Key Concepts and Formulas

1. Leibniz Integral Rule (Fundamental Theorem of Calculus, Part 1): For a function $G(x) = \int_{a(x)}^{b(x)} f(t) dt$, its derivative is $G'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$. If the limits are constants, the derivative is zero. If the upper limit is $x$ and the lower limit is a constant $c$, then $\frac{d}{dx} \int_c^x f(t) dt = f(x)$.
2. Product Rule for Differentiation: If $y = u(x)v(x)$, then $y' = u'(x)v(x) + u(x)v'(x)$.
3. Integral with Identical Limits: $\int_a^a f(t) dt = 0$.

We are given the function: $\phi(x)=\frac{1}{\sqrt{x}} \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, \quad x>0$ Our objective is to find the value of $\phi^{\prime}\left(\frac{\pi}{4}\right)$.

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Step-by-Step Solution

Step 1: Differentiate $\phi(x)$ using the Product Rule and Leibniz Rule

The given function $\phi(x)$ is a product of two functions: $u(x) = \frac{1}{\sqrt{x}}$ and $v(x) = \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$. We will use the Product Rule: $\phi'(x) = u'(x)v(x) + u(x)v'(x)$.

First, find the derivative of $u(x)$: $u(x) = x^{-1/2}$ $u'(x) = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}}$

Next, find the derivative of $v(x)$ using the Leibniz Integral Rule. The integrand is $f(t) = 4 \sqrt{2} \sin t-3 \phi^{\prime}(t)$. The upper limit is $x$ (whose derivative is 1) and the lower limit is $\frac{\pi}{4}$ (a constant, whose derivative is 0). So, $v'(x) = f(x) \cdot \frac{d}{dx}(x) - f\left(\frac{\pi}{4}\right) \cdot \frac{d}{dx}\left(\frac{\pi}{4}\right)$ $v'(x) = \left(4 \sqrt{2} \sin x-3 \phi^{\prime}(x)\right) \cdot 1 - f\left(\frac{\pi}{4}\right) \cdot 0$ $v'(x) = 4 \sqrt{2} \sin x-3 \phi^{\prime}(x)$

Now, apply the Product Rule to find $\phi'(x)$: $\phi^{\prime}(x) = \left(-\frac{1}{2x^{3/2}}\right) \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t + \left(\frac{1}{\sqrt{x}}\right) \left(4 \sqrt{2} \sin x-3 \phi^{\prime}(x)\right)$

Step 2: Evaluate $\phi'(x)$ at $x = \frac{\pi}{4}$

Substitute $x = \frac{\pi}{4}$ into the expression for $\phi'(x)$: $\phi^{\prime}\left(\frac{\pi}{4}\right) = -\frac{1}{2\left(\frac{\pi}{4}\right)^{3/2}} \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{4}}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t + \frac{1}{\sqrt{\frac{\pi}{4}}} \left(4 \sqrt{2} \sin \left(\frac{\pi}{4}\right)-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$

Simplify each term:

• The integral term: $\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{4}}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t = 0$ because the upper and lower limits are identical.
• The second term:
  • $\frac{1}{\sqrt{\frac{\pi}{4}}} = \frac{1}{\frac{\sqrt{\pi}}{2}} = \frac{2}{\sqrt{\pi}}$
  • $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Substitute these values back into the equation for $\phi^{\prime}\left(\frac{\pi}{4}\right)$: $\phi^{\prime}\left(\frac{\pi}{4}\right) = -\frac{1}{2\left(\frac{\pi}{4}\right)^{3/2}} \cdot 0 + \frac{2}{\sqrt{\pi}} \left(4 \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) - 3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$ $\phi^{\prime}\left(\frac{\pi}{4}\right) = 0 + \frac{2}{\sqrt{\pi}} \left(4 - 3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$ $\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{8}{\sqrt{\pi}} - \frac{6}{\sqrt{\pi}} \phi^{\prime}\left(\frac{\pi}{4}\right)$

Step 3: Solve the algebraic equation for $\phi^{\prime}\left(\frac{\pi}{4}\right)$

We have an equation with $\phi^{\prime}\left(\frac{\pi}{4}\right)$ on both sides. Let $y = \phi^{\prime}\left(\frac{\pi}{4}\right)$ for simplicity. $y = \frac{8}{\sqrt{\pi}} - \frac{6}{\sqrt{\pi}} y$ Move all terms containing $y$ to one side: $y + \frac{6}{\sqrt{\pi}} y = \frac{8}{\sqrt{\pi}}$ Factor out $y$: $y \left(1 + \frac{6}{\sqrt{\pi}}\right) = \frac{8}{\sqrt{\pi}}$ Combine the terms in the parenthesis: $y \left(\frac{\sqrt{\pi} + 6}{\sqrt{\pi}}\right) = \frac{8}{\sqrt{\pi}}$ Solve for $y$: $y = \frac{8}{\sqrt{\pi}} \times \frac{\sqrt{\pi}}{\sqrt{\pi} + 6}$ $y = \frac{8}{6 + \sqrt{\pi}}$ Therefore, $\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$.

This result matches option (D).

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Common Mistakes & Tips

• Incorrect application of the Product Rule: Ensure both parts of the product are differentiated correctly.
• Forgetting the Leibniz Rule: The derivative of the integral term is crucial and requires careful application of the rule.
• Ignoring the integral with identical limits: This simplifies the calculation significantly, as the entire first term in Step 2 becomes zero.
• Algebraic errors: Double-check all fractional arithmetic and rearrangement of terms.

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Summary

This problem requires a combined application of the Product Rule and the Leibniz Integral Rule. By differentiating the given functional equation and then evaluating at the specific point $x = \frac{\pi}{4}$, we obtain an algebraic equation for $\phi^{\prime}\left(\frac{\pi}{4}\right)$. Solving this equation yields the desired value. The key insight is that the integral term vanishes when the limits of integration are the same.

The final answer is $\boxed{\frac{8}{6+\sqrt{\pi}}}$.