How do you differentiate #f(x) = (3x ^2 -3x + 8) ^4# using the chain rule?

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Answer 1 · 2018-04-11T00:51:17

#4(6x-3)(3x^2 -3x + 8)^3#

Let #u= 3x^2 -3x +8#
Let #v= u^4#

The differential of #u#

#u^'= 3(2x) -3 = 6x-3 #

The differential of #v#

#v^' = 4u^3#

You will now multiply the differentials

#(6x-3)(4u^3)#

Replace #u# with the original function

#4(6x-3)(3x^2 -3x +8)^3#

Answer 2 · 2018-04-11T05:35:21

#4(6x-3)(3x^2-3x+8)^3#

The chain rule states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=3x^2-3x+8,:.(du)/dx=6x-3#.

Then #y=u^4,dy/(du)=4u^3#.

So, we get,

#dy/dx=4u^3(6x-3)#

Substituting back our original terms, we get,

#=4(3x^2-3x+8)^3(6x-3)#

Personally, I'll make this neater, and rearrange it into

#=4(6x-3)(3x^2-3x+8)^3#