If x = sin 2theta cos 3theta and y = sin 3the | vedclass

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

Vedclass · Accepted Answer

$\frac{3\cos 3	heta \cos 2	heta - 2\sin 3	heta \sin 2	heta}{2\cos 2	heta \cos 3	heta - 3\sin 2	heta \sin 3	heta}$

Vedclass · Answer

(B) Given $x = \sin 2	heta \cos 3	heta$ and $y = \sin 3	heta \cos 2	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = \frac{d}{d	heta}(\sin 2	heta \cos 3	heta) = \cos 2	heta(2) \cos 3	heta + \sin 2	heta(-\sin 3	heta)(3) = 2\cos 2	heta \cos 3	heta - 3\sin 2	heta \sin 3	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = \frac{d}{d	heta}(\sin 3	heta \cos 2	heta) = \cos 3	heta(3) \cos 2	heta + \sin 3	heta(-\sin 2	heta)(2) = 3\cos 3	heta \cos 2	heta - 2\sin 3	heta \sin 2	heta$. Using the chain rule, $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3\cos 3	heta \cos 2	heta - 2\sin 3	heta \sin 2	heta}{2\cos 2	heta \cos 3	heta - 3\sin 2	heta \sin 3	heta}$. Thus, the correct option is $B$.

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

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