Den deriverte til eksponentialfunksjonen

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

$$ \begin{array}{rcl}f\left(x\right) & =& x^5·5^x\\ f^{\text{'}}\left(x\right) & =& {\left(x^5\right)}^{\text{'}}·5^x+x^5·{\left(5^x\right)}^{\text{'}}\\ & =& 5x^4·5^x+x^5·5^x\ln 5\\ & =& x^4·5^x\left(5+x\ln 5\right)\end{array}$$

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

Vi bruker kjerneregelen:

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

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

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

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

$$ \begin{array}{rcl}f\left(x\right) & =& 3\sqrt{x}+12e^{0,5x}\\ & & \\ f\left(x\right) & =& 3x^{\frac{1}{2}}+12e^{0,5x} u=0,5x\\ & & \\ f^{\text{'}}\left(x\right)& =& 3·\frac{1}{2}{x}^{-\frac{1}{2}}+12e^u·u^{\text{'}}\\ & & \\ & =& \frac{3}{2}·\frac{1}{\sqrt{x}}+12e^{0,5x}·0,5\\ & & \\ & =& \frac{3}{2\sqrt{x}}+6e^{0,5x}\end{array}$$

Vi ser at $$ f^{\text{'}}\left(0\right)$$ ikke er definert siden vi har 0 i nevneren.

Innbyggertallet i Akersund om $$ x$$ år er

$$ f\left(x\right)=76 538·1,{023}^x$$

$$ \begin{array}{rcl}f\left(10\right) & =& 76 538·1,{023}^{10}\\ & & \\ f\left(10\right) & =& 96 080,10\end{array}$$

Man forventer at innbyggertallet i Akersund er $$ 96 080$$ om $$ 10$$ år.

$$\begin{gathered} $ \begin{array}{rcl}f\left(x\right)& = & 76 538·1,{023}^x\\ & & \\ 76 538·1,{023}^x& =& 100 000\\ & & \\ \ln \left(1,{023}^x\right)& =& \ln \left(\frac{100 000}{76 538}\right)\\ & & \\ x& =& \frac{\ln \left(\frac{100 000}{76 538}\right)}{\ln 1,023}\\ & & \\ x& \approx & 11,76\\ & & \\ & & \\ & & \\ & & \\ & & \\ & & \end{array}$ \\ \end{gathered}$$

Vi bruker CAS i GeoGebra:

Det forventes at innbyggertallet i Akersund stiger til over $$ 100 000$$ innbyggere om $$ 12$$ år.

$$ \begin{array}{rcl}f\left(x\right) & =& 76 538·1,{023}^x\\ f^{\text{'}}\left(x\right) & =& 76 538·{\left(1,{023}^x\right)}^{\text{'}}\\ & =& 76 538·\ln 1,023·1,{023}^x\\ & =& 1 740·1,{023}^x\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& 1 950\\ & & \\ 76 538·1,{023}^x·\ln 1,023 & =& 1 950\end{array}$$

Vi bruker CAS i GeoGebra. Funksjonen til $$ f^{\text{'}}\left(x\right)$$ ligger allerede inne i GeoGebra.

Vi ser at innbyggertallet øker med 1950 om fem år.

Vi bruker GeoGebra og finner $$ f^{\text{'}}\left(3\right)$$:

Innbyggertallet i Akersund øker med $$ 1 863$$ innbyggere i året om $$ 3$$ år.