If x = 2 cos^3 theta and y = 3 sin^2 theta,th | vedclass

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Question

(A) Given $x = 2 \cos^3 	heta$ and $y = 3 \sin^2 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = 2 \cdot 3 \cos^2 	heta

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$-\sec 	heta$

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(A) Given $x = 2 \cos^3 	heta$ and $y = 3 \sin^2 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = 2 \cdot 3 \cos^2 	heta \cdot (-\sin 	heta) = -6 \cos^2 	heta \sin 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = 3 \cdot 2 \sin 	heta \cdot \cos 	heta = 6 \sin 	heta \cos 	heta$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{6 \sin 	heta \cos 	heta}{-6 \cos^2 	heta \sin 	heta}$. Simplifying the expression: $\frac{dy}{dx} = \frac{1}{-\cos 	heta} = -\sec 	heta$.

If $x = 2 \cos^3 \theta$ and $y = 3 \sin^2 \theta$,then $\frac{dy}{dx} = $

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