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 $n = 2$.
$\blacksquare$