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find the equation of the tangent to the curve
exy=x+(In y)2 at the point (0,e)
using the chain rule ....
exy(y+xdy/dx)=1 + 2.ln y.1/y.dy/dx
yexy-1=[2(ln y)/y-xexy] dy/dx
i.e. {y(yexy-1)/(2ln y-yxexy)}=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
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