Stelling
Laat f:D$\to$R en g:D$\to$R differentieerbaar zijn in a. Dan geldt voor alle $\alpha$,$\beta\in$R is $\alpha$f+$\beta$g differentieerbaar in a met afgeleide:
($\alpha$f+$\beta$g)'(a)=$\alpha$f'(a)+$\beta$g'(a)
Bewijs
$
\eqalign{
& \mathop {\lim }\limits_{x \to a} \frac{{\left( {\alpha f + \beta g} \right)\left( x \right) - \left( {\alpha f + \beta g} \right)\left( a \right)}}
{{x - a}} = \cr
& \mathop {\lim }\limits_{x \to a} \frac{{\alpha f\left( x \right) + \beta g\left( x \right) - \alpha f\left( a \right) - \beta g\left( a \right)}}
{{x - a}} = \cr
& \mathop {\lim }\limits_{x \to a} \left[ {\frac{{\alpha f\left( x \right) - \alpha f\left( a \right)}}
{{x - a}} + \frac{{\beta g\left( x \right) - \beta g\left( a \right)}}
{{x - a}}} \right] = \cr
& \alpha f'(x) + \beta g'(a) \cr}
$
bron: Almering e.d. pag.140
