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$\frac{d}{dx}a\,f(x)=a\,f'(x)$, for every constant a.

Prerequisites

Limit definition of the derivative, $f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Proof

Let $g(x)=a\,f(x)$ for some constant a. By the limit definition of the derivative:

$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$a\,f'(x)=a\,\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{a\,f(x+h)-a\,f(x)}{h}$

To prove the proposition, it suffices to show that $g'(x)=a\,f'(x)$.

$g'(x)=\lim_{h \to 0}\frac{g(x+h)-g(x)}{h}$
$g'(x)=\lim_{h \to 0}\frac{a\,f(x+h)-a\,f(x)}{h}=a\,f'(x)$

QED