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If $f'(g(x))$ and $g'(x)$ both exist, then

$$f'(g(x))=\lim_{\Delta g(x)\to 0}\frac{\Delta f(g(x))}{\Delta g(x)}\stackrel{(1)}\implies \frac{\Delta f(g(x))}{\Delta g(x)}=f'(g(x))+\alpha(\Delta x),$$

where $\lim_{\Delta x\to 0}\alpha(\Delta x)=0$.

$$\Delta f(g(x))=f'(g(x))\cdot \Delta g(x)+\alpha(\Delta x)\cdot \Delta g(x)$$

$$\frac{\Delta f(g(x))}{\Delta x}=f'(g(x))\cdot \frac{\Delta g(x)}{\Delta x}+\alpha(\Delta x)\cdot \frac{\Delta g(x)}{\Delta x}$$

Now take the limit as $\Delta x\to 0$ of both sides:

$$(f\circ g)'(x)=f'(g(x))g'(x)+0\cdot g'(x)=f'(g(x))g'(x)\ \ \ \square$$

$(1)$ is a known fact I've proved. Would you leave all the notation and math details as it is or would you change anything? $\Delta f(g(x))$ is a bit suspicious. Is everything fine?

user89167
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    How do you take care of $\Delta g(x)$ possibly being zero at times where $\Delta x$ is not? Maybe you already have thought about that, I am just asking you to clarify. After all, your expression for $\Delta f(g(x))$ depends on $\Delta g(x)$ being non-zero as I read it from (1). – String Mar 24 '15 at 12:38
  • But perhaps you are already thinking along the same lines as I did in this answer? – String Mar 24 '15 at 12:41
  • @String Indeed I haven't thought about that case. This proof is mostly the proof in my textbook, only simplified somewhat to my taste (textbook uses $y_x'$ to denote $f'(x)$ for $y=f(x)$, etc. I don't like the notation). So the proof seems to be flawed. – user89167 Mar 27 '15 at 02:48
  • I do not know whether you have been true to the original idea of the proof or not, but in the current version, the LHS of $$\frac{\Delta f(g(x))}{\Delta g(x)}=f'(g(x))+\alpha(\Delta x)$$ may not be defined for all $\Delta x\neq 0$. On the other hand, you CAN actually assign meaning to $$\Delta f(g(x))=f'(g(x))\Delta g(x)+\alpha(\Delta x)\Delta g(x)$$ by letting $\alpha(\Delta x)$ be defined by the former equation whenever $\Delta g(x)\neq 0$. Then since $\Delta g(x)=0\implies \Delta f(g(x))=0$ you can assign whatever value you want to $\alpha(\Delta x)$ in those cases. – String Mar 27 '15 at 10:01
  • But this still does not explain very well why $\alpha(\Delta x)\rightarrow 0$ for $\Delta x\rightarrow 0$ since we may possibly have infinitely many values of $\alpha(\Delta x)$ that are not defined by the $\dfrac{\Delta f(g(x))}{\Delta g(x)}$-equation. What if for instance $g(x)$ is constant? – String Mar 27 '15 at 10:03

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