If `x = 3 cos theta - 2 cos^3 theta,y = 3 sin theta - 2 sin^3 theta`, then `dy/dx` is

Explanation

To find \( \frac{dy}{dx} \) given the parametric equations \( x = 3 \cos \theta - 2 \cos^3 \theta \) and \( y = 3 \sin \theta - 2 \sin^3 \theta \), we will follow these steps:

### Step 1: Differentiate \( x \) with respect to \( \theta \)

Given:
\[ x = 3 \cos \theta - 2 \cos^3 \theta \]

Differentiating \( x \) with respect to \( \theta \):
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(3 \cos \theta) - \frac{d}{d\theta}(2 \cos^3 \theta) \]
Using the chain rule:
\[ \frac{dx}{d\theta} = 3(-\sin \theta) - 2 \cdot 3 \cos^2 \theta (-\sin \theta) \]
\[ = -3 \sin \theta + 6 \cos^2 \theta \sin \theta \]
\[ = \sin \theta (6 \cos^2 \theta - 3) \]

### Step 2: Differentiate \( y \) with respect to \( \theta \)

Given:
\[ y = 3 \sin \theta - 2 \sin^3 \theta \]

Differentiating \( y \) with respect to \( \theta \):
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(3 \sin \theta) - \frac{d}{d\theta}(2 \sin^3 \theta) \]
Using the chain rule:
\[ \frac{dy}{d\theta} = 3 \cos \theta - 2 \cdot 3 \sin^2 \theta \cos \theta \]
\[ = 3 \cos \theta - 6 \sin^2 \theta \cos \theta \]
\[ = \cos \theta (3 - 6 \sin^2 \theta) \]

### Step 3: Find \( \frac{dy}{dx} \)

Using the formula:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\cos \theta (3 - 6 \sin^2 \theta)}{\sin \theta (6 \cos^2 \theta - 3)} \]

### Step 4: Simplify the expression

Factoring out common terms:
\[ = \frac{3 \cos \theta - 6 \sin^2 \theta \cos \theta}{6 \cos^2 \theta - 3} = \frac{3 \cos \theta (1 - 2 \sin^2 \theta)}{3(2 \cos^2 \theta - 1)} \]
\[ = \frac{\cos \theta (1 - 2 \sin^2 \theta)}{2 \cos^2 \theta - 1} \]

Using the identity \( 1 - 2 \sin^2 \theta = \cos 2\theta \) and \( 2 \cos^2 \theta - 1 = \cos 2\theta \):
\[ = \frac{\cos \theta \cos 2\theta}{\cos 2\theta} \]

Thus, we can cancel \( \cos 2\theta \) (assuming \( \cos 2\theta \neq 0 \)):
\[ \frac{dy}{dx} = \cos \theta \]

### Final Answer:
\[ \frac{dy}{dx} = \cos \theta \]