Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=2cos^(2)theta,y=2sin^(2)theta`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 2 \cos^2 \theta\) and \(y = 2 \sin^2 \theta\), we will differentiate both equations 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 \(\theta\)**:
\[ x = 2 \cos^2 \theta \]
Using the chain rule, we differentiate:
\[ \frac{dx}{d\theta} = 2 \cdot 2 \cos \theta \cdot (-\sin \theta) = -4 \cos \theta \sin \theta \]

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = 2 \sin^2 \theta \]
Again, using the chain rule, we differentiate:
\[ \frac{dy}{d\theta} = 2 \cdot 2 \sin \theta \cdot \cos \theta = 4 \sin \theta \cos \theta \]

3. **Find \(\frac{dy}{dx}\)**:
We use the relationship:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{4 \sin \theta \cos \theta}{-4 \cos \theta \sin \theta} \]

4. **Simplify the expression**:
The \(4\), \(\sin \theta\), and \(\cos \theta\) terms cancel out:
\[ \frac{dy}{dx} = \frac{4 \sin \theta \cos \theta}{-4 \cos \theta \sin \theta} = -1 \]

Thus, the final result is:
\[ \frac{dy}{dx} = -1 \]