If $x = e^{\cos 2t}, y = e^{\sin 2t}$, then $\frac{dy}{dx}$ equals to

$-\frac{y\log_ex}{x\log_ey}$

The correct answer is Option (3) → $-\frac{y\log_ex}{x\log_ey}$

Given:

\(y = e^{\sin 2t}\)

\(\Rightarrow \ln y = \sin 2t\)

Also,

\[ \frac{dy}{dx} = - e^{\sin 2t - \cos 2t} \cdot \cot 2t = - \frac{e^{\sin 2t}}{e^{\cos 2t}} \cdot \cot 2t = - \frac{y}{e^{\cos 2t}} \cdot \cot 2t \]

From \(x = e^{\cos 2t}\),

\(\ln x = \cos 2t\)

So,

\[ \frac{dy}{dx} = - \frac{y}{x} \cdot \cot 2t \]

Express \(\cot 2t\) in terms of \(\sin 2t = \ln y\) and \(\cos 2t = \ln x\):

\[ \cot 2t = \frac{\cos 2t}{\sin 2t} = \frac{\ln x}{\ln y} \]

Therefore,

\({\frac{dy}{dx} = - \frac{y}{x} \cdot \frac{\ln x}{\ln y}}\)