If ϕ ( x ) = 1 x ∫ _ (π)/(4) ^ x (4 2 t - 3 ϕ ^ ' ( t ))dt, x > 0...

Given,

$ϕ ( x ) = \dfrac{1}{\sqrt{x}}{∫}_{\pi / 4}^{x}\left(4\sqrt{2}\sin t - 3{ϕ}^{'} ( t )\right)dt$

Differentiate both sides w.r.t. $x$ we get,

${ϕ}^{'} ( x ) = \dfrac{- 1}{2{x}^{3 / 2}}{∫}_{\pi / 4}^{x}\left(4\sqrt{2}\sin t - 3{ϕ}^{'} ( t )\right)dt + \dfrac{1}{\sqrt{x}}\left(4\sqrt{2}\sin x - 3{ϕ}^{'} ( x )\right)$

Put $x = \dfrac{\pi}{4}$

${ϕ}^{'} ( \dfrac{\pi}{4} ) = \dfrac{- 1}{2{\left(\dfrac{\pi}{4}\right)}^{3 / 2}} \times 0 + \sqrt{\dfrac{4}{\pi}}\left(4\sqrt{2} \times \dfrac{1}{\sqrt{2}} - 3{ϕ}^{'} ( \dfrac{\pi}{4} )\right)$

$⇒ \sqrt{\dfrac{\pi}{4}}{ϕ}^{'} ( \dfrac{\pi}{4} ) + 3{ϕ}^{'} ( \dfrac{\pi}{4} ) = 4$

$⇒ {ϕ}^{'}\left(\dfrac{\pi}{4}\right)\left(3 + \sqrt{\dfrac{\pi}{4}}\right) = 4$

$⇒ {ϕ}^{'}\left(\dfrac{\pi}{4}\right) = \dfrac{8}{6 + \sqrt{\pi}}$