Some fun with definitions of derivatives
The easiest to interpret definition of an derivative $f'$ of an function $f: \mathbb{R}\to\mathbb{R}$ at a point $x_0$ is
$$ f(x)=f(x_0)+ f'(x_0) (x-x_0) + \varphi( x-x_0) $$
such that $\lim_{x\rightarrow x_0} \varphi(x-x_0)/ ||x-x_0||=0$. In particular the definition tells us, that derivatives are a way to approximate functions by linear functions.
More generally the idea, that we can approximate functions by polynomials, is known as Taylor's theorem. In the case of an analytic function $f(x)=\sum_{k=0}^{\infty} a_{k}x^{k}$, we have
$$ \frac{1}{m!} \frac{\partial^m}{\partial x^m} f(0) = a_m $$
and consequently the Taylor expansion up to second order around a point $x_0$
$$ f(x)\approx f(x_0) + f'(x_0) (x-x_0) + 1/2 f''(x_0) x^2 + \mathcal{O}(x^3). $$
We can then argue that the analytic functions are dense in the space of functions to gain a very useful approximation technique from this.
Until yesterday I believed that one could also prove the second function directly from the first definition. However, something interesting happens, let $\varphi$ the error term from the definition, $\varphi_2$ the error term of the error term, and assume $\varphi$ is differentiable:
$$ \varphi(x)=\varphi(x_0)+\varphi'(x_0) (x-x_0) + \varphi_2 (x-x_0)\\ = -f''(x_0) (x-x_0)^2 + \varphi_2 (x-x_0) $$
and following directly
$$ f(x)=f(x_0) + f'(x_0) (x-x_0) - f''(x_0) (x-x_0)^2 + \varphi_2 (x-x_0) $$
which when we compare that with the second order Taylor expansion, we find that
$$ f(x)-\varphi_2 (x-x_0) =\left[f(x_0) + f'(x_0) (x-x_0) + 1/2 f''(x_0) x^2\right]\\ -\frac{3}{2} f''(x_0) (x-x_0)^2 \sim \mathcal{O}(x^2) $$
where the square brackets delimit the Taylor expansion. So $\varphi_2$ is only the right function up to second order (which is precisely the order we need for the definition of the derivative), and therefore we can not get Taylor approximation beyond first order from this definition directly. (It is of course still possible, we just need a better characterization of $\varphi_2$ than we get from the definition.)