Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=a(costheta-cos2theta),y=a(sintheta-sin2theta)`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = a(\cos\theta - \cos 2\theta)\) and \(y = a(\sin\theta - \sin 2\theta)\), we will differentiate both \(x\) and \(y\) with respect to \(\theta\) and then use the chain rule to find \(\frac{dy}{dx}\).

### Step-by-Step Solution:

1. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = a(\cos\theta - \cos 2\theta) \]
Using the chain rule, we differentiate:
\[ \frac{dx}{d\theta} = a\left(-\sin\theta + 2\sin 2\theta\right) \]
Here, we used the derivative of \(\cos\theta\) which is \(-\sin\theta\) and the derivative of \(-\cos 2\theta\) which is \(2\sin 2\theta\) (using the chain rule).

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = a(\sin\theta - \sin 2\theta) \]
Again, using the chain rule:
\[ \frac{dy}{d\theta} = a\left(\cos\theta - 2\cos 2\theta\right) \]
Here, we used the derivative of \(\sin\theta\) which is \(\cos\theta\) and the derivative of \(-\sin 2\theta\) which is \(-2\cos 2\theta\).

3. **Find \(\frac{dy}{dx}\)**:
Using the relationship \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\):
\[ \frac{dy}{dx} = \frac{a(\cos\theta - 2\cos 2\theta)}{a(-\sin\theta + 2\sin 2\theta)} \]
The \(a\) cancels out:
\[ \frac{dy}{dx} = \frac{\cos\theta - 2\cos 2\theta}{-\sin\theta + 2\sin 2\theta} \]

4. **Final Result**:
Therefore, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta} \]