How do you find tangent line equation for 1+ln(xy)=e^(x-y) in point (1,1)?

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##y=1,## which is a spiritless verse.

We foreclosure that, the increase of tgt. at pt. ##(h,k)=[dy/dx]_((h.k))##

Hence, diff. twain sides of the attached eqn. of flexion,

##d/dx[1+ln(xy)]=d/dx[e^(x-y)],## ##:. d/dx[1+lnx+lny]=e^(x-y)*d/dx(x-y)##.........[] ##:. 0+1/x+(1/y)dy/dx=e^(x-y){1-dy/dx}.##....[by Implicit diff. ##d/dx(lny)=d/dylny*dy/dx##] ##:. 1/x+(1/y)dy/dx=e^(x-y)-e^(x-y)*dy/dx## ##:. (1/y)dy/dx+e^(x-y)*dy/dx=e^(x-y)-1/x## ##:.{(1+ye^(x-y))/y}dy/dx={xe^(x-y)-1}/x## ##:. dy/dx=(y(xe^(x-y)-1))/(x(1+ye^(x-y))##

##:.## The increase of tgt. at ##(1,1)=[dy/dx]_(x=1,y=1)=0/2=0.##

##:.## The eqn. of tgt., using slop-point formula, is attached by, ##y-1=0(x-1),## i.e., ##y=1,## which is a spiritless verse.

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