# Find $\frac{dy}{dx}$ when $x=\cos^3{\theta}$, $y=\sin^3{\theta}$

How did they get the 1st step in the 1st place?

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$$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$$ –  Ｊ. Ｍ. Nov 18 '11 at 7:36

For the sake of having an answer...

They used the chain rule,

$$\frac{dy}{dx} = \frac{dy}{d\theta} \frac{d\theta}{dx} = \frac{dy}{d\theta} \Big/ \frac{dx}{d\theta} = \frac{3\sin^2\theta \cos\theta}{-3\cos^2\theta\sin\theta}$$

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