1. Given equation

We have

$$\phi(x)=\frac{1}{\sqrt{x}}\int_{\pi/4}^{x}\left(4\sqrt2\sin t-3\phi'(t)\right)\,dt,\qquad x>0.$$

We need to find $\phi'\left(\frac{\pi}{4}\right)$.

---

2. Multiply by $\sqrt{x}$

Let

$$I(x)=\int_{\pi/4}^{x}\left(4\sqrt2\sin t-3\phi'(t)\right)\,dt.$$

Then

$$\sqrt{x}\,\phi(x)=I(x).$$

Differentiate both sides w.r.t. $x$.

Left side:

$$\frac{d}{dx}\big(\sqrt{x}\,\phi(x)\big)=\frac{\phi(x)}{2\sqrt{x}}+\sqrt{x}\,\phi'(x).$$

Right side, by FTC:

$$I'(x)=4\sqrt2\sin x-3\phi'(x).$$

So,

$$\frac{\phi(x)}{2\sqrt{x}}+\sqrt{x}\,\phi'(x)=4\sqrt2\sin x-3\phi'(x).$$

Hence,

$$\left(\sqrt{x}+3\right)\phi'(x)=4\sqrt2\sin x-\frac{\phi(x)}{2\sqrt{x}}.$$

Therefore,

$$\phi'(x)=\frac{4\sqrt2\sin x-\dfrac{\phi(x)}{2\sqrt{x}}}{\sqrt{x}+3}.$$

---

3. Find $\phi\left(\frac{\pi}{4}\right)$

Substitute $x=\frac{\pi}{4}$ in the original equation:

$$\phi\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{\pi/4}}\int_{\pi/4}^{\pi/4}\left(4\sqrt2\sin t-3\phi'(t)\right)\,dt=0.$$

So,

$$\phi\left(\frac{\pi}{4}\right)=0.$$

---

4. Now evaluate $\phi'\left(\frac{\pi}{4}\right)$

Using

$$\phi'(x)=\frac{4\sqrt2\sin x-\dfrac{\phi(x)}{2\sqrt{x}}}{\sqrt{x}+3},$$

put $x=\frac{\pi}{4}$:

$$\phi'\left(\frac{\pi}{4}\right)=\frac{4\sqrt2\sin(\pi/4)-0}{\sqrt{\pi/4}+3}.$$

Now,

$$\sin\frac{\pi}{4}=\frac{1}{\sqrt2},$$

so

$$4\sqrt2\sin\frac{\pi}{4}=4\sqrt2\cdot\frac{1}{\sqrt2}=4.$$

Also,

$$\sqrt{\frac{\pi}{4}}=\frac{\sqrt\pi}{2}.$$

Thus,

$$\phi'\left(\frac{\pi}{4}\right)=\frac{4}{3+\frac{\sqrt\pi}{2}}.$$

Multiply numerator and denominator by $2$:

$$\phi'\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt\pi}.$$

---

5. Match with options

$$\phi'\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt\pi}$$

which is Option D.

---

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.