Find `(dy)/(dx)`, when
`x=sqrt(sin 2theta), y=sqrt(cos 2theta)`

Answer

To find \(\frac{dy}{dx}\) when \(x = \sqrt{\sin 2\theta}\) and \(y = \sqrt{\cos 2\theta}\), we will use the chain rule of differentiation. Here’s the step-by-step solution:

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

Given:
\[ y = \sqrt{\cos 2\theta} \]

Using the chain rule, we differentiate \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = \frac{1}{2\sqrt{\cos 2\theta}} \cdot \frac{d}{d\theta}(\cos 2\theta) \]
Now, we differentiate \(\cos 2\theta\):
\[ \frac{d}{d\theta}(\cos 2\theta) = -\sin 2\theta \cdot 2 = -2\sin 2\theta \]
Thus,
\[ \frac{dy}{d\theta} = \frac{1}{2\sqrt{\cos 2\theta}} \cdot (-2\sin 2\theta) = -\frac{\sin 2\theta}{\sqrt{\cos 2\theta}} \]

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

Given:
\[ x = \sqrt{\sin 2\theta} \]

Using the chain rule again, we differentiate \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = \frac{1}{2\sqrt{\sin 2\theta}} \cdot \frac{d}{d\theta}(\sin 2\theta) \]
Now, we differentiate \(\sin 2\theta\):
\[ \frac{d}{d\theta}(\sin 2\theta) = \cos 2\theta \cdot 2 = 2\cos 2\theta \]
Thus,
\[ \frac{dx}{d\theta} = \frac{1}{2\sqrt{\sin 2\theta}} \cdot (2\cos 2\theta) = \frac{\cos 2\theta}{\sqrt{\sin 2\theta}} \]

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

Using the chain rule, we find \(\frac{dy}{dx}\) as follows:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \frac{-\frac{\sin 2\theta}{\sqrt{\cos 2\theta}}}{\frac{\cos 2\theta}{\sqrt{\sin 2\theta}}} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sin 2\theta}{\cos 2\theta} \cdot \frac{\sqrt{\sin 2\theta}}{\sqrt{\cos 2\theta}} = -\tan 2\theta \cdot \sqrt{\frac{\sin 2\theta}{\cos 2\theta}} = -\tan 2\theta \cdot \sqrt{\tan 2\theta} \]

### Final Result

Thus, the final result is:
\[ \frac{dy}{dx} = -\tan 2\theta \cdot (\tan 2\theta)^{1/2} = -(\tan 2\theta)^{3/2} \]