real analysis - Prove that $f$ is differentiable at $x_0$ if and only if $g$ is differentiable at $x_0$

Suppose $f: D \rightarrow \mathbb{R}$ and $g: E \rightarrow \mathbb{R}$ and that $x_0$ is in the domain of both functions and an accumulation point of their intersection. Suppose further that there exists $\mu >0$ such that the intersection of the $\mu$-neighborhood of $x_0$ with $D$ is equal to the intersection of the $\mu$-neighborhood with $E$ and for all $x$ in that intersection, $f(x)=g(x)$. Prove that $f$ is differentiable at $x_0$ if and only if $g$ is differentiable at $x_0$

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The concept of this proof is not really clicking with me,Given most of my theorems and lemmas of differentiability state the use of one function,I'm lost on how to approach the proof

Answer

Hint: write down the definition of $f'(x_0)$ and $g'(x_0)$.

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Denote the "$\mu$-neighborhood" of $x_0$ as $H$. Then $S:=H\cap D=H\cap E$. Also $f=g$ on $S$.

Now
$$f'(x_0)=\lim_{\substack{x\to x_0\\x\in S\setminus\{x_0\}}}\frac{f(x)-f(x_0)}{x-x_0} =\lim_{\substack{x\to x_0\\x\in S\setminus\{x_0\}}}\frac{g(x)-g(x_0)}{x-x_0}$$