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Is my proof correct?

show: $(g\circ f)'(x_0)=g'(y_0)f'(x_0)$

Since $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

and Since $g'(y_0) = \lim_{y\to y_0} \frac{g(y)-g(y_0)}{y-y_0}$

Multiply them to get: $$g'(y_0)f'(x_0) = \lim_{y\to y_0} \frac{g(y)-g(y_0)}{y-y_0} \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$

since $y=f(x)$ and $y_0 = f(x_0)$, we have: $$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} \lim_{x\to x_0} \frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)} \frac{f(x)-f(x_0)}{x-x_0}$$

$$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} \lim_{x\to x_0} \frac{g(f(x))-g(f(x_0))}{x-x_0}$$

$$g'(f(x_0))f'(x_0) = \lim_{f(x)\to f(x_0)} (g\circ f)'(x_0)$$

not sure how to get rid of that limit.

proof: (2nd try) To prove the chain rule, I use Newton's approximation.

We need to show that: $$\forall \epsilon >0\text{ }\exists \delta \text{ such that } |g(f(x))-g(f(x_0))-f'(x_0)g'(y_0)(x-x_0)-|\leq \epsilon |x-x_0| \text{ whenever }|x-x_0|\leq \delta\text{and} x\in X$$

Since $g(y)$ is differentiable at $y_0$, we Know: $$\forall \epsilon_1>0 \text{ }\exists \delta_1 \text{such that} |g(y)-g(y_0)-g'(y_0)(y-y_0)|\leq\epsilon_{1}|y-y_0| \text{ whenever }|y-y_0|\leq \delta_1\text{and} y\in Y$$ for $x\neq x_0$ choose $\delta_1 =f'(x_0)(x-x_0)$ and $\epsilon_1 = \frac{\epsilon }{f'(x_0)(x-x_0)} $

Proof (3rd try): $$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{g(y)-g(y_0)}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{f(x)-f(x_0)}{x-x_0}$$ $$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{g(y)-g(y_0)}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{y-y_0}{x-x_0}$$ $$RHS = lim_{y\to y_0,y\in Y-\{y_o\}}\frac{y-y_0}{y-y_0}lim_{x\to x_0,x\in X-\{x_o\}}\frac{g(f(x))-g(f(x_0))}{x-x_0}$$ $$RHS = lim_{x\to x_0,x\in X-\{x_o\}}\frac{g(f(x))-g(f(x_0))}{x-x_0}=LHS$$

Is this correct?

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    $\begingroup$ Related. $\endgroup$ Commented Dec 20, 2014 at 20:20

3 Answers 3

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No, $\lim_{f(x)\to f(x_0)}$ doesn't make any sense and if $f$ is constant, you divide by $0$ in your proof.

One way of proving this is to set $$w(x)=\begin{cases}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}&\text{if }f(x)\ne f(x_0)\\g'(f(x_0))&\text{else}\end{cases}$$ Let's prove the continuity of $w(x)$ at $x_0$. We know $g'(f(x_0))$ exists, so for all $\varepsilon>0$ there's $\delta>0$ such that $|y-f(x_0)|<\delta\implies\left|\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}-g'(f(x_0))\right|<\varepsilon$ for any $y\ne f(x_0)$.

So for any sequence $x_n\to x_0$ from the continuity of $f(x)$ at $x_0$ there's some $N$ such that for all $n>N$ it's true that $|f(x_n)-f(x_0)|<\delta$. Now either $f(x)=f(x_0)$ and $w(x_n)=g'(f(x_0))$, or $f(x)\ne f(x_0)$ and $|w(x_n)-g'(f(x_0))|<\varepsilon$.

Now it's easy to make your proof rigorous: \begin{align*}\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{x-x_0}&=\lim_{x\to x_0}w(x)\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}w(x)\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\\&=g'(f(x_0))f'(x_0)\end{align*}

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  • $\begingroup$ Thanks for reply. for the case where f(x)=f(x0), can I say that g'(f(x0))=0. then rhs is zero and lhs is also zero. $\endgroup$ Commented Dec 22, 2014 at 7:09
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    $\begingroup$ at the end of my post, I have written another proof based on Newton's approximation. Can you please give it a look? $\endgroup$ Commented Dec 22, 2014 at 7:11
  • $\begingroup$ I edited my post again to include a 3rd proof. your Help is appreciated. $\endgroup$ Commented Dec 22, 2014 at 7:51
  • $\begingroup$ @MAS I'm sorry, but I can't assist you with that. It's not mathematical. Maybe you should start with easier proofs such as why $\lim f+g=\lim f + \lim g$ and $\lim fg=\lim f \cdot\lim g$. $\endgroup$ Commented Dec 22, 2014 at 13:28
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taking $\lim_{x \to x_0}$ implies $\lim_{y \to y_0}$ you don't need this extra limit to begin with

This would be like adding an extra degree of freedom to the problem.

You can simply say

$g'(y_0)f'(x_0) = \lim_{x \to x_0}{[\frac{g(f(x)) - g(f(x_0))}{y - y_0} \frac{f(x) -f(x_0)}{x - x_0}]}$

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Here's a neat well-written one I just found: http://kruel.co/math/chainrule.pdf

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