Question

Find an equation of the tangent line to the graph of the equation ln x xey = 1 at the point (1, 0).

Answer 1

Ln x + xe^y = 1
1/x + xf'(x)e^y + e^y = 0
f'(x) = (-e^y - 1/x)/xe^y
f'(1, 0) = (-e^0 - 1/1)/e^0 = (-1 - 1)/1 = -2

Let the required equation of tangent be y = mx + c, where y = 0, m = -2, x = 1
0 = -2(1) + c = -2 + c
c = 2
Therefore, required equation is y = -2x + 2