find the equation of the tangent to the curveexy=x+(In y)2 at the point (0,e)

using the chain rule ....

e^xy(y+xdy/dx)=1 + 2.ln y.1/y.dy/dx

ye^xy-1=[2(ln y)/y-xe^xy] dy/dx

i.e. {y(ye^xy-1)/(2ln y-yxe^xy)}=dy/dx

now putting the values x=o and y=e we get...

e(e-1)/2=dy/dx            [ ln e=1 ]

therefore the tangent to the curve at (0,e) is e(e-1)/2

Approved