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 \(x\) and \(y\)

Given:
\[ x = e^{\cos 2t} \implies \log x = \cos 2t \]
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