Differentiation Question 5
Question 5 - 31 January - Shift 1
Let
$y=f(x)=\sin ^{3}(\frac{\pi}{3}(\cos (\frac{\pi}{3 \sqrt{2}}(-4 x^{3}+5 x^{2}+1)^{\frac{3}{2}})))$. Then, at $x=1$,
(1) $2 y^{\prime}+\sqrt{3} \pi^{2} y=0$
(2) $2 y^{\prime}+3 \pi^{2} y=0$
(3) $\sqrt{2} y^{\prime}-3 \pi^{2} y=0$
(4) $y^{\prime}+3 \pi^{2} y=0$
Show Answer
Answer: (2)
Solution:
Formula: Basic theorems
$y=\sin ^{3}(\pi / 3 \cos g(x))$
$g(x)=\frac{\pi}{3 \sqrt{2}}(-4 x^{3}+5 x^{2}+1)^{3 / 2}$
$g(1)=2 \pi / 3$
$\begin{aligned} y^{\prime}=3 \sin ^{2}(\frac{\pi}{3} \cos g(x)) & \times \cos (\frac{\pi}{3} \cos g(x)) \times \frac{\pi}{3}(-\sin g(x)) g^{\prime}(x)\end{aligned}$
$y^{\prime}(1)=3 \sin ^{2}(-\frac{\pi}{6}) \cdot \cos (\frac{\pi}{6}) \cdot \frac{\pi}{3}(-\sin \frac{2 \pi}{3}) g^{\prime}(1)$
$g^{\prime}(x)=\frac{\pi}{3 \sqrt{2}}(-4 x^{3}+5 x^{2}+1)^{1 / 2}(-12 x^{2}+10 x)$
$g^{\prime}(1)=\frac{\pi}{2 \sqrt{2}}(\sqrt{2})(-2)=-\pi$
$y^{\prime}(1)=\frac{\not p}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\pi}{\not p}(\frac{-\sqrt{3}}{2})(-\pi)=\frac{3 \pi^{2}}{16}$
$y(1)=\sin ^{3}(\pi / 3 \cos 2 \pi / 3)=-\frac{1}{8}$
$2 y^{\prime}(1)+3 \pi^{2} y(1)=0$
