1.

If `xlogy-ylogx=1` then `((dy)/(dx))_(x=1)=`

Answer» `logy+x*1/y*dy/dx-logx*dy/dx-y*1/x=0`
`logy-y/x+dy/dx(x/y-logx)=0`
`dy/dx=((y/x-logy)/(x/y-logx))`
`(dy/dx)_(x=1,y=e)=((e/1-loge)/(1/e-log1))`
`=(e-1)/(1/e)=e(e-1)`
Option C is correct.


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