Derivative of Curve at Point

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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 $h \to 0$.

$\blacksquare$


Sources