I am looking for a proof that the derivative is the best linear approximation that makes this property "feel" fundamental. I have a proof below that somewhat gets close to what I am going for. But, first, I will explain my problems with it.

Ideally, the best linear approximation would mean that, for a given point and for any linear approximation at this point, the derivative gives an at-least-as-good approximation (ideally, strictly better except at the given point) than this linear approximation everywhere within some neighborhood of this point. But, the proof below only shows that, for any linear function, there exists a neighborhood of the point such that, in this neighborhood, the maximum error of the derivative as an approximation is less than the maximum error of this linear approximation.

Also, the following proof does not make this property feel fundamental. A large part of this is that this proof depends on the derivative being unique. Ideally, it would be nice to understand how this property makes the derivative unique.

Here is the proof: Let $f$ be a differentiable function from $\mathbb R$ to $\mathbb R$ at a point $x_0$. Then, there exists a function $\phi$ such that $$ f(x) = f(x_0) + f'(x_0) \cdot (x-x_0) + \phi(x) $$ and $\lim_{x \to x_0} \frac{\phi(x)}{x-x_0} = 0$. Let $L$ be a real number, and let $\psi$ be the function $f(x) - (f(x_0) + L \cdot (x - x_0))$ so that $$ f(x) = f(x_0) + L \cdot (x - x_0) + \psi (x). $$ Since the derivative is unique, the limit of $\frac{\psi (x)}{x - x_0}$ is not equal to $0$, that is, there exists a positive real number $\epsilon$ such that, for any positive real number $\delta$, $$ \vert \frac{\psi (x)}{x - x_0} \vert \geq \epsilon $$ for some $x$ that is $\delta$-near $x_0$. Take $\delta$ to be such that $$ \vert \frac{\phi(x)}{x-x_0} \vert < \epsilon $$ for every $x$ that is $\delta$-near $x_0$. Thus, the derivative is a better approximation than $L$ in the ball of radius $\delta$ around $x_0$.


marked as duplicate by Michael Hoppe, YuiTo Cheng, Leucippus, Shogun, Thomas Shelby Jun 14 at 5:38

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