Brøkregelen for derivasjon

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

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

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

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

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