If $x = {e^{\cos 2t}}$ and $y = {e^{\sin 2t}}$, then prove that $\frac{{dy}}{{dx}} = - \frac{{y\log x}}{{x\log y}}$

and $y = {e^{\sin 2t}}$
therefore,$\frac{{dx}}{{dt}} = \frac{d}{{dt}}{e^{\cos 2t}} = {e^{\cos 2t}} \cdot \frac{d}{{dt}}\cos 2t$

$= {e^{\cos 2t}} \cdot ( - \sin 2t) \cdot \frac{d}{{dt}}(2t)$
$\frac{{dx}}{{dt}} = - 2{e^{\cos 2t}} \cdot \sin 2t$ …….(i)

and $\frac{{dy}}{{dt}} = \frac{d}{{dt}}{e^{\sin 2t}} = {e^{\sin 2t}} \cdot \frac{d}{{dt}}\sin 2t$

$= {e^{\sin 2t}}\cos 2t \cdot \frac{d}{{dt}}2t$
$= 2{e^{\sin 2t}} \cdot \cos 2t$ …….(ii)

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

$= \frac{{{e^{\sin 2t}} \cdot \cos 2t}}{{{e^{\cos 2t}} \cdot \sin 2t}}$ …….(iii)

We know that, $\log x = \cos 2t \cdot \log e = \cos 2t$ ……(iv)

and $\log y = \sin 2t \cdot \log e = \sin 2t$ …….(v)

therefore, $\frac{{dy}}{{dx}} = \frac{{ - y\log x}}{{x\log y}}$

[using Eqs. (iv) and (v) in Eq. (iii) and $x = {e^{\cos 2t}},y = {e^{\sin 2t}}$]

Hence proved.