Question

How do you differentiate `g(y) =(x^2 - 2x + 1)^4 (4x^6 + 5)` using the product rule?

Answer 1

Treat the two terms as a normal product rule, but then use the chain rule on the first function when differentiating to get the full solution: `g'(x)=8(x-1)^7(7x^6-3x^5+5)`

Explanation:

First, let's designate the two pieces of the function as `f_1(x)` and `f_2(x)`:

`f_1(x)=(x^2-2x+1)^4`
`f_2(x)=4x^6+5`

For a given function, the derivative using the product rule is:

`(dy)/(dx)(f_1(x)f_2(x))=f_1'(x)*f_2(x)+f_1(x)*f_2'(x)`

This means we'll need to know the derivatives of each `f` function. The first function requires a chain rule expansion:

`(dy)/(dx)f(g(x))=f'(g(x))*g'(x)`

Deriving `f_1'(x)`:

`(dy)/(dx)(x^2-2x+1)^4=4(x^2-2x+1)^3*(2x-2)`

`f_1'(x)=(x^2-2x+1)^3(8x-8)`

Deriving `f_2'(x)`:

`(dy)/(dx)(4x^6+5)=24x^5`

Now, we reassemble (and simplify):

`g'(x)=(x^2-2x+1)^3(8x-8)(4x^6+5)+24x^5(x^2-2x+1)^4`

`g'(x)=8((x^2-2x+1)^3(x-1)(4x^6+5)+3x^5(x^2-2x+1)^4)`

Note that the factored form of `x^2-2x+1` is `(x-1)^2`

`g'(x)=8(((x-1)^2)^3(x-1)(4x^6+5)+3x^5((x-1)^2)^4)`

`g'(x)=8((x-1)^6(x-1)(4x^6+5)+3x^5(x-1)^8)`

`g'(x)=8((x-1)^7(4x^6+5)+3x^5(x-1)^8)`

`g'(x)=8(x-1)^7((4x^6+5)+3x^5(x-1))`

`g'(x)=8(x-1)^7(4x^6+5+3x^6-3x^5)`

`color(green)(g'(x)=8(x-1)^7(7x^6-3x^5+5)`