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


We have, $x=e^{\cos 2 t}$ and $y=e^{\sin 2 t}$

$\Rightarrow \frac{d x}{d t}=\frac{d}{d t}\left(e^{\cos 2 t}\right)$ and $\frac{d y}{d t}=\frac{d}{d t}\left(e^{\sin 2 t}\right)$

$\Rightarrow \frac{d x}{d t}=e^{\cos 2 t} \frac{d}{d t}(\cos 2 t)$ and $\frac{d y}{d t}=e^{\sin 2 t} \frac{d}{d t}(\sin 2 t)$

$\Rightarrow \frac{d x}{d t}=e^{\cos 2 t}(-\sin 2 t) \frac{d}{d t}(2 t)$ and $\frac{d y}{d t}=e^{\sin 2 t}(\cos 2 t) \frac{d}{d t}(2 t)$

$\Rightarrow \frac{d x}{d t}=-2 \sin 2 t e^{\cos 2 t}$ and $\frac{d y}{d t}=2 \cos 2 t e^{\sin 2 t}$

$\because \frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 \cos 2 t e^{\sin 2 t}}{-2 \sin 2 t e^{\cos 2 t}}$

$\Rightarrow \frac{d y}{d x}=-\frac{y \log x}{x \log y}$            $\left[\begin{array}{c}\because x=e^{\cos 2 t} \Rightarrow \log x=\cos 2 t \\ y=e^{\sin 2 t} \Rightarrow \log y=\sin 2 t\end{array}\right]$

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