If x=e^cos 2x , y=e^sin 2t ,then prove that dy/dx=-y log x/x log y

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$$\begin{gathered} \mathrm{As} \mathrm{x}={\mathrm{e}}^{\cos 2\mathrm{t}} \\ \Rightarrow \ln \left(\mathrm{x}\right)=\ln \left({\mathrm{e}}^{\cos 2\mathrm{t}}\right) \\ \mathrm{As} \ln \left({\mathrm{e}}^{\mathrm{a}}\right)=\mathrm{a} \\ \Rightarrow \ln \left(\mathrm{x}\right)=\cos 2\mathrm{t} ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{y}={\mathrm{e}}^{\sin 2\mathrm{t}} \\ \Rightarrow \ln \left(\mathrm{y}\right)=\ln \left({\mathrm{e}}^{\sin 2\mathrm{t}}\right) \\ \Rightarrow \ln \left(\mathrm{y}\right)=\sin 2\mathrm{t} ;\mathrm{equation}\left(\mathrm{ii}\right) \\ \mathrm{We} \mathrm{have} \\ \mathrm{x}={\mathrm{e}}^{\cos 2\mathrm{t}} \\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}={\mathrm{e}}^{\cos 2\mathrm{t}}.\frac{\mathrm{d}}{\mathrm{dt}}\left(\cos 2\mathrm{t}\right)=-2{\mathrm{e}}^{\cos 2\mathrm{t}}\sin 2\mathrm{t} \\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-2\mathrm{xsin}2\mathrm{t} \left[\mathrm{As} \mathrm{x}={\mathrm{e}}^{\cos 2\mathrm{t}}\right] \\ \mathrm{Using} \mathrm{equation}\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}. \\ \frac{\mathrm{dx}}{\mathrm{dt}}=-2\mathrm{x}.\ln \left(\mathrm{y}\right) ;\mathrm{equation}\left(\mathrm{iii}\right) \\ \mathrm{y}={\mathrm{e}}^{\sin 2\mathrm{t}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{e}}^{\sin 2\mathrm{t}}.\frac{\mathrm{d}}{\mathrm{dt}}\left(\sin 2\mathrm{t}\right)=2{\mathrm{e}}^{\sin 2\mathrm{t}}\cos 2\mathrm{t} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=2\mathrm{ycos}2\mathrm{t} \left[\mathrm{As} \mathrm{y}={\mathrm{e}}^{\sin 2\mathrm{t}}\right] \\ \mathrm{Using} \mathrm{equation}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}. \\ \frac{\mathrm{dy}}{\mathrm{dt}}=2\mathrm{y}.\ln \left(\mathrm{x}\right) ;\mathrm{equation}\left(\mathrm{iv}\right) \\ \mathrm{equation}\left(\mathrm{iv}\right)/\mathrm{equation}\left(\mathrm{iii}\right) \\ \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=-\frac{2\mathrm{y}.\ln \left(\mathrm{x}\right)}{2\mathrm{x}.\ln \left(\mathrm{y}\right)} \\ \mathbf{\Rightarrow }\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}\mathbf{.}\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}\mathbf{.}\mathbf{ln}\left(\mathbf{y}\right)}\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{proved}\right) \end{gathered}$$

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