If (x)=1x _( )/(4)^x(4√2 t-3 ^'(t))dt,x0, then ^'(( )/(4)) is equal to :

Question

If $\phi (x)=\frac{1}{\sqrt{x}}\int _{\frac{\pi }{4}}^{x}(4\sqrt{2}\sin t-3{\phi }^{'}(t))dt,x>0$,

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

Accepted Answer

$\phi (x)=\frac{1}{\sqrt{x}}{\int }_{x}^{\pi /4}(4\sqrt{2}\sin t-3{\phi }^{'}(t))dt$

$\Rightarrow {\phi }^{'}(x)=\frac{-1}{2{x}^{3/2}}{\int }_{x}^{\pi /4}(4\sqrt{2}\sin t-3{\phi }^{'}(t))dt$

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

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

${\phi }^{'}(\frac{\pi }{4})=\frac{-1}{2{(\frac{\pi }{4})}^{3/2}}	imes 0+\sqrt{\frac{4}{\pi }}(4\sqrt{2}	imes \frac{1}{\sqrt{2}}-3{\phi }^{'}(\frac{\pi }{4}))$

$\Rightarrow {\phi }^{'}(\frac{\pi }{4})[1+\frac{6}{\sqrt{\pi }}]=\frac{2}{\sqrt{\pi }}	imes 4$

$\Rightarrow {\phi }^{'}(\frac{\pi }{4})=\frac{8}{6+\sqrt{x}}$

Solutions · 1