`x=sqrt(sin 2t),y=sqrt(cos 2 t)`

Answer

To find \(\frac{dy}{dx}\) for the given parametric equations \(x = \sqrt{\sin(2t)}\) and \(y = \sqrt{\cos(2t)}\), we will use the chain rule and the derivatives with respect to the parameter \(t\).

### Step-by-Step Solution:

1. **Differentiate \(x\) with respect to \(t\)**:
\[ x = \sqrt{\sin(2t)} = (\sin(2t))^{1/2} \]
Using the chain rule:
\[ \frac{dx}{dt} = \frac{1}{2} (\sin(2t))^{-1/2} \cdot \frac{d}{dt}(\sin(2t)) \]
The derivative of \(\sin(2t)\) is \(2\cos(2t)\):
\[ \frac{dx}{dt} = \frac{1}{2} (\sin(2t))^{-1/2} \cdot 2\cos(2t) = \frac{\cos(2t)}{\sqrt{\sin(2t)}} \]

2. **Differentiate \(y\) with respect to \(t\)**:
\[ y = \sqrt{\cos(2t)} = (\cos(2t))^{1/2} \]
Again, using the chain rule:
\[ \frac{dy}{dt} = \frac{1}{2} (\cos(2t))^{-1/2} \cdot \frac{d}{dt}(\cos(2t)) \]
The derivative of \(\cos(2t)\) is \(-2\sin(2t)\):
\[ \frac{dy}{dt} = \frac{1}{2} (\cos(2t))^{-1/2} \cdot (-2\sin(2t)) = -\frac{\sin(2t)}{\sqrt{\cos(2t)}} \]

3. **Find \(\frac{dy}{dx}\)** using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substitute the derivatives we found:
\[ \frac{dy}{dx} = \frac{-\frac{\sin(2t)}{\sqrt{\cos(2t)}}}{\frac{\cos(2t)}{\sqrt{\sin(2t)}}} \]
Simplifying this gives:
\[ \frac{dy}{dx} = -\frac{\sin(2t)}{\sqrt{\cos(2t)}} \cdot \frac{\sqrt{\sin(2t)}}{\cos(2t)} = -\frac{\sin(2t) \cdot \sqrt{\sin(2t)}}{\cos(2t) \cdot \sqrt{\cos(2t)}} \]

4. **Express \(\frac{dy}{dx}\) in terms of tangent**:
Recognizing that \(\tan(2t) = \frac{\sin(2t)}{\cos(2t)}\), we can write:
\[ \frac{dy}{dx} = -\tan(2t) \cdot \sqrt{\frac{\sin(2t)}{\cos(2t)}} = -\tan(2t) \cdot \sqrt{\tan(2t)} \]
Thus, we have:
\[ \frac{dy}{dx} = -\tan(2t) \cdot \sqrt{\tan(2t)} \]

### Final Answer:
\[ \frac{dy}{dx} = -\tan(2t) \]