- Thread starter
- #1

Question:

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

f

^{n}(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))