1) Deriver funksjonen g(x)=x*(lnx)^2
2)Løs disse ligningene ved regning: I) 4*lnx=1
II) 3*ln(2x-1)=1
Trenger hjelp!
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
1)
[tex] g(x)=x(\ln x)^2 \\ \ \\ g^,(x)=2\ln x+(\ln x)^2 [/tex]
2.1)
[tex] 4\ln x=1\\ \ \\ \ln x=\frac{1}{4}\\ \ \\ e^{\ln x}=e^{\frac{1}{4}}\\ \ \\ x=e^{\frac{1}{4}}\\ \ \\ x \approx 1.2840254[/tex]
2.2)
[tex]3\ln(2x-1)=1\\ \ \\ \ln(2x-1)=\frac{1}{3}\\ \ \\ 2x-1=e^{\frac{1}{3}}\\ \ \\ 2x=e^{\frac{1}{3}}+1\\ \ \\ x=\frac{e^{\frac{1}{3}}+1}{2}\\ \ \\ x \approx 1.1978062 [/tex]
[tex] g(x)=x(\ln x)^2 \\ \ \\ g^,(x)=2\ln x+(\ln x)^2 [/tex]
2.1)
[tex] 4\ln x=1\\ \ \\ \ln x=\frac{1}{4}\\ \ \\ e^{\ln x}=e^{\frac{1}{4}}\\ \ \\ x=e^{\frac{1}{4}}\\ \ \\ x \approx 1.2840254[/tex]
2.2)
[tex]3\ln(2x-1)=1\\ \ \\ \ln(2x-1)=\frac{1}{3}\\ \ \\ 2x-1=e^{\frac{1}{3}}\\ \ \\ 2x=e^{\frac{1}{3}}+1\\ \ \\ x=\frac{e^{\frac{1}{3}}+1}{2}\\ \ \\ x \approx 1.1978062 [/tex]
Geogebra: http://www.geogebra.org/cms/
Utfordringer: http://projecteuler.net/index.php?section=problems
[tex]M_{2147483647}[/tex] er ikke et primtall. 295257526626031 deler det.
Utfordringer: http://projecteuler.net/index.php?section=problems
[tex]M_{2147483647}[/tex] er ikke et primtall. 295257526626031 deler det.
Kan jo spesifisere hva knuta har gjort her:
[tex]g(x) = x(ln(x))^2[/tex]
Bruker produktregel og kjerneregel:
[tex]g`(x) = x`*(ln(x))^2 + x*(ln(x))`*((ln(x))^2)`[/tex]
[tex]g`(x) = 1*(ln(x))^2 + x*\frac {1}{x}*2*ln(x)[/tex]
[tex]g`(x) = (ln(x))^2 + 2ln(x)[/tex]
[tex]g(x) = x(ln(x))^2[/tex]
Bruker produktregel og kjerneregel:
[tex]g`(x) = x`*(ln(x))^2 + x*(ln(x))`*((ln(x))^2)`[/tex]
[tex]g`(x) = 1*(ln(x))^2 + x*\frac {1}{x}*2*ln(x)[/tex]
[tex]g`(x) = (ln(x))^2 + 2ln(x)[/tex]
