Hvordan bestemme den deriverte i et punkt algebraisk

$\begin{array}{rcl}f\left(x+∆x\right)& =& 3\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+\mathrm{\Delta }x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{3-3}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{0}{\mathrm{\Delta }x}\\ & =& 0\end{array}$

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

$\begin{array}{rcl}f\left(2\right)& =& 2\\ & & \\ f\left(2+∆x\right)& =& 2+∆x\\ & & \\ f^{\text{'}}\left(2\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(2+\mathrm{\Delta }x\right)-f\left(2\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{2}+∆x-\cancel{2}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{∆x}}{\cancel{\mathrm{\Delta }x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }1\\ & =& 1\end{array}$

At den deriverte er lik 1, betyr at stigningen i punktet  $x=2$  er $1$, at stigningstallet til tangenten til grafen i punktet  $x=2$  er $1$, og at den momentane vekstfarten i punktet  $x=2$  er $1$.

Siden funksjonen er lineær, kunne vi ha lest stigningstallet direkte ut fra funksjonsuttrykket.

$\begin{array}{rcl}f\left(2\right)& =& 3·2=6\\ & & \\ f\left(2+∆x\right)& =& 3(2+∆x)=6+3∆x\\ & & \\ f^{\text{'}}\left(2\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(2+\mathrm{\Delta }x\right)-f\left(2\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{6}+3∆x-\cancel{6}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{3\cancel{∆x}}{\cancel{\mathrm{\Delta }x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }3\\ & =& 3\\ & & \end{array}$

Stigningen i punktet  $x=2$  er $3$. Det betyr at stigningstallet til tangenten til grafen i punktet  $x=2$  er $3$. Den momentane vekstfarten i punktet  $x=2$  er $3$.

Siden vi vet at grafen er ei rett linje, ser vi at stigningstallet er lik 3.

$\begin{array}{rcl}f\left(2\right)& =& 2^2=4\\ & & \\ f\left(2+∆x\right)& =& {\left(2+∆x\right)}^2=4+4∆x+{\left(∆x\right)}^2\\ & & \\ f^{\text{'}}\left(2\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(2+\mathrm{\Delta }x\right)-f\left(2\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{4}+4∆x+{\left(∆x\right)}^2-\cancel{4}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{4\cancel{∆x}+{\left(∆x\right)}^{\cancel{2}}}{\cancel{\mathrm{\Delta }x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(4+∆x\right)\\ & =& 4\end{array}$

Stigningen i punktet  $x=2$  er $4$. Stigningstallet til tangenten til grafen i punktet  $x=2$  er $4$. Den momentane vekstfarten i punktet  $x=2$  er $4$.

$\begin{array}{rcl}f\left(2\right)& =& 2·2^2+4=12\\ & & \\ f\left(2+∆x\right)& =& 2{\left(2+∆x\right)}^2+4\\ & =& 2(2^2+2·2·∆x+(∆x{)}^2)+4\\ & =& 12+8∆x+2(∆x{)}^2\\ & & \\ f^{\text{'}}\left(2\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(2+\mathrm{\Delta }x\right)-f\left(2\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{12+8∆x+2(∆x{)}^2-12}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{12}+8∆x+2{\left(∆x\right)}^2+\cancel{12}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{8∆x+2(∆x{)}^2}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{∆x}\left(8+2∆x\right)}{\cancel{\mathrm{\Delta }x}}\\ & =& 8\end{array}\begin{array}{}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}$

Stigningen i punktet  $x=2$  er $8$. Stigningstallet til tangenten til grafen i punktet  $x=2$  er $8$. Den momentane vekstfarten i punktet  $x=2$  er $8$.

Den generelle førstegradsfunksjonen er $f\left(x\right)=ax+b.$

$\begin{array}{rcl}f\left(x+∆x\right)& =& a\left(x+∆x\right)+b=ax+a∆x+b\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+∆x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{ax+a∆x+b-\left(ax+b\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{ax}+a∆x+\cancel{b}-\cancel{ax}-\cancel{b}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a\cancel{∆x}}{\cancel{\mathrm{\Delta }x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }a\\ & =& a\end{array}$

Den generelle andregradsfunksjonen:

$f\left(x\right)=a{x}^2+bx+c$

$\begin{array}{rcl}f\left(x+∆x\right)& =& a{\left(x+∆x\right)}^2+b\left(x+∆x\right)+c\\ & =& a\left(x^2+2x∆x+{\left(∆x\right)}^2\right)+bx+b∆x+c\\ & =& a{x}^2+2ax∆x+a{\left(∆x\right)}^2+bx+b∆x+c\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+\mathrm{\Delta }x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a{x}^2+2ax∆x+a{\left(∆x\right)}^2+bx+b∆x+c-\left(a{x}^2+bx+c\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{2ax∆x+a{\left(∆x\right)}^2+b∆x}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{2ax\cancel{∆x}+a{\left(∆x\right)}^{\cancel{2}}+b\cancel{∆x}}{\cancel{\mathrm{\Delta }x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(2ax+a\left(∆x\right)+b\right)\\ & =& 2ax+b\end{array}$

$\begin{array}{rcl}f\left(x+∆x\right)& =& \frac{1}{x+∆x}\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+\mathrm{\Delta }x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\frac{1}{x+∆x}-{\displaystyle \frac{1}{x}}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\frac{1·x}{(x+∆x)·x}-\frac{1·(x+∆x)}{x(x+∆x)}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{{\displaystyle \frac{x-(x+∆x)}{x(x+∆x)}}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{{\displaystyle \frac{\cancel{x}-\cancel{x}-∆x}{x(x+∆x)}}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{{\displaystyle \frac{∆x}{x(x+∆x)}}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(\frac{-∆x}{x(x+∆x)}·\frac{1}{∆x}\right)\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(-\frac{\cancel{∆x}}{x\left(x+∆x\right)\cancel{∆x}}\right)\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(-\frac{\cancel{∆x}}{x\left(x+∆x\right)\cancel{∆x}}\right)\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(-\frac{1}{x^2+x∆x}\right)\\ & =& -\frac{1}{x^2}\end{array}$

$\begin{array}{rcl}f\left(x+∆x\right)& =& \frac{1}{x+∆x-1}\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+\mathrm{\Delta }x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\frac{1}{x+∆x-1}-{\displaystyle \frac{1}{x-1}}}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(\frac{1}{x+∆x-1}-\frac{1}{x-1}\right)·\frac{1}{∆x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\left(x-1\right)-\left(x+∆x-1\right)}{\left(x+∆x-1\right)·\left(x-1\right)}·\frac{1}{∆x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{x}-\cancel{1}-\cancel{x}-∆x+\cancel{1}}{\left(x+∆x-1\right)·\left(x-1\right)}·\frac{1}{∆x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{-∆x}{\left(x+∆x-1\right)·\left(x-1\right)}·\frac{1}{∆x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{-\cancel{∆x}}{\left(x+∆x-1\right)·\left(x-1\right)}·\frac{1}{\cancel{∆x}}\\ & =& -\frac{1}{{\displaystyle {\left(x-1\right)}^2}}\end{array}$

$\begin{array}{rcl}f\left(5\right)& =& 5^2+3=28\\ f\left(5+∆x\right)& =& {\left(5+∆x\right)}^2+3\\ & =& 5^2+2·5∆x+{\left(∆x\right)}^2+3\\ & =& 28+10∆x+{\left(∆x\right)}^2\\ f^{\text{'}}\left(5\right)& =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f\left(x+\mathrm{\Delta }x\right)-f\left(x\right)}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{28+10∆x+{\left(∆x\right)}^2-28}{\mathrm{\Delta }x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cancel{28}+10∆x+{\left(∆x\right)}^2+\cancel{28}}{∆x}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{10\cancel{∆x}+{\left(∆x\right)}^{\cancel{2}}}{\cancel{∆x}}\\ & =& \underset{\mathrm{\Delta }x\to 0}{\lim }\left(10+∆x\right)\\ & =& 10\end{array}$

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