[tex]f(x) = ({\rm{1 + x}}\sqrt x )[/tex] men jeg skjønner ikke hvor feilen ligger, svaret skal blir[tex]3({x^2} + \sqrt x )[/tex]
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Derivasjon
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Andreas345
- Grothendieck
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[tex]f(x)=(1+x\cdot sqrt{x})^2[/tex]
Så forenkler du [tex]x\cdot sqrt{x}=x^{1}\cdot x^{\frac{1}{2}}=x^{\frac{3}{2}}[/tex] og du står igjen med
[tex](1+x^{\frac{3}{2}})^2[/tex]
Bruker kjerneregelen, og voila:
[tex]f\prime(x)=2\cdot (1+x\cdot sqrt{x})\cdot \frac{3}{2}\cdot sqrt{x}[/tex]
[tex]f\prime(x)=3\cdot sqrt{x}(1+x\cdot sqrt{x})=3\dot (sqrt{x}+x^2)[/tex]
Så forenkler du [tex]x\cdot sqrt{x}=x^{1}\cdot x^{\frac{1}{2}}=x^{\frac{3}{2}}[/tex] og du står igjen med
[tex](1+x^{\frac{3}{2}})^2[/tex]
Bruker kjerneregelen, og voila:
[tex]f\prime(x)=2\cdot (1+x\cdot sqrt{x})\cdot \frac{3}{2}\cdot sqrt{x}[/tex]
[tex]f\prime(x)=3\cdot sqrt{x}(1+x\cdot sqrt{x})=3\dot (sqrt{x}+x^2)[/tex]
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Nebuchadnezzar
- Fibonacci
- Posts: 5648
- Joined: 24/05-2009 14:16
- Location: NTNU
[tex] f\left( x \right) = {\left( {1 + x\sqrt x } \right)^2} [/tex]
[tex] \frac{d}{{dx}}p\left( {g\left( x \right)} \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right) [/tex]
[tex] p\left( x \right) = \sqrt {g\left( x \right)} [/tex]
[tex] p\left( x \right) = 1 + x\sqrt x [/tex]
[tex] f\left( x \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right) [/tex]
[tex] g\left( x \right) = 1 + x\sqrt x = 1 + {x^1} \cdot {x^{1/2}} = 1 + {x^{3/2}} [/tex]
[tex] g^{\prime}\left( x \right) = \frac{3}{2}{x^{1/2}} [/tex]
[tex] f\left( x \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right)[/tex]
[tex] f\left( x \right) = 2\left( {1 + x\sqrt x } \right) \cdot \frac{3}{2}{x^{1/2}} [/tex]
[tex] f\left( x \right) = 3\left( {1 + {x^{3/2}}} \right){x^{1/2}} [/tex]
[tex] \underline{\underline {f\left( x \right) = 3\left( {\sqrt x + {x^2}} \right)}}[/tex]
[tex] \frac{d}{{dx}}p\left( {g\left( x \right)} \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right) [/tex]
[tex] p\left( x \right) = \sqrt {g\left( x \right)} [/tex]
[tex] p\left( x \right) = 1 + x\sqrt x [/tex]
[tex] f\left( x \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right) [/tex]
[tex] g\left( x \right) = 1 + x\sqrt x = 1 + {x^1} \cdot {x^{1/2}} = 1 + {x^{3/2}} [/tex]
[tex] g^{\prime}\left( x \right) = \frac{3}{2}{x^{1/2}} [/tex]
[tex] f\left( x \right) = \frac{d}{{dx}}p\left( x \right) \cdot \frac{d}{{dx}}g\left( x \right)[/tex]
[tex] f\left( x \right) = 2\left( {1 + x\sqrt x } \right) \cdot \frac{3}{2}{x^{1/2}} [/tex]
[tex] f\left( x \right) = 3\left( {1 + {x^{3/2}}} \right){x^{1/2}} [/tex]
[tex] \underline{\underline {f\left( x \right) = 3\left( {\sqrt x + {x^2}} \right)}}[/tex]
Man kan også gange ut først og så derivere leddvis på vanlig måte.
[tex]f(x) = (1 + x^{\frac{3}{2}})^2 = 1 + 2x^{\frac{3}{2}} + x^3[/tex]
[tex]f^{\tiny\prime}(x) = 2(\frac{3}{2})x^{\frac{1}{2}} + 3x^2 = 3x^{\frac{1}{2}} + 3x^2[/tex]
[tex]f(x) = (1 + x^{\frac{3}{2}})^2 = 1 + 2x^{\frac{3}{2}} + x^3[/tex]
[tex]f^{\tiny\prime}(x) = 2(\frac{3}{2})x^{\frac{1}{2}} + 3x^2 = 3x^{\frac{1}{2}} + 3x^2[/tex]
An ant on the move does more than a dozing ox.
Lao Tzu
Lao Tzu
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Andreas345
- Grothendieck
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- Joined: 13/10-2007 00:33
Voldsom til interesse for denne oppgaven 
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Nebuchadnezzar
- Fibonacci
- Posts: 5648
- Joined: 24/05-2009 14:16
- Location: NTNU
Men dette er vell en altfor lett oppgave for universitetsstudenter
De burde vell heller få en oppgave alla: "finn den [tex]n[/tex]`te deriverte til
[tex] f(x) \, = \, (1+x\sqrt(x))^2 [/tex] "
De burde vell heller få en oppgave alla: "finn den [tex]n[/tex]`te deriverte til
[tex] f(x) \, = \, (1+x\sqrt(x))^2 [/tex] "