sin x \cos x - 1$ Now, divide by $\cos^2 x$. Since $x \in (0, \pi/2)$, $\cos x \neq 0$, so $\cos^2 x \neq 0$: $f(\cos x) = \frac{3\sin x \cos x - 1}{\cos^2 x}$ Separate the terms: $f(\cos x) = \frac{3\sin x \cos x}{\cos^2 x} - \frac{1}{\cos^2 x}$ $f(\cos x) = 3\frac{\sin x}{\cos x} - \sec^2 x$ $f(\cos x) = 3\tan x - \sec^2 x$

Step 3: Differentiate $f(\cos x)$ with respect to $x$ to find $f'(\cos x)(-\sin x)$. We need to find $f'(1/\sqrt{3})$, so we differentiate the expression for $f(\cos x)$ with respect to $x$. Using the chain rule on the LHS, $\frac{d}{dx} f(\cos x) = f'(\cos x) \cdot \frac{d}{dx}(\cos x) = f'(\cos x) (-\sin x)$. Differentiate the RHS: $\frac{d}{dx} (3\tan x - \sec^2 x) = \frac{d}{dx}(3\tan x) - \frac{d}{dx}(\sec^2 x)$ $= 3\sec^2 x - (2\sec x \cdot \frac{d}{dx}(\sec x))$ $= 3\sec^2 x - (2\sec x \cdot \sec x \tan x)$ $= 3\sec^2 x - 2\sec^2 x \tan x$ Equating the derivatives of both sides: $f'(\cos x) (-\sin x) = 3\sec^2 x - 2\sec^2 x \tan x$

Step 4: Determine the values of trigonometric functions when $\cos x = 1/\sqrt{3}$. We need to evaluate the expression at a point where $\cos x = \frac{1}{\sqrt{3}}$. Since $x \in (0, \pi/2)$, $\sin x > 0$. Using $\sin^2 x + \cos^2 x = 1$: $\sin^2 x = 1 - \cos^2 x = 1 - \left(\frac{1}{\sqrt{3}}\right)^2 = 1 - \frac{1}{3} = \frac{2}{3}$ $\sin x = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$ Now we find $\sec x$ and $\tan x$: $\sec x = \frac{1}{\cos x} = \frac{1}{1/\sqrt{3}} = \sqrt{3}$ $\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{2}/\sqrt{3}}{1/\sqrt{3}} = \sqrt{2}$

Step 5: Substitute these values into the differentiated equation and solve for $f'(1/\sqrt{3})$. Substitute $\cos x = \frac{1}{\sqrt{3}}$, $\sin x = \frac{\sqrt{2}}{\sqrt{3}}$, $\sec x = \sqrt{3}$, and $\tan x = \sqrt{2}$ into the equation from Step 3: $f'\left(\frac{1}{\sqrt{3}}\right) \left(-\frac{\sqrt{2}}{\sqrt{3}}\right) = 3(\sqrt{3})^2 - 2(\sqrt{3})^2 (\sqrt{2})$ $f'\left(\frac{1}{\sqrt{3}}\right) \left(-\frac{\sqrt{2}}{\sqrt{3}}\right) = 3(3) - 2(3)(\sqrt{2})$ $f'\left(\frac{1}{\sqrt{3}}\right) \left(-\frac{\sqrt{2}}{\sqrt{3}}\right) = 9 - 6\sqrt{2}$ Now, solve for $f'\left(\frac{1}{\sqrt{3}}\right)$: $f'\left(\frac{1}{\sqrt{3}}\right) = \frac{9 - 6\sqrt{2}}{-\frac{\sqrt{2}}{\sqrt{3}}} = -\frac{\sqrt{3}(9 - 6\sqrt{2})}{\sqrt{2}}$

Step 6: Calculate the final expression $\frac{1}{\sqrt{3}}f'\left(\frac{1}{\sqrt{3}}\right)$. Multiply the expression for $f'\left(\frac{1}{\sqrt{3}}\right)$ by $\frac{1}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} \left( -\frac{\sqrt{3}(9 - 6\sqrt{2})}{\sqrt{2}} \right)$ $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = -\frac{9 - 6\sqrt{2}}{\sqrt{2}}$ Distribute the negative sign and divide by $\sqrt{2}$: $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = \frac{-9 + 6\sqrt{2}}{\sqrt{2}}$ $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = -\frac{9}{\sqrt{2}} + \frac{6\sqrt{2}}{\sqrt{2}}$ $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = -\frac{9}{\sqrt{2}} + 6$ Rearranging the terms to match the options: $\frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = 6 - \frac{9}{\sqrt{2}}$

Common Mistakes & Tips

• Incorrect application of Leibniz Rule: Ensure you correctly identify $g(t)$, $a(x)$, $b(x)$, and their derivatives. Pay close attention to the signs when subtracting $g(a(x)) \cdot a'(x)$.
• Algebraic errors: Simplification of trigonometric expressions and fractions can lead to errors. Double-check each step, especially when dealing with square roots.
• Chain Rule application: When differentiating $f(\cos x)$, remember to multiply by the derivative of the inner function, $-\sin x$.

Summary

The problem involves differentiating an integral with variable limits and then evaluating the derivative of the function $f$. We used the Leibniz Integral Rule to differentiate the given equation, which allowed us to find an expression for $f(\cos x)$. By differentiating this expression again (using the chain rule), we obtained an equation involving $f'(\cos x)$. Finally, by substituting the appropriate trigonometric values for $\cos x = 1/\sqrt{3}$, we were able to solve for the required expression $\frac{1}{\sqrt{3}}f'\left(\frac{1}{\sqrt{3}}\right)$.

The final answer is $\boxed{6 - \frac{9}{\sqrt 2 }}$.