Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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`