STYGG derivasjonsoppgave

[Razzy]

[tex]$$I = \left( {x + {1 \over 2}} \right) \cdot \ln \left( {1 + x + {x^2}} \right) - 2x + \sqrt 3 \cdot \arctan \left( {{{2x + 1} \over {\sqrt 3 }}} \right) + C$$[/tex]

[tex]$$I^\prime = \left( {1 \cdot \ln \left( {1 + x + {x^2}} \right) + \left( {x + {1 \over 2}} \right) \cdot {1 \over {\left( {1 + x + {x^2}} \right)}} \cdot \left( {1 + 2x} \right)} \right) - 2 + \left( {0 \cdot \arctan \left( {{{2x + 1} \over {\sqrt 3 }}} \right) + \sqrt 3 \cdot \frac{1}{\left ( \frac{2x+1}{\sqrt{3}} \right )^{2}+1} \cdot {2 \over {\sqrt 3 }}} \right)$$[/tex]

Jeg har hittil brukt produktregelen to ganger og kjerneregelen to ganger.

Resten skal være å "bare" trekke sammen.

Hva sier dere?! hehe

[kimjonas]

Hva er problemet? Det er vel rett ut

[Razzy]

Hva er problemet? Det er vel rett ut

Det kan hende det, men for en litt uerfaren person som meg så var det ikke bare bare! Nå gjelder det å evt forkorte ned...


Løsning:

[tex]$$I = \left( {x + {1 \over 2}} \right) \cdot \ln \left( {1 + x + {x^2}} \right) - 2x + \sqrt 3 \cdot \arctan \left( {{{2x + 1} \over {\sqrt 3 }}} \right) + C$$[/tex]

[tex]I^\prime = \left( {1 \cdot \ln \left( {1 + x + {x^2}} \right) + \left( {x + {1 \over 2}} \right) \cdot {1 \over {\left( {1 + x + {x^2}} \right)}} \cdot \left( {1 + 2x} \right)} \right) - 2 + \left( {0 \cdot \arctan \left( {{{2x + 1} \over {\sqrt 3 }}} \right) + \sqrt 3 \cdot \frac{1}{\left ( \frac{2x+1}{\sqrt{3}} \right )^{2}+1} \cdot {2 \over {\sqrt 3 }}} \right)[/tex]

Herfra løser vi bare algebraisk...

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {x + {1 \over 2}} \right) \cdot \left( {1 + 2x} \right)} \over {\left( {{x^2} + x + 1} \right)}} + \left ( \frac{2}{\left ( \frac{2x+1}{\sqrt{3}} \right )^{2}+1} \right )-2$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {\left( {x + {1 \over 2}} \right) + 2x\left( {x + {1 \over 2}} \right)} \right)} \over {\left( {{x^2} + x + 1} \right)}} + \left ( \frac{2}{\frac{\left ( 2x+1 \right )^{2}+3}{3}} \right )-2$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {x + {1 \over 2} + 2{x^2} + x} \right)} \over {\left( {{x^2} + x + 1} \right)}} + \left( {{6 \over {{{\left( {2x + 1} \right)}^2} + 3}}} \right) - 2$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {2{x^2} + 2x + {1 \over 2}} \right)} \over {\left( {{x^2} + x + 1} \right)}} + {{6 - 2\left( {{{\left( {2x + 1} \right)}^2} + 3} \right)} \over {{{\left( {2x + 1} \right)}^2} + 3}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {2{x^2} + 2x + {1 \over 2}} \right)} \over {\left( {{x^2} + x + 1} \right)}} + {{6 - 2\left( {{{\left( {2x + 1} \right)}^2} + 3} \right)} \over {{{\left( {2x + 1} \right)}^2} + 3}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {2{x^2} + 2x + {1 \over 2}} \right)} \over {\left( {{x^2} + x + 1} \right)}} + {{6 - \left( {2{{\left( {2x + 1} \right)}^2} + 6} \right)} \over {\left( {4{x^2} + 4x + 1} \right) + 3}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {2{x^2} + 2x + {1 \over 2}} \right)} \over {\left( {{x^2} + x + 1} \right)}} + {{6 - 2{{\left( {2x + 1} \right)}^2} - 6} \over {4\left( {{x^2} + x + 1} \right)}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {2{x^2} + 2x + {1 \over 2}} \right)} \over {\left( {{x^2} + x + 1} \right)}} - {{{{\left( {2x + 1} \right)}^2}} \over {2\left( {{x^2} + x + 1} \right)}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{2\left( {2{x^2} + 2x + {1 \over 2}} \right) - {{\left( {2x + 1} \right)}^2}} \over {\left( {{x^2} + x + 1} \right)}}$$[/tex]

[tex]$$I^\prime = \ln \left( {1 + x + {x^2}} \right) + {{\left( {4{x^2} + 4x + 1} \right) - {{\left( {2x + 1} \right)}^2}} \over {\left( {{x^2} + x + 1} \right)}}$$[/tex]

[tex]$$\underline{\underline {I^\prime = \ln \left( {1 + x + {x^2}} \right)}} $$[/tex]