Derivative of Curve at Point

Theorem
Let $f: \R \to \R$ be a real function.

Let the graph $G$ of $f$ be depicted on a Cartesian plane.

Then the derivative of $f$ at $x = \xi$ is equal to the tangent to $G$ at $x = \xi$.

Proof
Let $f: \R \to \R$ be a real function.


 * DerivativeOfCurve.png

Let the graph $G$ of $f$ be depicted on a Cartesian plane.

Let $A = \tuple {\xi, \map f \xi}$ be a point on $G$.

Consider the secant $AB$ to $G$ where $B = \tuple {\xi + h, \map f {\xi + h} }$.

From Slope of Secant, the slope of $AB$ is given by:
 * $\dfrac {\map f {x + h} - \map f x} h$

By taking $h$ smaller and smaller, the secant approaches more and more closely the tangent to $G$ at $A$:
 * $\map {f'} x = \ds \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h$

where $\lim$ denotes the limit as $h \to 0$.