Let f(x) be a differentiable function such that $f'(x)=\sin x+\sin 4 x \cos x$. Then, $f'\left(2 x^2+\frac{\pi}{2}\right)$ at $x=\sqrt{\frac{\pi}{2}}$ is equal to

$-2 \sqrt{2 \pi}$

We have,

$f'(x)=\sin x+\sin 4 x \cos x$  for all  x

$\Rightarrow \frac{d}{d x}(f(x)) d x=\sin x+\sin 4 x \cos x$  for all  x       .....(i)

Now,

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

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

$=\left\{\sin \left(2 x^2+\frac{\pi}{2}\right)+\sin \left(8 x^2+2 \pi\right) \cos \left(2 x^2+\frac{\pi}{2}\right)\right\} \times 4 x$               [Using (i)]

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

∴   $\left\{f'\left(2 x^2+\frac{\pi}{2}\right)\right\}_{\text {at } x=\sqrt{\frac{\pi}{2}}}=4 \sqrt{\frac{\pi}{2}}\{\cos \pi-\sin 4 \pi \sin \pi \}$

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