# Proof of the Chain Rule

1. Feb 15, 2014

### Bashyboy

Hello everyone,

I am reading a proof of the chain rule given in this link: http://kruel.co/math/chainrule.pdf

Here is the portion I am troubled with:

"We know use these equations to rewrite f(g(x+h)). In particular, use the first equation to obtain

f(g(x+h)) = f(g(x) + [g'(x) + v]h),

and use the second equation applied to the right-hand-side with k = [g'(x) + v]h..."

How do they arrive at this, k = [g'(x) + v]h. Based above previous equations and definitions, I don't see how it is possible to write k in terms of the derivative of g(x), v, and h.

Could someone help me?

2. Feb 15, 2014

### pasmith

$$f(y + k) = f(y) + k(f'(y) + w) \\ g(x + h) = g(x) + h(g'(x) + v)$$

We want to calculate $f(g(x+h)) - f(g(x))$, so first we need $f(g(x+h))$. From the second equation,
$$f(g(x + h)) = f(g(x) + h(g'(x) + v))$$
and now we apply the first equation with $y = g(x)$ and $k = h(g'(x) + v)$.

Last edited: Feb 15, 2014
3. Feb 15, 2014

### HallsofIvy

What "previous" equation or definition did they have involving k? It looks to me like they are defining k to be [g'(x)+ v]h. Had they already defined it as something else?

4. Feb 16, 2014

### Bashyboy

Well, if they are defining k as [g'(x) + v]h, that would seem awfully arbitrary. What is the justification for such a definition?

5. Feb 16, 2014

### pasmith

Go back to here:
$$f(g(x+h))=f(g(x)+h(g′(x)+v))$$
We also have
$$f(y + k) = f(y) + k(f'(y) + w))$$
which holds for all $y$ and for all $k$.

Thus, to express $f(g(x+h))$ in the form $f(g(x)) + (\mbox{something involving x and h})$, which is what we must do to attain our ultimate goal of finding $f(g(x+h))- f(g(x))$, we need to choose $y$ and $k$ subject to
$$y + k = g(x) + h(g'(x) + v).$$
What choice for $y$ and $k$ would you make here?