Derivative of Square Function

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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

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

$\blacksquare$


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