If x = sin theta and y = sin^3 theta,then the | vedclass

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(C) Given $x = \sin 	heta$ and $y = \sin^3 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{d	heta} = 3 \sin^2 	heta \cos 	het

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$3$

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(C) Given $x = \sin 	heta$ and $y = \sin^3 	heta$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{d	heta} = 3 \sin^2 	heta \cos 	heta$ $\frac{dx}{d	heta} = \cos 	heta$ $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3 \sin^2 	heta \cos 	heta}{\cos 	heta} = 3 \sin^2 	heta$. Now,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{dx} (3 \sin^2 	heta) = \frac{d}{d	heta} (3 \sin^2 	heta) \cdot \frac{d	heta}{dx}$. Since $\frac{dx}{d	heta} = \cos 	heta$,then $\frac{d	heta}{dx} = \frac{1}{\cos 	heta}$. $\frac{d^2 y}{dx^2} = (6 \sin 	heta \cos 	heta) \cdot \frac{1}{\cos 	heta} = 6 \sin 	heta$. At $	heta = \frac{\pi}{6}$: $\frac{d^2 y}{dx^2} = 6 \sin \left( \frac{\pi}{6} ight) = 6 \cdot \frac{1}{2} = 3$.

If $x = \sin \theta$ and $y = \sin^3 \theta$,then the value of $\frac{d^2 y}{d x^2}$ at $\theta = \frac{\pi}{6}$ is:

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