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 $\cos(2x^2) - \sin(8x^2)\sin(2x^2)$
 A: 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 $$\displaystyle \bullet\; \sin (2\pi+\phi) = \sin \phi$$
A: This seems to be the closest approximation to the question that actually makes sense:
Suppose $$f'(x) = \sin x + \sin 4x \cdot \cos x.$$
Let $g(x) = f(2x^2 + \frac\pi2)$.  Then
$$g'(x) = 4x \big(\!\cos(2x^2) - \sin(8x^2)\sin(2x^2) \big).$$
That would make two typos in the question: the lack of parentheses around the expression multiplied by $4x$, and the mistaken use of $f'(2x^2+\frac\pi2)$ to refer to $g'(x)$ (I think you correctly computed the former but the answer suggests the latter).
A: $$f\'(x) = \sin x + \sin 4x \cos x$$
We need to find $f\'(2x^2+\pi/2)$.
Now
$$f\'(2x^2+π/2) = f\'(2x^2+π/2).Derivative of (2x^2+π/2)$$
$$ = {\sin (2x^2+π/2) + \sin 4(2x^2+π/2).Cos(2x^2+π/2)}.4x$$
$$ = 4x{\cos(2x^2) - \sin(8x^2).\sin(2x^2)}$$
A: $$f\left(2x^2+\dfrac\pi2\right)=\sin\left(2x^2+\dfrac\pi2\right)+\sin4\left(2x^2+\dfrac\pi2\right)\cos\left(2x^2+\dfrac\pi2\right)$$
Now $\sin\left(\dfrac\pi2+y\right)=+\cos y$ and $\cos\left(\dfrac\pi2+y\right)=-\sin y$
Now use chain rule like $$\dfrac{d\{\cos(2x^2)\}}{dx}=\dfrac{d\{\cos(2x^2)\}}{d(2x^2)}\cdot\dfrac{d(2x^2)}{dx}=-\sin(2x^2)\cdot4x$$
A: @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 :)
