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If x= e^(cos2t) and y = e^(sin2t), the...

If `x= e^(cos2t)` and `y = e^(sin2t)`, then move that `(dy)/(dx) = -(ylogx)/(xlogy)`.

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`x=e^(cos2t)and y=e^(sin 2t)`
` cos 2t= log x and sin 2t = log y`
`therefore" "cos^(2) 2t +sin^(2) 2t = (log x)^(2) + (log y)^(2)`
`rArr" "(log x)^(2)+(log y)^(2)=1`
Differentiating both sides w.r.t. x, we get
`2log x(1)/(x)+2 log y (1)/(y)(dy)/(dx)=0`
`rArr" "(dy)/(dx)=(-y log x)/(x log y)`
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