If $$\phi(x) = \dfrac{1}{\sqrt{x}} \int_{\pi/4}^{x} \left(4\sqrt{2}\sin t - 3\phi'(t)\right) dt$$, $$x > 0$$ then $$\phi'\left(\dfrac{\pi}{4}\right)$$ is equal to ______.

$$\sqrt{x} \phi(x) = \int_{\pi/4}^{x} \left( 4\sqrt{2}\sin t - 3\phi'(t) \right) dt$$

Differentiating both sides with respect to $$x$$:

$$\frac{1}{2\sqrt{x}} \phi(x) + \sqrt{x} \phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$$

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

$$\phi\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\pi/4}} \int_{\pi/4}^{\pi/4} \left( 4\sqrt{2}\sin t - 3\phi'(t) \right) dt = 0$$

$$0 + \sqrt{\frac{\pi}{4}} \phi'\left(\frac{\pi}{4}\right) = 4\sqrt{2}\sin\left(\frac{\pi}{4}\right) - 3\phi'\left(\frac{\pi}{4}\right)$$

$$\frac{\sqrt{\pi}}{2} \phi'\left(\frac{\pi}{4}\right) = 4\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) - 3\phi'\left(\frac{\pi}{4}\right)$$

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

$$\left(\frac{\sqrt{\pi} + 6}{2}\right) \phi'\left(\frac{\pi}{4}\right) = 4 \implies \phi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$$