If `x= e^(cos2t)` and `y = e^(sin2t)`, then prove that `(dy)/(dx) = -(ylogx)/(xlogy)`.

Explanation

To prove that \(\frac{dy}{dx} = -\frac{y \log x}{x \log y}\) given \(x = e^{\cos 2t}\) and \(y = e^{\sin 2t}\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

1. **Differentiate \(x\)**:
\[ x = e^{\cos 2t} \]
Using the chain rule:
\[ \frac{dx}{dt} = e^{\cos 2t} \cdot \frac{d}{dt}(\cos 2t) = e^{\cos 2t} \cdot (-\sin 2t) \cdot 2 = -2 \sin 2t \cdot e^{\cos 2t} \]

2. **Differentiate \(y\)**:
\[ y = e^{\sin 2t} \]
Similarly, using the chain rule:
\[ \frac{dy}{dt} = e^{\sin 2t} \cdot \frac{d}{dt}(\sin 2t) = e^{\sin 2t} \cdot \cos 2t \cdot 2 = 2 \cos 2t \cdot e^{\sin 2t} \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the relationship \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\[ \frac{dy}{dx} = \frac{2 \cos 2t \cdot e^{\sin 2t}}{-2 \sin 2t \cdot e^{\cos 2t}} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\cos 2t \cdot e^{\sin 2t}}{\sin 2t \cdot e^{\cos 2t}} \]

### Step 3: Substitute \(e^{\sin 2t}\) and \(e^{\cos 2t}\)

Since \(y = e^{\sin 2t}\) and \(x = e^{\cos 2t}\), we can substitute these into our expression:
\[ \frac{dy}{dx} = -\frac{\cos 2t \cdot y}{\sin 2t \cdot x} \]

### Step 4: Express \(\cos 2t\) and \(\sin 2t\) in terms of \(\log x\) and \(\log y\)

From the definitions of \(x\) and \(y\):
- Taking logarithms:
\[ \log x = \cos 2t \quad \text{(since \(x = e^{\cos 2t}\))} \]
\[ \log y = \sin 2t \quad \text{(since \(y = e^{\sin 2t}\))} \]

### Step 5: Substitute back into \(\frac{dy}{dx}\)

Now substituting \(\cos 2t\) and \(\sin 2t\):
\[ \frac{dy}{dx} = -\frac{\log x \cdot y}{\log y \cdot x} \]

### Conclusion

Thus, we have shown that:
\[ \frac{dy}{dx} = -\frac{y \log x}{x \log y} \]
Hence proved.