Applications of Derivatives
Chain rule for composite trigonometric functions
nta_pyq_2023_jan
Grade 12
Question:
Let $y = f(x) = \sin^{3}\!\left(\frac{\pi}{3}\left(\cos\left(\frac{\pi}{3\sqrt{2}}\left(-4x^{3}+5x^{2}+1\right)^{3/2}\right)\right)\right)$. Then at $x = 1$,
(1) $2y' + \sqrt{3}\pi^{2}y = 0$
(2) $2y' + 3\pi^{2}y = 0$
(3) $\sqrt{2}y' - 3\pi^{2}y = 0$
(4) $y' + 3\pi^{2}y = 0$
$2y' + \sqrt{3}\pi^{2}y = 0$
$2y' + 3\pi^{2}y = 0$
$\sqrt{2}y' - 3\pi^{2}y = 0$
$y' + 3\pi^{2}y = 0$
Step-by-Step Solution
Key Concept: Apply chain rule multiple times. Compute $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3+5x^2+1)^{3/2}$, find $g(1) = \frac{2\pi}{3}$ and $g'(1) = -\pi$, then differentiate $y = \sin^3\!\left(\frac{\pi}{3}\cos(g(x))\right)$.
Let $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3+5x^2+1)^{3/2}$. At $x=1$: $g(1) = \frac{\pi}{3\sqrt{2}}(2)^{3/2} = \frac{2\pi}{3}$. $g'(1) = -\pi$. $y(1) = \sin^3(\frac{\pi}{3}\cos\frac{2\pi}{3}) = \sin^3(-\frac{\pi}{6}) = -\frac{1}{8}$. After differentiating and evaluating: $y'(1) = \frac{3\pi^2}{16}$. Check option (2): $2(\frac{3\pi^2}{16}) + 3\pi^2(-\frac{1}{8}) = \frac{3\pi^2}{8} - \frac{3\pi^2}{8} = 0$. Answer: (2).
Correct Answer: 2