If $f’(x) = \sin x + (\sin4x)(\cos x)$, then $f’(2x^2 + \pi/2) $is? If $$f'(x) = \sin x + \sin4x \cdot \cos x,$$ then $$f'(2x^2 + \pi/2)$$ is?

Given answer: $$4x\cos(2x^2) – 4x\sin(8x^2) \sin(2x^2)$$

I tried and I'm getting the answer as $\cos(2x^2) - \sin(8x^2)\sin(2x^2)$
If $$f(x) = \sin x+\sin 4x\cdot \cos x\;,$$ Then put $$\displaystyle x = 2x^2+\frac{\pi}{2}$$

So $$\displaystyle f\left(2x^2+\frac{\pi}{2}\right) = \sin \left(2x^2+\frac{\pi}{2}\right)+\sin \left(8x^2+4\pi\right)\cdot \cos \left(2x^2+\frac{\pi}{2}\right)$$

So $$\displaystyle f\left(2x^2+\frac{\pi}{2}\right) = \cos 2x^2-\sin 4x^2\cdot \sin 2x^2$$

Above we have used $$\displaystyle \bullet\; \sin \left(\frac{\pi}{2}+\phi\right) = \cos \phi$$ and $$\displaystyle \bullet\; \cos \left(\frac{\pi}{2}+\phi\right) = -\sin \phi$$

and $$\displaystyle \bullet\; \sin (2\pi+\phi) = \sin \phi$$
@N Nair ...I was solving arihant today and came across the same problem (the reason I am here :p lol) and I think I have got a relevant answer to the question but have some doubts too.
 f'(2x^2+π/2) does not mean df(2x^2+π/2)/dx....(i)
It rather means ... df(2x^2+π/2)/d(2x^2+π/2) .....(ii)
so, case (i) which gives answer 4x{cos(2x^2)-sin(8x^2)sin(2x^2)} is WRONG and so I think the answer provided is WRONG.
and case (ii) will yield the answer cos(2x^2)-sin(8x^2)sin(2x^2) which according to me is CORRECT and thus YOUR ANSWER is CORRECT.
Any arguement to this? Please let me know :)