If the prime sign (') represents differentiation w.r.t. $x$ and $f^{'}=\sin x+\sin 4x.\cos x$, then $f^{'}\left ( 2x^{2}+\cfrac{\pi }{2} \right )$ at $x=\sqrt{\dfrac{\pi }{2}}$ is equal to

Given $f^{'}(x)=\sin x+\sin 4x.\cos x$

Now, $f^{'}\left ( 2x^{2}+\dfrac{\pi }{2} \right )$

$=f^{ ' }\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \dfrac { d }{ dx } \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right)$

$=\left[ \sin \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) +\sin 4\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) .\cos \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \right] 4x$

$=\left[ \cos 2x^{ 2 }-\sin 8x^{ 2 }.\sin 2x^{ 2 } \right] 4x$

$=\left[ \cos \pi -\sin 4\pi .\sin \pi \right] 4\sqrt { \dfrac { \pi }{ 2 } }$

$=-2\sqrt{2\pi}$