Let $y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2

Question

Let $y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$. Then, at $x=1$,

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(c) : Let $-4 x^{3}+5 x^{2}+1=t$, then $$ \begin{aligned} & y=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}} t^{3 / 2}ight)ight)ight) \ & \frac{d t}{d x}=-12 x^{2}+10 x \ & 	herefore \frac{d y}{d x}=3 \sin ^{2}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}} t^{3 / 2}ight)ight)ight) \ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} t^{3 / 2}ight)ight) \cdot \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}} t^{3 / 2}ight)ight) \ & \frac{\pi}{3 \sqrt{2}} \cdot \frac{3}{2} t^{1 / 2} \cdot \frac{d t}{d x} \ & \Rightarrow \frac{d y}{d x}=-3 \sin ^{2}\left(\frac { \pi } { 3 } \left(\operatorname { c o s } \left(\frac { \pi } { 3 \sqrt { 2 } } \left(-4 x^{3}ight.ight.ight.ight. \ & \left.\left.\left.+5 x^{2}+1ight)^{3 / 2}ight)ight) \ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1ight)ight)^{3 / 2}ight) \ & \frac{\pi}{3} \sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1ight)^{3 / 2}ight) \ & \left(-12 x^{2}+10 xight) \end{aligned} $$ $$ \begin{aligned} & 	ext { At } x=1, t=-4+5+1=2 \ & \begin{array}{l} \frac{d y}{d x}(x=1) \ =-3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2^{3 / 2}ight)ight) \ \cdot \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2^{3 / 2}ight)ight) \end{array} \ & \frac{\pi}{3} \sin \left(\frac{\pi}{3 \sqrt{2}} 2^{3 / 2}ight) \cdot \frac{\pi}{2 \sqrt{2}} 2^{1 / 2} \cdot(-2) \ & =\pi^{2} \sin ^{2}\left(\frac{\pi}{3}\left(\cos \frac{2 \pi}{3}ight)ight) \cdot \cos \left(\frac{\pi}{3} \cos \frac{2 \pi}{3}ight) \ & =\pi^{2} \sin ^{2}\left(-\frac{\pi}{3} \cdot \frac{1}{2}ight) \cdot \cos \left(-\frac{\pi}{3} \cdot \frac{1}{2}ight) \frac{\sqrt{3}}{2} \ & =\frac{\sqrt{3}}{2} \pi^{2} \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2}=\frac{3}{16} \pi^{2} \ & 	ext { at } x=1, y=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \frac{\pi}{3 \sqrt{2}}(2)^{3 / 2}ight)ight) \ & =\sin ^{3}\left(\frac{\pi}{3}\left(\cos \frac{2 \pi}{3}ight)ight) \ & =\sin ^{3}\left(-\frac{\pi}{3} 	imes \frac{1}{2}ight)=-\frac{1}{8} \ & \Rightarrow 2 y^{\prime}+3 \pi^{2} y=2 	imes \frac{3}{16} \pi^{2}+3 \pi^{2} 	imes\left(-\frac{1}{8}ight)=0 \end{aligned} $$