I created this proof of the chainrule. Being a (relative) beginner at math I have a few questions.
- Is the proof below correct? I was especially in doubt about the use of $h$ on both sides.
- Is the (Langrange?) notation correct this way?
- How to write the same proof using Leibniz's notation? I wrestled writing this proof in Leibniz notation, because what would in that case be the meaning of $dg$? Is it $g(x+h)-g(x)$ or $k$ or $h$?
To be proved:
If $f(u)$ is differentiable at $u=g(x)$, and $g(x)$ is differentiable at $x$ then:
$$f(g(x))'\stackrel{?}{=}f'(g(x))g'(x)$$
Or similarly
$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}\stackrel{?}{=}\lim \limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$
Case 1: if $h$ has a value such that $g(x+h)=g(x)$ then:
$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=0$$
And
$$\lim \limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}=0$$
Both sides of the equation to prove equal zero, therefore the equation holds in this case.
Case 2: if $h$ has a value such that $g(x+h)\ne g(x)$ then:
We multiply the lefthandside by $\frac{g(x+h)-g(x)}{g(x+h)-g(x)}$
$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$
Taking
$$u=g(x)$$
$$k=g(x+h)-g(x)$$
We get
$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{h \to 0}\frac{f(u+k)-f(u)}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$
And as $h\to 0, k\to 0$, therefore
$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{k \to 0}\frac{f(u+k)-f(u)}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$
Thus
$$f(g(x))'=f'(u)g'(x)=f'(g(x))g'(x) \tag*{$\blacksquare$}$$
Best Answer
Your case 1 is irreparably flawed.$ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $ In that case you claimed that if $g(x+h)=g(x)$ for some $h$ then $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{h} = 0$. That is false. For example let $f$ be the identity function, and $g = \sin$ and $x = 0$ and $h = π$. Then $g(x+h) = g(x)$ but $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{h}$ $= \lim_{h\to0} \lfrac{\sin(h)-\sin(0)}{h} = 1$, contradicting your claim.
Your case 2 is also completely broken for the same logical reason, because even if $g(x+h) \ne g(x)$ for some $h$ it does not mean that $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{g(x+h)-g(x)}$ exists, so your first line in that case is already wrong. For example let $f$ be the identity function again, and $g(t) = |t-1|+|t+1|$ for every real $t$, and $x = 0$. Then $g(2) \ne g(0)$ but $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{g(x+h)-g(x)}$ does not exist because $g(x+h)-g(x) = 0$ for every $h \in [-1,1]$. Note that $g'(0) = 0$ and $f'(2) = 1$, and the chain-rule still holds, but the limit you wrote down does not exist.