What is the equation of the tangent line given the curve:

Question

y = (3+lnx)^2 at x = 1/e

dy/dx = 2[3 + Ln(x))]*1/x

at x = 1/e this is: 4*e

y(1/e) = 4

The equation of the tangent line is thus given by:

4e(x-1/e) + 4