Nokre derivasjonsreglar

$$ f\left(x\right)$$ er ein konstant, som gir

$$ f^{\text{'}}\left(x\right)=0$$

$$ f^{\text{'}}\left(x\right)=0$$ (Hugs at $$ \pi $$ er ein konstant!)

$$ f^{\text{'}}\left(x\right)=0$$

Vi skriv $$ f\left(x\right)=a=a·x^0$$ sidan $$ x^0=1$$ for alle verdiar av $$ x\ne 0$$. Då får vi

$$ f^{\text{'}}\left(x\right)=a·0·x^{0-1}=0$$

$$ f^{\text{'}}\left(x\right)=3$$

$$ f^{\text{'}}\left(x\right)=-2$$

$$ f^{\text{'}}\left(x\right)=5$$

$$ f^{\text{'}}\left(x\right)=5x^{5-1}=5x^4$$

$$ f^{\text{'}}\left(x\right)=7x^{7-1}=7x^6$$

$$ f^{\text{'}}\left(x\right)=3·6x^{6-1}=3·6x^5=18x^5$$

Vi deriverer ledd for ledd.

$$ f^{\text{'}}\left(x\right)=3x^{3-1}-2·2x^{2-1}+0=3x^2-4x$$

$$ f^{\text{'}}\left(t\right)=4·2t^{2-1}-3-0=8t-3$$

$$ f^{\text{'}}\left(x\right)=2·3x^{3-1}-5·2x^{2-1}+4-0=6x^2-10x+4$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& 2·\left(-3\right)x^{-3-1}-5·\left(-2\right)x^{-2-1}+4-0\\ & =& -6x^{-4}+10x^{-3}+4\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \frac{3}{2}{x}^{\frac{3}{2}-1}-\frac{1}{2}{x}^{\frac{1}{2}-1}+4-0\\ & =& \frac{3}{2}{x}^{\frac{3}{2}-\frac{2}{2}}-\frac{1}{2}{x}^{\frac{1}{2}-\frac{2}{2}}+4\\ & =& \frac{3}{2}{x}^{\frac{1}{2}}-\frac{1}{2}{x}^{-\frac{1}{2}}+4\end{array}$$

$$ f^{\text{'}}\left(x\right)=2·\frac{3}{2}{x}^{\frac{3}{2}-1}-10·\frac{1}{2}{x}^{\frac{1}{2}-1}+4-0=3x^{\frac{1}{2}}-5x^{-\frac{1}{2}}+4$$

Vi skriv $$ f\left(x\right)$$ som ein potensfunksjon.

$$ f\left(x\right)=\sqrt{x}=x^{\frac{1}{2}}$$

$$ f^{\text{'}}\left(x\right)=x^{\frac{1}{2}-1}=\frac{1}{2}·x^{-\frac{1}{2}}=\frac{1}{2}·\frac{1}{x^{{\displaystyle \frac{1}{2}}}}=\frac{1}{2\sqrt{x}}$$

Vi set $$ p\left(x\right)=h\left(x\right)+i\left(x\right)$$. Då får vi at

$$ f\left(x\right)=g\left(x\right)+p\left(x\right) \Rightarrow f^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)+p^{\text{'}}\left(x\right)$$

Vidare får vi at

$$ p\left(x\right)=h\left(x\right)+i\left(x\right) \Rightarrow p^{\text{'}}\left(x\right)=h^{\text{'}}\left(x\right)+i^{\text{'}}\left(x\right)$$

Set vi inn for $$ p^{\text{'}}\left(x\right)$$ i likninga for $$ f^{\text{'}}\left(x\right)$$, får vi

$$ f^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)+h^{\text{'}}\left(x\right)+i^{\text{'}}\left(x\right)$$

som var det vi skulle vise.

$$ {\left(2x^3-5x^2+4x-9\right)}^{\text{'}}=6x^2-10x+4$$

$$ {\left(t^3-t^2+4t-9\right)}^{\text{'}}=3t^2-2t+4$$

$$ {\left(2x^5-10x^3+19x-100\mathrm{\pi }\right)}^{\text{'}}=10x^4-30x^2+19$$

$$ {\left(2tx\right)}^{\text{'}}=2x$$

Hugs at $$ x$$ her er ein konstant når vi skal derivere med omsyn på $$ t$$.

$$ {\left(2t^x\right)}^{\text{'}}=2·x·t^{x-1}$$

$$ {\left(ab{t}^{cx}+2b{x}^3\right)}^{\text{'}}=ab·cx·t^{cx-1}$$

Det andre leddet i uttrykket inneheld ikkje variabelen $$ t$$ og er derfor eit konstandledd når vi deriverer med omsyn på $$ t$$.