Proving equality of a function

[aruwin]

I have the solution to the question, but I don't understand the first step so I am just going to paste the first step here.I need people to explain to me.

Question:
For the function f(x) = (logx)/x, prove the following equality.

fⁿ (1) = (-1)^(n-1) n![1+ 1/2 +1/3 +...+1/n]

First step of the solution(The bolded ones are what I don't understand):
Let y(x) = 1 - x and g(x) = f(y(x)) = log(1 - x) / (1 - x). Then:

g(y(x)) = log(1 - (1 - x)) / (1 - (1 - x)) = log(x) / x = f(x)

Thus:

f'(x) = g'(y(x)) * y'(x) ... (chain rule)
= g'(y(x)) * -1
= -g'(y(x))

f''(x) = -g''(y(x)) * y'(x)
= -g''(y(x)) * -1
= g''(y(x))

So on, by induction, we can see that:

f^(n)(x) = (-1)^n * g^(n)(y(x))
===> f^(n)(1) = (-1)^n * g^(n)(y(1)) = (-1)^n * g^(n)(0)

Ok, first off, why and how do we know to put y(x) = 1 - x?
Secondly, the chain rule part. Where does he get y(x) in that chain rule?
I thought f'(x)=g'(y(x))

 

[arildno]

As to your first question: "How and why?", that is a misfocus!
Instead, you should focus on understanding that introducing the auxiliary function y(x) is a LEGITIMATE mathematical operation.
Do you agree that it is?

 

[aruwin]

As to your first question: "How and why?", that is a misfocus!
Instead, you should focus on understanding that introducing the auxiliary function y(x) is a LEGITIMATE mathematical operation.
Do you agree that it is?

Yes, but that's not my question. I want to know how y(x) is 1- x, not why y(x) is introduced.

 

[chiro]

With regard to the chain rule, the derivative of y(z(x)) with respect to x is z'(x)y'(z(x)) and not y'(z(x)) which you stated in your last sentence which needs to be taken into account.