Why we call the definition of differentiability the First principle Definition of differentiability (First principle) is 
$$
f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}
$$
So, dose there exist something like the second principle definition of differentiable?
 A: I would guess that the phrasing is the "derivative from first principles."  Generally speaking, a computation from first principles is done without reliance on deduced results, but depends only on the definitions and axioms (with some wiggle room about how deep we want to chase the rabbit down the hole vis-a-vis what we consider definitional or axiomatic).
For example, the derivative:

Let $f$ be a function.  Then the derivative of $f$ at $x$, denoted $f'(x)$, is defined to be
  $$ f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}, $$
  assuming that this limit exists.

If we compute a derivative using this definition directly, then we are computing a derivative from first principles.  If $f(x) = 2x^3 + 3x - 1$, we might compute the derivative $\frac{\mathrm{d}}{\mathrm{d}x}f(x)$ from first principles as follows:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} f(x)
    &= \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \\
    &= \lim_{h\to 0} \frac{(2(x+h)^3 + 3(x+h) - 1) - (2x^3 + 3x - 1)}{h} \\
    &= \lim_{h\to 0} \frac{2x^3 + 6hx^2 + 6h^2x + 2h^3 + 3x + 3h - 1 - 2x^3 - 3x + 1}{h} \\
    &= \lim_{h\to 0} \frac{6hx^2 + 6h^2x + 2h^3 + 3h}{h} \\
    &= \lim_{h\to 0} (6x^2 + 6hx + 2h^2 + 3) \\
    &= 6x^2 + 3.
\end{align}
(Note that I am assuming a lot about limits already---it could be argued that this isn't really a proof from first principles, as it relies on several results about limits.  Maybe I should be arguing via $\varepsilon$-$\delta$.  In this case, we probably don't want to chase that rabbit...)
As we move further along into the theory, we start to deduce computational rules for computing derivatives.  For example,


*

*For any constant $C$,
$$ \frac{\mathrm{d}}{\mathrm{d}x} C = 0 $$
(the derivative of a constant is zero);

*assuming that the derivatives of $f$ and $g$ exist, then
$$
\frac{\mathrm{d}}{\mathrm{d}x} (f+g)(x)
= \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x} g(x)
$$
(that is, the derivative is linear);

*if $C$ is any constant, then
$$ \frac{\mathrm{d}}{\mathrm{d}x} (Cf(x))
 = C \left( \frac{\mathrm{d}}{\mathrm{d}x} f(x) \right)
$$
(the derivative plays nice with scalar multiplication); and
and

*for any natural number $n$, we have
$$ \frac{\mathrm{d}}{\mathrm{d}x} (x^n) = n x^{n-1} $$
(the power rule).


All of these statements require some proof, but once they are proved, we can use them to compute derivatives more quickly.  For example, going back to my original function, we have
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} (2x^3 + 3x - 1)
&= \frac{\mathrm{d}}{\mathrm{d}x} (2x^3) + \frac{\mathrm{d}}{\mathrm{d}x} (3x) - \frac{\mathrm{d}}{\mathrm{d}x}(1) \\
&= 2\frac{\mathrm{d}}{\mathrm{d}x} (x^3) + 3 \frac{\mathrm{d}}{\mathrm{d}x}(x) \\
&= 2(3x^2) + 3(1) \\
&= 6x^2 + 3.
\end{align}
This second computation is much easier (in my opinion), but relies on results that are proved from the definition of the derivative.  Hence this is not a computation from first principles, but a computation that relies on subsequently proved theorems.
