Question

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

Answer 1

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)`