<s>[INST] 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?
 [/INST] 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 $x=1$, $\large\frac{y'}y=\large\frac{\frac{3\pi^2}{16}}{-\frac18}=-\frac{3\pi^2}2\implies 2y'+3\pi^2y=0.$
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