[Math] Chain rule proof doubt

Question

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

Answer 1

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 $(f \circ g)'(x) = 0$.