${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} },\,{\left(y_{n}\right)}_{n\in \mathbb {N} }}$

${\displaystyle {}{\left(z_{n}\right)}_{n\in \mathbb {N} }}$

${\displaystyle x_{n}\leq y_{n}\leq z_{n}{\text{ für alle }}n\in \mathbb {N} }$

und ${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}$ und ${\displaystyle {}{\left(z_{n}\right)}_{n\in \mathbb {N} }}$

${\displaystyle {}a}$

${\displaystyle {}{\left(y_{n}\right)}_{n\in \mathbb {N} }}$

${\displaystyle {}a}$

${\displaystyle {}z,w\in \mathbb {C} }$

${\displaystyle {}\exp \left(z+w\right)=\exp z\cdot \exp w\,.}$

${\displaystyle {}a<b}$

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

eine stetige, auf ${\displaystyle {}]a,b[}$ differenzierbare Funktion. Dann gibt es ein ${\displaystyle {}c\in {]a,b[}}$ mit

${\displaystyle {}f'(c)={\frac {f(b)-f(a)}{b-a}}\,.}$

${\displaystyle {}D\subseteq {\mathbb {K} }}$

${\displaystyle {}a\in D}$

${\displaystyle f,g\colon D\longrightarrow {\mathbb {K} }}$

Funktionen, die beide in ${\displaystyle {}a}$ differenzierbar seien und wobei ${\displaystyle {}g}$ keine Nullstelle in ${\displaystyle {}D}$ besitze. Dann ist ${\displaystyle {}f/g}$ differenzierbar in ${\displaystyle {}a}$ mit

${\displaystyle (f/g)'(a)={\frac {f'(a)g(a)-f(a)g'(a)}{(g(a))^{2}}}.}$