Derivative of Square Function

From ProofWiki
Jump to: navigation, search

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

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

$\blacksquare$


Sources