# Derivative of Square Function

## Theorem

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

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

Then the derivative of $f$ is given by:

$f' \paren x = 2 x$

## Proof

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

$\blacksquare$