If `x=e^(cos2t)` and `y=e^(sin2t)` , 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: Take the logarithm of both sides
Starting with the given equations:
\[ x = e^{\cos 2t} \quad \text{and} \quad y = e^{\sin 2t} \]
Taking the logarithm of both sides, we have:
\[ \log x = \cos 2t \quad \text{and} \quad \log y = \sin 2t \]

### Step 2: Differentiate both sides with respect to \(t\)
Now, we differentiate both equations with respect to \(t\):
1. For \(\log x\):
\[ \frac{d}{dt}(\log x) = \frac{1}{x} \frac{dx}{dt} = -2 \sin 2t \]
Thus, we have:
\[ \frac{dx}{dt} = -2x \sin 2t \]

2. For \(\log y\):
\[ \frac{d}{dt}(\log y) = \frac{1}{y} \frac{dy}{dt} = 2 \cos 2t \]
Thus, we have:
\[ \frac{dy}{dt} = 2y \cos 2t \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \frac{2y \cos 2t}{-2x \sin 2t} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{y \cos 2t}{x \sin 2t} \]

### Step 4: Substitute back \(\cos 2t\) and \(\sin 2t\)
From our earlier steps, we know:
\[ \cos 2t = \log x \quad \text{and} \quad \sin 2t = \log y \]
Substituting these into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{y \log x}{x \log y} \]

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