Aus Online Mathematik Brückenkurs 2

By using the rule for differentiation

\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}

the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain

\displaystyle \begin{align} & {f}'\left( x \right)=\frac{d}{dx}\left( x^{2}-3x+1 \right)=\frac{d}{dx}x^{2}-3\frac{d}{dx}x^{1}+\frac{d}{dx}1 \\ & =2x^{2-1}-3\centerdot 1x^{1-1}+0=2x-3 \\ \end{align}