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

find the equation of the tangent to the curve


exy=x+(In y)2  at the point (0,e)


 

Grade:

1 Answers

arka mitra
18 Points
10 years ago

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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