Derivative of a functional and taylor series

Question

In the book Quantum Field Theory for the Gifted Amateur, the authors used the derivation shown below:

I have trouble understanding the step in equation 1.16. Shouldn't the term $\frac{dg(f')}{df'} $ replaced by $\frac{dg(f')}{dx} $. The authors used $\frac{dg(f)}{dx} $ in a similar derivation below.

Answer 1

The idea is that $g = g(f')$. To make it a bit clearer, call $\omega = f'(x)$, that means that

$$ g(\omega) = g(\omega_0) + (\omega - \omega_0)\left.\color{red}{\frac{{\rm d} g(\omega)}{{\rm d}\omega}}\right|_{\omega_0} + \cdots $$

That means that the linear term in the expansion is actually

$$ \color{red}{\frac{{\rm d} g(\omega)}{{\rm d}\omega}} = \frac{{\rm d} g(f')}{{\rm d}f'} $$

Of course you use the chain rule and further expand it in terms of $x$

Answer 2

The confusion arises because of a printing error earlier in the book.

Relevant literature erratum:

Page 13, line before eqn 1.13: the definition of $g‘$ should read $g‘=\mathrm{d}g/\mathrm{d}f$ (rather than $g‘=\mathrm{d}g/\mathrm{d}x$). (14/01/15)

https://tomlancaster.webspace.durham.ac.uk/qftgabook/errata/