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Step-by-Step Solution
Step 1: Express the given integral and identify the approach
We are given that
$$
\int_{\cos x}^{1} t^2 f(t)\,dt \;=\; \sin^3 x + \cos x.
$$
To find $f'\!\bigl(\tfrac{1}{\sqrt{3}}\bigr)$, we will differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus.
Step 2: Differentiate both sides with respect to x
Using the rule for differentiating a definite integral with variable limits, if
$$
I(x) = \int_{g(x)}^{h(x)} F(t)\,dt,
$$
then
$$
\frac{d}{dx}\,I(x) = F\bigl(h(x)\bigr)\cdot h'(x)\;-\;F\bigl(g(x)\bigr)\cdot g'(x).
$$
Here, the upper limit is 1 (constant) and the lower limit is $\cos x$. Thus,
$$
\frac{d}{dx} \int_{\cos x}^{1} t^2 f(t)\,dt
= \;0 \;-\;\bigl( (\cos x)^2 f(\cos x)\bigr)\times \frac{d}{dx}\bigl(\cos x\bigr).
$$
Since $\frac{d}{dx}(\cos x) = -\sin x$, the differentiation gives
$$
\frac{d}{dx} \int_{\cos x}^{1} t^2 f(t)\,dt
= \; \sin x\,\cos^2 x\, f(\cos x).
$$
The right-hand side, $\sin^3 x + \cos x$, differentiates to
$$
\frac{d}{dx}\bigl(\sin^3 x + \cos x\bigr)
= 3\sin^2 x\,\cos x - \sin x.
$$
Step 3: Equate the derivatives
Equating the two results from differentiation, we have:
$$
\sin x\,\cos^2 x\, f(\cos x)
= 3\sin^2 x\,\cos x - \sin x.
$$
Step 4: Solve for f(cos x)
Factor out $\sin x$ from the right side (noting $\sin x > 0$ for $x \in (0,\tfrac{\pi}{2})$):
$$
\sin x \,\cos^2 x\, f(\cos x)
= \sin x\Bigl(3\sin x\,\cos x - 1\Bigr).
$$
Divide by $\sin x \cos^2 x$ (assuming $\sin x \neq 0$, $\cos x \neq 0$):
$$
f(\cos x)
= \frac{3\sin x\,\cos x - 1}{\cos^2 x}
= 3\,\tan x\;-\;\sec^2 x.
$$
Step 5: Differentiate f(cos x) with respect to x to find f'(cos x)
Let $y = \cos x$. Then $\frac{d}{dx}f(y) = f'(y)\,\frac{dy}{dx} = f'(\cos x)\cdot(-\sin x)$. Meanwhile,
$$
\frac{d}{dx}\bigl(3\,\tan x - \sec^2 x\bigr)
= 3\,\sec^2 x \;-\; 2\,\sec^2 x\,\tan x
= \sec^2 x \bigl(3 \;-\; 2\,\tan x\bigr).
$$
Hence,
$$
f'(\cos x)\,\bigl(-\sin x\bigr)
= 3\,\sec^2 x \;-\; 2\,\sec^2 x\,\tan x.
$$
Step 6: Substitute cos x = 1 / sqrt(3)
Given $\cos x = \tfrac{1}{\sqrt{3}}$, we find:
$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \tfrac{1}{3}} = \sqrt{\tfrac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{\tfrac{\sqrt{2}}{\sqrt{3}}}{\tfrac{1}{\sqrt{3}}} = \sqrt{2}\,.$
$\sec x = \frac{1}{\cos x} = \sqrt{3},\quad \sec^2 x = 3.$
Plug these into the differentiated equation:
$$
f'\Bigl(\tfrac{1}{\sqrt{3}}\Bigr)\;\bigl(-\sin x \bigr)
= 3 \times 3 \;-\; 2 \times 3 \times \sqrt{2}
= 9 \;-\; 6\sqrt{2}.
$$
Since $\sin x = \tfrac{\sqrt{2}}{\sqrt{3}}$, we have:
$$
f'\Bigl(\tfrac{1}{\sqrt{3}}\Bigr)\,\Bigl(- \tfrac{\sqrt{2}}{\sqrt{3}}\Bigr)
= 9 - 6\sqrt{2}.
$$
Step 7: Solve for (1 / sqrt(3)) * f'(1 / sqrt(3))
Rearrange to isolate $f'\bigl(\tfrac{1}{\sqrt{3}}\bigr)$ and then multiply by $\tfrac{1}{\sqrt{3}}$:
$$
f'\Bigl(\tfrac{1}{\sqrt{3}}\Bigr)
= \frac{9 - 6\sqrt{2}}{-\,\tfrac{\sqrt{2}}{\sqrt{3}}}
\quad\Longrightarrow\quad
\tfrac{1}{\sqrt{3}}\;f'\Bigl(\tfrac{1}{\sqrt{3}}\Bigr)
= 6 \;-\; \frac{9}{\sqrt{2}}.
$$
This shows that the required value is
$$
6 - \frac{9}{\sqrt{2}}.
$$
Final Answer
$$
\frac{1}{\sqrt{3}}\,f'\bigl(\tfrac{1}{\sqrt{3}}\bigr)
\;=\;
6 \;-\; \frac{9}{\sqrt{2}}.
$$
Relevant Figure (unchanged as provided):
