

InterviewSolution
Saved Bookmarks
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. |
|