Proof that derivative of a function at a point is the slope of the tangent at the point

Solution 1:

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.