How do you differentiate # f(x)=1/(e^((lnx-2)^2 ))# using the chain rule.?

1 Answer
Jul 7, 2018

Answer:

#f'(x) = -2x^3(lnx - 2)e^(-((ln x)^2+4)#

or
#f'(x) = -2x^3(lnx - 2)/e^((ln x)^2+4#

Explanation:

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#