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If `x^(y)=y^(x)`, fin `(dy)/(dx).` |
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Answer» Given : `x^(y)=y^(x)` `rArr ylog x=x logy." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `y.(d)/(dx)(logx)+(logx).(d)/(dx)(y)=x.(d)/(dx)(logy)+(logy).(d)/(dx)(x)` `rArry.(1)/(x)+(logx).(dy)/(dx)=x.(1)/(y).(dy)/(dx)=x.(1)/(y).(dy)/(dx)+(logy).1` `rArr(logx-(x)/(y))(dy)/(dx)=(logy-(y)/(x))` `rArr((ylogx-x))/(y).(dy)/(dx)=((xlogy-y))/(x)` `rArr(dy)/(dx)=(y(xlogy-y))/(x(logx-x)).` |
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