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If $x$ and $y$ are connected parametrically by the given equations, without eliminating the parameter. Find $\frac{d y}{d x}$
$x=\cos \theta-\cos 2 \theta, \quad y=\sin \theta-\sin 2 \theta$
$x=\cos \theta-\cos 2 \theta, \quad y=\sin \theta-\sin 2 \theta$
Solution:
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Verified Answer
$\frac{d x}{d \theta}=-\sin \theta-(-\sin 2 \theta) \cdot 2=2 \sin 2 \theta-\sin \theta$
$\frac{d y}{d \theta}=\cos \theta-2 \cos 2 \theta \quad \therefore \frac{d y}{d x}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}$
$\frac{d y}{d \theta}=\cos \theta-2 \cos 2 \theta \quad \therefore \frac{d y}{d x}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}$
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