Derivative of Square Function/Proof 1

Theorem

Let $f: \R \to \R$ be the square function:

$\forall x \in \R: \map f x = x^2$

Then the derivative of $f$ is given by:

$\map {f'} x = 2 x$

Proof

 $\ds \map {f'} x$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\paren {x + h}^2 - x^2} h$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {x^2 + 2 x h + h^2 - x^2} h$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {2 x h + h^2} h$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} 2 x + h$ $\ds$ $=$ $\ds 2 x$

$\blacksquare$