Derivative of $y = \sin^3(\frac\pi 3(\cos(\frac\pi{3\sqrt2}(-4x^3 + 5x^2 + 1)^{3/2})))$ at $x=1$ Today  I came across a problem:

If  $y = \sin^3\left(\frac\pi 3\left(\cos\left(\frac\pi{3\sqrt2}\left(-4x^3 + 5x^2 + 1\right)^{3/2}\right)\right)\right)$, then at $x=1$ which of the following option is correct?

*

*$2y'+\sqrt{3} \pi^2 y = 0$

*$2y' +3 \pi^2 y = 0$

*$\sqrt2 y' - 3\pi^2 y = 0 $

*$y' + 3\pi^2 y = 0$

This question is from JEE Main 2023 examination.

My attempt:
Firstly I used the identity $\sin^3(x) = \frac34 \sin(x) - \frac14 \sin(3x)$ and then by applying a brute force chain rule, I got that $f'(1) = \frac3{16} \pi^2$ and $f(1) = -\frac18$.
So, second option i.e. $2y' + 3\pi^2 y = 0$ is correct.
But I don't think that this question is designed to be solved in this manner. Is there something which I'm missing?

It is good to know the answer and solve. You never see your mistakes. I tried to be quick:
$y=\sin^3A$ and $A=\frac\pi 3\cos B$ and $B=\frac{\pi}{3\sqrt 2}C^{3/2}$ and $C=-4x^3+5x^2+1$.
$y'=3\sin^2A\cos A\frac\pi 3(-\sin B)\frac{\pi}{3\sqrt2}\frac32C^{1/2}(-12x^2+10x).$
At $x=1$, $C=2$, $B=\frac{2\pi}3$, $A=-\frac\pi 6$.
At $x=1$, $y'=3(-\frac12)^2\frac{\sqrt{3}}{2}\frac{\pi}3(-\frac{\sqrt 3}{2})\frac{\pi}{3\sqrt2}\frac32\sqrt 2(-2)=\frac{3\pi^2}{16}$.
At $x=1$, $y=\sin^3(-\frac\pi 6)=(-\frac12)^2=-\frac18$.
At $x=1$, $\large\frac{y'}y=\large\frac{\frac{3\pi^2}{16}}{-\frac18}=-\frac{3\pi^2}2\implies 2y'+3\pi^2y=0.$
