Question

If \[x = e^{\cos 2 t} \text{ and y }= e^{\sin 2 t} ,\] prove that \[\frac{dy}{dx} = - \frac{y \log x}{x \log y}\] ?

Answer 1

\[\text{ We have, x  }= e^{\cos2 t} \text{ and y } = e^{ \sin 2 t }\]

\[\Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left( e^{\cos2t} \right) \text{ and }\frac{dy}{dt} = \frac{d}{dt}\left( e^{\sin2t} \right)\]
\[ \Rightarrow \frac{dx}{dt} = e^{\cos2t} \frac{d}{dt}\left( \cos2t \right) \text{ and } \frac{dy}{dt} = e^{ \sin2t } \frac{d}{dt}\left( \sin2t \right)\]
\[ \Rightarrow \frac{dx}{dt} = e^{ \cos2t } \left( - \sin2t \right)\frac{d}{dt}\left( 2t \right) and \frac{dy}{dt} = e^{ \sin2t } \left( \cos2t \right)\frac{d}{dt}\left( 2t \right) \]
\[ \Rightarrow \frac{dx}{dt} = - 2\sin 2t e^{ \cos2t }\text{  and } \frac{dy}{dt} = 2\cos2t e^{ \sin2t }\]
\[ \because \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 \cos 2t e^{ \sin2t }}{- 2\sin2t e^{ \cos2t }}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{y \log x}{x \log y} .........[{ \because x = e^{\cos2t } \Rightarrow \log x = \cos2t, y = e^{\sin2t} \Rightarrow \log y = \sin 2t}]\]