How do you differentiate # f(x)=e^((6x-2)^2 # using the chain rule.?

1 Answer
Apr 6, 2016

Answer:

Answer:
#h'(x)= 12(6x-2)*e^((6x-2)^2)=(72x-24)*e^((6x-2)^2)#

Explanation:

Given : #f(x)=e^((6x-2)^2#

Required: The derivative of #f(x)#

Definition and principles: The chain rule
Let #h(x)=(f@g)(x) # then the derivative of h(x) is

#h'(x)= f'(g(x))g'(x)#

Solution Strategy: let #f(x)=(6x-2)^2# and #g(x)=e^(f(x))#
A) Evaluate #f'(g(x)) and #g'(x)# B) Put it all together - #f'(g(x))*g'(x) #

A) I am going to write the chain rule as:
let #g(x) = (6x-2)^2#, then #g'(x)= 12(6x-2)#
now #f(g(x))= e^(g(x))# and #f'(g(x)) = e^g(x)=color(red)(e^((6x-2)^2)#
B) #f'(g(x))*g'(x) = color(red)(e^((6x-2)^2)*color(blue)(12(6x-2))#

Answer:
#h'(x)= 12(6x-2)*e^((6x-2)^2)=(72x-24)*e^((6x-2)^2)#

Remark: for the chain rule you can use the notation given below which I find a little simpler to follow and is the notation most frequently used especially in physics community.

#(dh(x))/(dx)=(df(u))/(du)* (du)/(dx)# and you let
#u=(6x-2)^2# and #f(u)=e^u# then
#(df(u))/(du)=e^u and (du)/(dx)=12(6x-2)#
and #(dh(x))/(dx)=[e^u*12(6x-2)]_(u=(6x-2)^2)=(72x-24)*e^((6x-2)^2)#