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 1

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

$\blacksquare$


Proof 2

From Power Rule for Derivatives:

$\map {\dfrac \d {\d x} } {x^n} = n x^{n - 1}$

The result follows by setting $n = 2$.

$\blacksquare$