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

Answer

N/a

Explanation

`:' x = e^(cos2t)` and ` y = e^(sin2t)`
`:. (dx)/(dt) = (d)/(dt)e^(cos2t) = e^(cos2t).(d)/(dt) cos 2t`
`= e^(cos2t).(-sin2t).(d)/(dt) (2t)`
`(dx)/(dt) = - 2e^(cos2t).sin2t"......"(i)`
and `(dy)/(dt) = (d)/(dt)e^(sin2t)=e^(sin2t).(d)/(dt) sin 2t`
`=e^(sin2t)cos2t.(d)/(dt)2t`
`= 2e^(sin2t).cos2t"......"(ii)`
`:. (dy)/(dx) = (dy//dt)/(dx//dt) = (2e^(sin2t).cos2t)/(-2e^(cos2t).sin2t)"`
`= (e^(sin2t).cos2t)/(e^(cos2t).sin2t)"....."(iii)`
We know that , `logx=cos2t.loge=cos2t"......(iv)`
and `logy = sin2t.loge=sin2t"....."(v)`
`:. (dy)/(dx) = (-ylogx)/(xlogy)`
[using Eqs. (iv) and (v) in Eq. (iii) and `x=e(cos2t), y = e^(sin2t)`]