ln(xy) - 2x =0
slope of the tangent line = derivative of the function
[ln(xy)]' = [2x]'
[1/(xy)] [y + xy'] = 2
y + xy' = 2(xy)
xy' = 2xy - y =y(2x-1)
y' = y(2x-1)/x
Now use x = -1 to find y and after to find y'
ln(xy) = 2x
x=-1
ln(-y) =-2
-y = e^-2
y = - e^-2
y' = [-e^-2][2(-1)-1]/(-1) = [e^-2](-2-1)= [e^-2](-3) = - 3e^-2
Answer: option 6. from the list
