If $\phi(x)=\frac{1}{\sqrt{x}} \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0$,

then $\emptyset^{\prime}\left(\frac{\pi}{4}\right)$ is equal to :

$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$ <br/><br/>$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$ <br/><br/>$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$ <br/><br/>At, $x=\frac{\pi}{4}$ <br/><br/>$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$ <br/><br/>$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$ <br/><br/>$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$