If `x=3costheta-cos^3theta` `y=3sintheta-sin^3theta` find `dy/dx`

Explanation

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

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

We have:
\[ x = 3\cos\theta - \cos^3\theta \]
To find \(\frac{dx}{d\theta}\), we differentiate \(x\):
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(3\cos\theta) - \frac{d}{d\theta}(\cos^3\theta) \]
Using the chain rule for the second term:
\[ \frac{dx}{d\theta} = 3(-\sin\theta) - 3\cos^2\theta(-\sin\theta) \]
This simplifies to:
\[ \frac{dx}{d\theta} = -3\sin\theta + 3\sin\theta\cos^2\theta \]
Factoring out \(-3\sin\theta\):
\[ \frac{dx}{d\theta} = -3\sin\theta(1 - \cos^2\theta) \]
Using the identity \(1 - \cos^2\theta = \sin^2\theta\):
\[ \frac{dx}{d\theta} = -3\sin\theta\sin^2\theta = -3\sin^3\theta \]

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

Now we differentiate \(y\):
\[ y = 3\sin\theta - \sin^3\theta \]
Differentiating \(y\):
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(3\sin\theta) - \frac{d}{d\theta}(\sin^3\theta) \]
Using the chain rule for the second term:
\[ \frac{dy}{d\theta} = 3\cos\theta - 3\sin^2\theta\cos\theta \]
Factoring out \(3\cos\theta\):
\[ \frac{dy}{d\theta} = 3\cos\theta(1 - \sin^2\theta) \]
Using the identity \(1 - \sin^2\theta = \cos^2\theta\):
\[ \frac{dy}{d\theta} = 3\cos\theta\cos^2\theta = 3\cos^3\theta \]

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

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{3\cos^3\theta}{-3\sin^3\theta} \]
The \(3\) cancels out:
\[ \frac{dy}{dx} = -\frac{\cos^3\theta}{\sin^3\theta} \]
This can be rewritten using the cotangent function:
\[ \frac{dy}{dx} = -\cot^3\theta \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\cot^3\theta \]

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