# proving derivative in real analysis

I have proved the following problem, can you help me check if there is any loopholes in my proof?

Let I be an open interval in R, let $c \in I$, and let $f, g\colon I\to \mathbb{R}$ be functions. Suppose that $f(c) = g(c)$, and that $f(x) \leq g(x)$ for all $x \in I$. Prove that if f and g are differentiable at c, then f'(c) = g'(c).

So the following is my proof:

Let $x \in I$,

if $x \lt c$, then $\lim\limits_{x\rightarrow c-}\frac{f(x) - f(c)}{x-c}$ = $\lim\limits_{x\rightarrow c-}\frac{f(c) - f(x)}{c-x} \geq \lim\limits_{x\rightarrow c-}\frac{g(c) - g(x)}{c-x} = \lim\limits_{x\rightarrow c-}\frac{g(x) - g(c)}{x-c}$

if $x \gt c$, then $\lim\limits_{x\rightarrow c+}\frac{f(x) - f(c)}{x-c} \leq \lim\limits_{x\rightarrow c+}\frac{g(x) - g(c)}{x-c}$

Since $f$ and $g$ are differentiable at $c$, we know that $\lim\limits_{x\rightarrow c-}\frac{f(x) - f(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{f(x) - f(c)}{x-c}=f'(c)$ and $\lim\limits_{x\rightarrow c-}\frac{g(x) - g(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{g(x) - g(c)}{x-c} = g'(c)$

Thus $\lim\limits_{x\rightarrow c-}\frac{f(x) - f(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{f(x) - f(c)}{x-c} = \lim\limits_{x\rightarrow c-}\frac{g(x) - g(c)}{x-c} = \lim\limits_{x\rightarrow c+}\frac{g(x) - g(c)}{x-c}$

Therefore $f'(c) = g'(c)$.

I can't find any problem in my proof, but for some reason I'm not feeling comfortable with it. However, if it's totally correct, just tell me :) Thanks!!

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+1 for showing your work and only seeking confirmation and critiques, not solutions. – Arturo Magidin Mar 11 '11 at 19:59

The final equality needs better justification; otherwise, it looks fine (modulo a couple of typos). To give a better justification for the final equality, do something like this: \begin{align*} g'(c) &= \lim_{x\to c}\frac{g(x)-g(c)}{x-c} = \lim_{x\to c-}\frac{g(x)-g(c)}{x-c}\\ &\leq \lim_{x\to c-}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)\\ &= \lim_{x\to c+}\frac{f(x)-f(c)}{x-c}\\ &\leq \lim_{x\to c+}\frac{g(x)-g(c)}{x-c} = \lim_{x\to c}\frac{g(x)-g(c)}{x-c} = g'(c). \end{align*} Thus, we have $g'(c)\leq f'(c)\leq g'(c)$, so equality holds.