# 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?

• The value of the derivative is usually taken to be the definition of the slope of the tangent, so there's nothing to prove. Alternatively, you can think of the derivative as giving the "best possible linear approximation". This one can prove (with the appropriate definition of "best possible linear approximation") and perhaps that's what you're looking for. – Simon S Dec 5 '14 at 20:24
• @SimonS But you have to prove that this tangent does exist and is unique. – AlexDan Dec 5 '14 at 20:31
• What does it mean for a line passing through a point $(x_0, f(x_0))$ of the graph of $y = f(x)$ to be the tangent? Give me your definition. The usual definition is: the tangent line is the line with this equation $$y - f(x_0) = f'(x_0)(x-x_0)$$ provided $f'(x_0)$ exists. Note that if $f'(x_0)$ exists then it is is well-defined, as it is unique, and therefore the equation of the line is well-defined. – Simon S Dec 5 '14 at 20:40
• My definition of a tangent of a function at a point x0 is a line which only intersect with this function at a single point x0. So to prove that the above equation is the equation of the tangent you have to prove that y is different from f(x) when x is different from x0. – AlexDan Dec 5 '14 at 20:48
• @SimonS To prove that equation is the equation of the tangent you have to prove that there exist an open interval ]x0-a,x0+a[\{x0} /a>0, for which every x that belong to that interval we have f(x) is different from y. – AlexDan Dec 5 '14 at 20:57

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.

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 $x$.