Grunnleggende derivasjonsregler

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{f\left(x+∆x\right)-f\left(x\right)}{∆x}\end{array}$

$\begin{array}{rcl}f\left(x\right)& = & \pi \\ & & \\ f\text{'}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{f\left(x+∆x\right)-f\left(x\right)}{∆x}\\ & =& \underset{∆x\to 0}{\lim }\frac{\pi -\pi }{∆x}=0\end{array}$

a) $\begin{array}{rcl}f\left(x\right) & =& 5\\ f\text{'}\left(x\right) & =& 0\end{array}$

b) $\begin{array}{rcl}y & =& e\\ y\text{'} & =& 0\end{array}$

c) $\begin{array}{rcl}g\left(x\right) & =& \pi +5\\ g\text{'}\left(x\right) & =& 0\end{array}$

d) $\begin{array}{rcl} h\left(x\right) & =& 5{\pi }^3\\ h\text{'}\left(x\right) & =& 0\end{array}$

e) $\begin{array}{rcl}i\left(x\right) & =& 2b\\ i\text{'}\left(x\right) & =& 0\end{array}$

f) $\begin{array}{rcl}j\left(x\right) & =& x+d\\ j\text{'}\left(x\right) & =& 1\end{array}$

g) $\begin{array}{rcl}k\left(x\right) & =& 3y+8\\ k\text{'}\left(x\right) & =& 0\end{array}$

Vi tegner $f\left(x\right)$ med digital graftegner:

Stigningstallet til $f\left(x\right)$ er lik 0. Det kan man finne med digital graftegner, se bildet over der $a=0$. Når stigningstallet er 0, har grafen ingen positiv eller negativ stigning. Den er 0 for hele $f\left(x\right).$

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

$\begin{array}{rcl}y\left(x\right) & =& x^5\\ & & \\ y\text{'}\left(x\right) & =& 5x^{5-1}\\ & =& 5x^4\end{array}$

$\begin{array}{rcl}g\left(x\right) & =& 5x^7\\ & & \\ g\text{'}\left(x\right) & =& 5·7x^{7-1}\\ & =& 35x^6\end{array}$

$\begin{array}{rcl}y\left(x\right) & =& x^{-5}\\ & & \\ y\text{'}\left(x\right) & =& -5x^{-5-1} \\ & =& -5x^{-6}=- \frac{5}{x^6}\end{array}$

$\begin{array}{rcl}g\left(x\right) & =& 3x^{-4}\\ & & \\ g\text{'}\left(x\right) & =& 3·\left(-4\right)x^{-4-1}\\ & =& -12x^{-5}=-\frac{12}{x^5}\end{array}$

$\begin{array}{rcl}f\left(t\right) & =& -6t^3\\ & & \\ f\text{'}\left(t\right) & =& -6·3t^{3-1}\\ & =& -18t^2\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& x^0\\ & =& 1\\ & & \\ f\text{'}\left(x\right) & =& 0\end{array}$

$\begin{array}{rcl}y\left(x\right) & =& \frac{1}{x^5}\\ & =& x^{-5}\\ & & \\ y\text{'}\left(x\right) & =& -5x^{-6}\\ & =& -\frac{5}{x^6}\end{array}$

$\begin{array}{rcl}g\left(x\right) & =& 5\sqrt{x}\\ & =& 5x^{\frac{1}{2}}\\ & & \\ g\text{'}\left(x\right) & =& 5·\frac{1}{2}·x^{\frac{1}{2}-1}\\ & =& \frac{5}{2}{x}^{-\frac{1}{2}}\\ & =& \frac{5}{2x^{{\displaystyle \frac{1}{2}}}}\\ & =& \frac{5}{2\sqrt{x}}\end{array}$

$\begin{array}{rcl}z\left(t\right) & =& \frac{1}{\sqrt{t}}\\ & =& t^{-\frac{1}{2}}\\ & & \\ z\text{'}\left(t\right) & =& -\frac{1}{2}{t}^{-\frac{1}{2}-\frac{2}{2}}\\ & =& -\frac{1}{2}{t}^{-\frac{3}{2}}\\ & =& -\frac{1}{2t^{{\displaystyle \frac{3}{2}}}}\\ & =& -\frac{1}{2t^{{\displaystyle \frac{2}{2}}}·t^{{\displaystyle \frac{1}{2}}}}\\ & =& -\frac{1}{2t·\sqrt{t}}\end{array}$

Husk at her skal vi derivere med hensyn på $t$.

$\begin{array}{rcl}f\left(t\right) & =& 3x^2\\ & & \\ f\text{'}\left(t\right) & =& 0\end{array}$

$\begin{array}{rcl}f\left(t\right) & =& 2t^3\\ & & \\ f\text{'}\left(t\right) & =& 2·3t^2\\ & =& 6t^2\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& 4x^3-7x\\ & & \\ f\text{'}\left(x\right) & =& 4·3x^{3-1}-7\\ & =& 12x^2-7\end{array}$

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

$\begin{array}{rcl}g\left(x\right) & =& \frac{1}{2}{x}^2-2x+7\\ & & \\ g\text{'}\left(x\right) & =& \frac{1}{2}·2·x^{2-1}-2\\ & =& x-2\end{array}$

$\begin{array}{rcl}g\left(t\right) & =& 2\left(2t^3+3\right)\\ & & \\ g\text{'}\left(t\right) & =& 2(2·3t^{3-1})\\ & =& 2\left(6t^2\right)\\ & =& 12t^2\end{array}$

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

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

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

$\begin{array}{rcl}f\left(x\right)& =& -3x^3+5x^2-2\\ & & \\ f\text{'}\left(x\right)& =& -3·3x^{3-1}+5·2x^{2-1}\\ & =& -9x^2+10x\\ & & \\ f\text{'}\left(0\right)& =& 0\\ & & \\ f\text{'}\left(2\right)& =& -9·2^2+10·2 \\ & & \\ & =& -16\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& 3x^3-\frac{1}{2}{x}^2+5x\\ & & \\ f\text{'}\left(x\right) & =& 3·3x^{3-1}-\frac{1}{2}·2x^{2-1}+5\\ & =& 9x^2-x+5\\ & & \\ f\text{'}\left(0\right) & =& 9·0^2-0+5\\ & =& 5\\ & & \\ f\text{'}\left(2\right) & =& 9·2^2-2+5\\ & =& 39\end{array}$

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

$\begin{array}{rcl}f\left(x\right) & =& 4x^2+5x\\ & & \\ f\text{'}\left(x\right) & =& 4·2x^{2-1}+5\\ & =& 8x+5\\ & & \\ f\text{'}\left(0\right) & =& 8·0+5 = 5\\ & & \\ f\text{'}\left(1\right) & =& 8·1+5 = 13\end{array}$

Vi bruker CAS til å løse oppgaven:

Vi skriver inn funksjonsuttrykket. Vi bruker kommandoen "Tangent()" og tegner tangenter som berører grafen i punktene $x=0$ og $x=1$. Vi bruker "Stigning()" og finner stigningen på tangentene. I punktet $x=0$ er stigningen 5, og i punktet $x=1$ er stigningen 13.

$f\text{'}\left(0\right)=5$ forteller oss at ved midnatt vokste bakteriekulturen med 5 bakterier i minuttet, mens $f\text{'}\left(1\right)=13$ forteller oss at kl. 00.01 om natten vokste bakteriekulturen med 13 bakterier i minuttet.

$\begin{array}{rcl}f\left(x\right) & =& 2x^4-x^2+\mathrm{\pi }\\ & & \\ f\text{'}\left(x\right) & =& 2·4x^{4-1}-2x^{2-1}\\ & =& 8x^3-2x\\ & & \\ f\text{'}\left(1\right) & =& 8·{\left(1\right)}^3-2·1 = 6\end{array}$

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

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

$$\begin{gathered} \begin{array}{rcl}f\left(t\right) & =& 2x+3t^2 \\ \\ & & \\ f\text{'}\left(t\right) & =& 0+3·2t\\ & =& 6t\\ & & \\ f\text{'}\left(-1,5\right) & =& 6·\left(-1,5\right)\\ & =& -9\end{array} \end{gathered}$$

Husk at $f\left(x\right)$ viser at vi skal derivere med hensyn på x.

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

Husk at $f\left(t\right)$ viser at vi skal derivere med hensyn på t.

$\begin{array}{rcl}f\left(t\right) & =& 2x^3-4t^2+t\\ & & \\ f\text{'}\left(t\right) & =& 0-4·2t+1\\ & =& -8t+1\end{array}$

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

$\begin{array}{rcl}g\left(z\right) & =& x^3-2z^2+x+\pi \\ & & \\ g\text{'}\left(z\right) & =& 0-2·2z+0+0\\ & =& -4z\end{array}$

$\begin{array}{rcl}(x^2+2z-2xz)\text{'} & =& 2x+0-2z\\ & =& 2x-2z\end{array}$

$\begin{array}{rcl}(x^2+2z-2xz)\text{'} & =& 0+2-2x\\ & =& 2-2x\end{array}$

$\begin{array}{rcl}(x^2+2z-2xz)\text{'} & =& \end{array}0+0-0 =0$

$\begin{array}{rcl}(3x^4-7x{y}^2+2xz-\frac{1}{3}{z}^3)\text{'} & =& 3·4x^{4-1}-7y^2+2z-0\\ & =& 12x^3-7y^2+2z\end{array}$

$\begin{array}{rcl}(3x^4-7x{y}^2+2xz-\frac{1}{3}{z}^3)\text{'} & =& 0-7x·2y+0-0\\ & =& -14xy\end{array}$

$\begin{array}{rcl}(3x^4-7x{y}^2+2xz-\frac{1}{3}{z}^3)\text{'} & =& 0-0+2x-\frac{1}{3}·3z^{3-1}\\ & =& 2x-z^2\end{array}$

$(3x^4-7x{y}^2+2xz-\frac{1}{3}{z}^3)\text{'}=0-0+0-0=0$