Proof that derivative of a function at a point is the slope of the tangent at the point Why is the derivative for a function at point A considered the slope of the tangent of the function at this point?
 A: Such a statement only makes sense if we have some precise notion of tangent to begin with. This notion should be geometrical and not involve a priori assumptions about the way the function under consideration is presented (e.g., as a polynomial, a series, etc.).
Assume that we are given a function $$f:\>[-h,h]\to{\mathbb R},\qquad x\mapsto y=f(x)$$ with  $f(0)=0$. The line $\ell:\>y=mx$ is called a tangent to the graph $\gamma$ of $f$ at $(0,0)$ if for any given  $m'<m<m''$ there is a $\delta>0$ such that $$m'x< f(x)<m'' x\qquad\bigl(|x|<\delta\bigr)\ .$$
Intuitively this means that, given any however narrow wedge enclosing $\ell$ , the graph of $f$  is ultimately within this wedge near $(0,0)$.
From the definition of derivative it is then obvious that $f'(0)=m$ ensures that $\ell$ is a tangent to $\gamma$ in the sense of this definition.
A: I'd go about it this way.
Take a functions $y = f(x).$
Now, take some positive $\Delta x$ and you can calculate the slope of the line between $(x,f(x))$ and $(x + \Delta x, f(x + \Delta x))$, which is
$$S = \frac{f(x+\Delta x)-f(x)}{x + \Delta x - x}.$$
Take the limit as $\Delta x \to 0$ and you have the limit of this slope, coming in from the positive side.
Similarly, do this for $x - \Delta x$, take the limit, and you'll have the limit of the slope from the negative side.
If these limits are the same, the limit exists.  This happens to be the derivative of $f(x)$ evaluated at $x$, a single point.  You know this is a single point because $f(x)$ is a function of $x$.
