Brøkregelen for derivasjon

$\begin{array}{l}f\left(x\right)=\frac{x-1}{x+2}\\ \\ f^{\text{'}}\left(x\right)=\frac{{\left(x-1\right)}^{\text{'}}·\left(x+2\right)-\left(x-1\right)·{\left(x+2\right)}^{\text{'}}}{{\left(x+2\right)}^2}\\ \\ =\frac{1·\left(x+2\right)-\left(x-1\right)·1}{{\left(x+2\right)}^2}\\ \\ =\frac{3}{{\left(x+2\right)}^2}\end{array}$

$\begin{array}{l}f\left(x\right)=\frac{x^2+1}{x}\\ \\ f^{\text{'}}\left(x\right)=\frac{{\left(x^2+1\right)}^{\text{'}}·x-\left(x^2+1\right)·x^{\text{'}}}{x^2}\\ \\ =\frac{2x·x-\left(x^2+1\right)·1}{x^2}\\ \\ =\frac{2x^2-x^2-1}{x^2}\\ \\ =\frac{x^2-1}{x^2}\end{array}$

$\begin{array}{l}g\left(x\right)=\frac{x^2-x-6}{x+2}\\ \\ g^{\text{'}}\left(x\right)=\frac{\left(2x-1\right)·\left(x+2\right)-\left(x^2-x-6\right)·1}{{\left(x+2\right)}^2}\\ \\ =\frac{2x^2+4x-x-2-x^2+x+6}{{\left(x+2\right)}^2}\\ \\ =\frac{x^2+4x+4}{{\left(x+2\right)}^2}\\ \\ =\frac{{\left(x+2\right)}^2}{{\left(x+2\right)}^2}\\ \\ =1\end{array}$

$\begin{array}{l}f\left(x\right)=3x^2-4x+\frac{\left(x+2\right)}{x+3}\\ \\ f^{\text{'}}\left(x\right)=3·2x-4+\frac{{\left(x+2\right)}^{\text{'}}·\left(x+3\right)-\left(x+2\right)·{\left(x+3\right)}^{\text{'}}}{{\left(x+3\right)}^2}\\ \\ =6x-4+\frac{1·\left(x+3\right)-\left(x+2\right)·1}{{\left(x+3\right)}^2}\\ \\ =6x-4+\frac{1}{{\left(x+3\right)}^2}\end{array}$

$\begin{array}{rcl}f\left(x\right) & = & \frac{x^2+2}{x}-\frac{x}{x-1}\\ & & \\ f^{\text{'}}\left(x\right) & =& \frac{{\left(x^2+2\right)}^{\text{'}}·x-\left(x^2+2\right)·x^{\text{'}}}{x^2}-\frac{x^{\text{'}}\left(x-1\right)-x·\left(x-1\right)}{{\left(x-1\right)}^2}\\ & & \\ & =& \frac{2x·x-\left(x^2+2\right)·1}{x^2}-\frac{1·\left(x-1\right)-x·1}{{\left(x-1\right)}^2}\\ & & \\ & =& \frac{x^2-2}{x^2}+\frac{1}{{\left(x-1\right)}^2}\end{array}$