Kjerneregelen

• Finn kjernen $u$.

• Finn deretter $g\left(u\right)$ og $u\left(x\right)$.

• Rekn ut $g^{\text{'}}\left(u\right)$ og $u^{\text{'}}\left(x\right)$.

• Set saman, og finn $f^{\text{'}}\left(x\right)$ til slutt.

$\begin{array}{l}\begin{array}{rcl}f\left(x\right) & =& {\left(4x-1\right)}^2 u=4x-1\\ & & \\ g\left(u\right) & =& u^2 g^{\text{'}}\left(u\right)=2u\\ & & \\ u\left(x\right) & =& 4x-1 u^{\text{'}}\left(x\right)=4\\ & & \\ f^{\text{'}}\left(x\right) & =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ & =& 2u·4\\ & & \\ & =& 8u\\ & & \\ & =& 8\left(4x-1\right)\end{array}\end{array}$

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

$\begin{array}{rcl}f\left(x\right) & =& {\left(2x^2-\frac{1}{4}x\right)}^2 u=\left(2x^2-\frac{1}{4}x\right)\\ & & \\ g\left(u\right) & =& {\left(u\right)}^2 g^{\text{'}}\left(u\right)=2u\\ & & \\ u\left(x\right) & =& 2x^2-\frac{1}{4}x u^{\text{'}}\left(x\right)=4x-\frac{1}{4}\\ & & \\ f^{\text{'}}\left(x\right) & =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ & =& 2u·\left(4x-\frac{1}{4}\right)\\ & & \\ & =& 2·\left(2x^2-\frac{1}{4}x\right)·\left(4x-\frac{1}{4}\right)\\ & & \\ & =& \left(4x^2-\frac{1}{2}x\right)·\left(4x-\frac{1}{4}\right)\\ & & \\ & =& \left(16x^3-x^2-2x^2+\frac{1}{8}x\right)\\ & & \\ & =& 16x^3-3x^2+\frac{1}{8}x\end{array}$

$\begin{array}{rcl}f\left(x\right)& = & 6\sqrt[3]{x^2+2}\\ & & \\ f\left(x\right)& =& 6·{\left(x^2+2\right)}^{\frac{1}{3}} u=x^2+2\\ & & \\ g\left(u\right)& =& 6u^{\frac{1}{3}} g^{\text{'}}\left(u\right)=6·\frac{1}{3}{\left(u\right)}^{\frac{1}{3}-1}\\ & & \\ u\left(x\right)& =& x^2+2 u^{\text{'}}\left(x\right)=2x\\ & & \\ f^{\text{'}}\left(x\right)& =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ & =& 6·\frac{1}{3}{\left(u\right)}^{\frac{1}{3}-1}·2x\\ & & \\ & =& 2{\left(u\right)}^{-\frac{2}{3}}·2x\\ & & \\ & =& 2{\left(x^2+2\right)}^{-\frac{2}{3}}·2x\\ & & \\ & =& \frac{4x}{{\left(\sqrt[3]{x^2+2}\right)}^2}\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& {\left(2x^2-4\right)}^3 u=2x^2-4\\ & & \\ g\left(u\right) & =& {\left(u\right)}^3 g^{\text{'}}\left(u\right)=3{\left(u\right)}^2\\ & & \\ u\left(x\right) & =& 2x^2-4 u^{\text{'}}\left(x\right)=4x\\ & & \\ f^{\text{'}}\left(x\right) & =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ & =& 3{\left(u\right)}^2·4x \\ & & \\ & =& 3{\left(2x^2-4\right)}^2·4x\\ & & \\ & =& 3\left(4x^4-2·2x^2·4+16\right)·4x\\ & & \\ & =& 12x\left(4x^4-16x^2+16\right)\\ & & \\ & =& 48x\left(x^4-4x^2+4\right)\end{array}$

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

$\begin{array}{rcl}f\left(x\right) & =& {\left(3x^2-2\right)}^{\frac{3}{2}} u=3x^2-2\\ & & \\ g\left(u\right) & =& {\left(u\right)}^{\frac{3}{2}} g^{\text{'}}\left(u\right)=\frac{3}{2}{\left(u\right)}^{\frac{1}{2}} \\ & & \\ u\left(x\right) & =& 3x^2-2 u^{\text{'}}\left(x\right)=6x\\ & & \\ f^{\text{'}}\left(x\right) & =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ f^{\text{'}}\left(x\right) & =& \frac{3}{2}{\left(u\right)}^{\frac{1}{2}}·6x\\ & & \\ & =& \frac{3}{2}{\left(3x^2-2\right)}^{\frac{1}{2}}·6x\\ & & \\ & =& 9x\sqrt{3x^2-2}\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& \sqrt{x^2-4} u=x^2-4\\ & & \\ g\left(u\right) & =& {\left(u\right)}^{\frac{1}{2}} g^{\text{'}}\left(u\right)=\frac{1}{2}{u}^{-\frac{1}{2}} \\ & & \\ u\left(x\right) & =& x^2-4 u^{\text{'}}\left(x\right)=2x\\ & & \\ f^{\text{'}}\left(x\right) & =& g^{\text{'}}\left(u\right)·u^{\text{'}}\left(x\right)\\ & & \\ & =& \frac{1}{2}{\left(x^2-4\right)}^{-\frac{1}{2}}·2x\\ & & \\ & =& \frac{x}{\sqrt{x^2-4}}\end{array}$

$f\left(x\right) = \frac{{\left(x^2+2\right)}^2}{x}=\frac{u\left(x\right)}{v\left(x\right)}$

der

$u\left(x\right)={\left(x^2+2\right)}^2 , v\left(x\right)=x$

Dette gir

$u^{\text{'}}\left(x\right)=2·\left(x^2+2\right)·2x=4x\left(x^2+2\right) , v^{\text{'}}\left(x\right)=1$

Vi brukte kjerneregelen i derivasjonen av $u\left(x\right)$. Vi bruker kvotientregelen og får

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

$\begin{array}{rcl}& & f\end{array}\begin{array}{rcl}& & \left(x\right)\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \frac{2x}{{\left(x-3\right)}^2}\end{array}$

Vi bruker kvotientregelen og kjerneregelen:

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

$\begin{array}{rcl}& & f\end{array}\begin{array}{rcl}& & \left(x\right)\end{array}\begin{array}{rcl}& & \end{array}=\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \frac{2x^2}{{\left(3x-2\right)}^2}\end{array}$

Vi bruker kvotientregelen og kjerneregelen:

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \frac{{\left(2x^2\right)}^{\text{'}}·{\left(3x-2\right)}^2-2x^2·{\left({\left(3x-2\right)}^2\right)}^{\text{'}}}{{\left({\left(3x-2\right)}^2\right)}^2}\\ & & \\ & =& \frac{4x·{\left(3x-2\right)}^{\cancel{2}}-2x^2·2\cancel{\left(3x-2\right)}·3}{{\left(3x-2\right)}^{\cancel{4 3}}}\\ & & \\ & =& \frac{4x·\left(3x-2\right)-12x^2}{{\left(3x-2\right)}^3}\\ & & \\ & =& \frac{12x^2-8x-12x^2}{{\left(3x-2\right)}^3}\\ & & \\ & =& \frac{-8x}{{\left(3x-2\right)}^3}\end{array}$