Given f(x)= 3e^(1-x^2)*ln(x), find the equation of the tangent line at x = 1.

Answer 1

f'(x) = -3/x e^(1-x^2) (2x^2 lnx - 1)

f(1) = 0
f'(1) = -3(1)(-1) = 3

So, now you have a point (1,0) and a slope (3) and the line is

y-0 = 3(x-1)