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

Given $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$.

When both variables depend on a parameter $t$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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For $x = e^{\cos 2t}$:

$\frac{dx}{dt} = e^{\cos 2t} \times \frac{d}{dt}(\cos 2t)$

$\frac{dx}{dt} = e^{\cos 2t} \times (-\sin 2t) \times 2$

$\frac{dx}{dt} = -2\sin 2t \cdot e^{\cos 2t}$

$\frac{dx}{dt} = -2x\sin 2t$

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For $y = e^{\sin 2t}$:

$\frac{dy}{dt} = e^{\sin 2t} \times \frac{d}{dt}(\sin 2t)$

$\frac{dy}{dt} = e^{\sin 2t} \times \cos 2t \times 2$

$\frac{dy}{dt} = 2y\cos 2t$

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$\frac{dy}{dx} = \frac{2y\cos 2t}{-2x\sin 2t}$

$\frac{dy}{dx} = \frac{-y\cos 2t}{x\sin 2t}$

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From $x = e^{\cos 2t}$:

$\ln x = \cos 2t$

$\cos 2t = \log_e x$

From $y = e^{\sin 2t}$:

$\ln y = \sin 2t$

$\sin 2t = \log_e y$

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$\frac{dy}{dx} = \frac{-y\cos 2t}{x\sin 2t}$

$\frac{dy}{dx} = \frac{-y \cdot \log_e x}{x \cdot \log_e y}$

Therefore, $\frac{dy}{dx} = \frac{-y\log_e x}{x\log_e y}$