I study differentiation, and trying to understand geometrical meaning of differentiability. My first guess was to use tangent lines, but geometrical meaning of tangent not obvious - it's usually said, that tangent - is local approximation of function, but this deffinition requires special definition of approximation, which differs from standart uniform approximation (like in Stone–Weierstrass theorem), so such defining of tangent realy looks like tautology (and metric used in this approximation actualy uses integration, which itself is oposite to differentiation).
My guess was, that tangent - is line, locally lying below or above of function, but it's totally wrong - i mean neither $\Rightarrow$ nor $\Leftarrow$ work with counterexamples given by $x^{3}$ and
$$\begin{cases} x\sin{\frac{1}{x}},\ x>0 \\
x,\ x\le 0
\end{cases}$$
My last guesses - are about "kissing" circles. So question - is it true, that function is differentiable at point iff there are two circles, wich intersect function graph at point, with one of them lying below of the graph, and other lying above of the graph. Another guess is - function is differentiable at point iff there is circle, unique up to inclusion, intersecting graph and lying below or above of it. I have no ideas how to proove or contradict such statements.
No. The map $$ \begin{array}{r|rcl} f\colon & \Bbb R & \longrightarrow & \Bbb R \\ & x & \longmapsto & \begin{cases} |x|^{\frac{3}{2}} \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0,\\ 0 & \text{if } x = 0 \end{cases} \end{array} $$ is differentiable at $0$ with $f'(0)=0$, but no circle passing through $(0,0)$ lies entirely (even locally around $0$) above or below the graph of $f$.
Intuitively, $|x|^{\frac{3}{2}}$ grows bigger than any circle around the origin. Indeed, such a circle should have centre on the vertical line (if not, it would cross the graph of $f$ at the origin). Say this centre is $(0,y_0)$. The equation of this circle, near $x=0$, is $y = y_0(1-\sqrt{1 - \frac{x^2}{y_0^2}})$. It is easy to show that such a function cannot be always above or below $y = \pm |x|^{\frac{3}{2}}$ around $0$ (the sign depending on whether $y_0 >0$ or $y_0 <0$).
Same thing occurs with straight lines though.