Given: #f(x) = 1/(e^((ln x - 2)^2)#
Move everything to the numerator by changing signs on the #e#:
#f(x) = 1/(e^((ln x - 2)^2)) = e^(-(ln x - 2)^2)#
Use the derivative rule: # (e^u)' = u' e^u#
Let #u = -(ln x - 2)^2; " "w = ln x - 2; " "k = -1#
#w' = 1/x -0 = 1/x " and " n = 2#
Using the chain rule: #(u^n)' = nk w^(n-1) w'#
#u' = -2(ln x - 2)^1 (1/x) = (-2(ln x - 2))/x#
Since # (e^u)' = u' e^u," "#then
#f'(x) = (-2(ln x - 2))/x e^(-(ln x - 2)^2)#
To simplify, #e^(-(ln x - 2)^2) = e^(-(ln)^2 + ln x^4 - 4)#
#= e^(-(ln)^2 - 4) e^(ln x^4)#
#= e^(-(ln)^2 - 4)x^4#
#f'(x) = (-2(ln x - 2))/x e^(-(ln)^2 - 4)x^4#
#f'(x) = -2(ln x - 2) e^(-(ln)^2 - 4)x^3#
Rearranging:
#f'(x) = -2x^3(lnx - 2)e^(-(ln x)^2-4)#
#f'(x) = -2x^3(lnx - 2)e^(-((ln x)^2+4)#
#f'(x) = -2x^3(lnx - 2)/e^((ln x)^2+4#
