Gitt U(ct) = -1/θ e^(-θc_t ) vis at U'(c_t )= e^(-θc_t )
Noen som kan hjelpe meg med grunnen til at det blir slik? litt ustø i derivasjonsregler....
Derivasjon
Moderators: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
[tex]U(c_t) = \frac{-1}{\theta} \ \cdot \ e^{-\theta c_t}[/tex]
Produktregel: (uv)' = u'v + uv'
[tex]U(c_t) = -\theta^{-1} \ \cdot \ e^{-\theta c_t}[/tex]
Antar at theta er en konstant.
[tex]U^,(c_t) = -\theta^{-1} \ \cdot \ (e^{-\theta c_t})^,[/tex]
[tex]U^,(c_t) = -\theta^{-1} \ \cdot \ (-\theta) \ \cdot \ e^{-\theta c_t}[/tex]
[tex]U^,(c_t) = \frac{1}{\cancel{-\theta}} \ \cdot \ \cancel{(-\theta )} \ \cdot \ e^{-\theta c_t}[/tex]
[tex]U^,(c_t) = e^{-\theta c_t} \ \ \text{QED}[/tex]
Produktregel: (uv)' = u'v + uv'
[tex]U(c_t) = -\theta^{-1} \ \cdot \ e^{-\theta c_t}[/tex]
Antar at theta er en konstant.
[tex]U^,(c_t) = -\theta^{-1} \ \cdot \ (e^{-\theta c_t})^,[/tex]
[tex]U^,(c_t) = -\theta^{-1} \ \cdot \ (-\theta) \ \cdot \ e^{-\theta c_t}[/tex]
[tex]U^,(c_t) = \frac{1}{\cancel{-\theta}} \ \cdot \ \cancel{(-\theta )} \ \cdot \ e^{-\theta c_t}[/tex]
[tex]U^,(c_t) = e^{-\theta c_t} \ \ \text{QED}[/tex]
