Chain rule proof doubt I was reading this pdf document that
shows a proof of the chain rule. My doubt is in the second slide I dont understand why the $k$ value is equal to $g'(x)$ plus $v$ all of this plus $h$. Sorry I dont write the equations, I dont know how to do it thats why I added the link to the pdf
 A: This is more a comment than an answer,
but whatever.
In Hardy's classic
"A Course of Pure Mathematics"
(the tenth edition (1960)
just happens to be sitting
on my shelf)
he writes
in chapter VI, section 114,
page 217
about the chain rule
that "The proof of this theorem
requires a little care."
The problem arises
when getting
$(F(f(x)))'$
if
$f'(x) = 0$
as Ian's answer mentions.
Hardy says in a footnote
that not considering the possibility
that $f'(x) = 0$
makes a number of published proofs
(including his own
in the first three editions)
incorrect.
This book,
which I highly recommend,
is available on Amazon
(in the tenth edition)
for $56 (US).
There are free versions available
on the web,
but these seems to all
be of the third edition
(1921).
This is also true
of the $16 (US) Kindle version
available on Amazon.
So, buyer beware.
My recommendation is that,
if you are interested,
look in a University library
and then decide if you want it.
If you do get it
and do all or most of the exercises,
you will have a good mastery
of the subject.
A: You have by definition
$$v=\frac{g(x+h)-g(x)}{h} - g'(x), \\
w=\frac{f(y+k)-f(y)}{k} - f'(y).
$$
You are originally dealing with
$$\frac{f(g(x+h)) - f(g(x))}{h}.$$
This means that the change in the input to $f$ should be
$$k=g(x+h)-g(x).$$
If this isn't clear, it's because 
$$\frac{f(g(x+h))-f(g(x))}{h} = \frac{f(g(x) + (g(x+h)-g(x)) - f(g(x))}{h}$$
so we're considering $y=g(x)$ and $k=g(x+h)-g(x)$.
Now we can rewrite our expression for $k$ as
$$k=h \left ( g'(x) + v \right )$$
by solving the equation which defines $v$ for $g(x+h)-g(x).$ This is just algebra:
$$v=\frac{g(x+h)-g(x)}{h} - g'(x) = \frac{k}{h} - g'(x) \\
v + g'(x) = \frac{k}{h} \\
k = h \left ( v + g'(x) \right ).$$
By the way, as a bonus, the proof at the end is correct provided $g'(x) \neq 0$, since it just needs $f'(g(x))$ and $1/g'(x)$ to be defined, along with the product rule for limits. I have usually seen the chain rule carefully proven by using that argument when $g'(x) \neq 0$ and then separately proving that if $g'(x)=0$ then $(f \circ g)'(x) = 0$.
