Ableitungen Exp./Logar.-Fkt. < Exp- und Log-Fktn < Analysis < Oberstufe < Schule < Mathe < Vorhilfe
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(Frage) beantwortet  | | Datum: | 09:48 Fr 21.03.2008 | | Autor: | hase-hh |
| Aufgabe | Bilde die 1. und 2. Ableitung
1. f(x) = ln(x+1)
2. f(x) = [mm] log_{2} [/mm] x + x + 2
3. f(x) = [mm] log_{3} [/mm] x
4. f(x) = 2*ln(2x)
5. f(t) = ln(1+kt)
6. h(t) = [mm] \bruch{1}{2}*ln(\bruch{t}{3} [/mm] -1)
7. f(x) = ln b * [mm] log_{b} [/mm] x
8. f(x) = [mm] e^{2*ln x} [/mm] + [mm] ln(e^{2x})
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9. f(t) = [mm] \bruch{1}{3}* [/mm] sin(ln t)
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Moin,
habe hier ein paar Aufgaben zu e-Funktionen und Logarithmus-Funktionen. Ist das so korrekt?
1. f(x) = ln(x+1)
f ' (x) = [mm] \bruch{1}{x+1}*1 =(x+1)^{-1}
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f '' (x) = - [mm] \bruch{1}{(x+1)^2}*1
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2. f(x) = [mm] log_{2} [/mm] x + x + 2
Kann ich [mm] log_{2} [/mm] x schreiben als [mm] 2^x [/mm] ?
Dann wäre
f(x) = [mm] 2^x [/mm] +x + 2
f ' (x) = ln [mm] 2*2^x [/mm] + 1
f '' (x) = ln 2*ln [mm] 2*2^x
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3. f(x) = [mm] log_{3} [/mm] x
f(x) = [mm] 3^x [/mm] + [mm] 3^x
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f ' (x) = 2*ln [mm] 3*3^x
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f '' (x) = 2*ln 3*ln [mm] 3*3^x
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4. f(x) = 2*ln(2x)
f ' (x) = [mm] 2*\bruch{1}{2x}*2 [/mm] = 2* [mm] \bruch{1}{x}
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f '' (x) = - [mm] 2*\bruch{1}{x^2}
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5. f(t) = ln(1+kt)
f ' (t) = [mm] \bruch{1}{1+kt}*k [/mm] = [mm] k*(1+kt)^{-1}
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f '' (t) = - [mm] k*\bruch{1}{(1+kt)^2}
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6. h(t) = [mm] \bruch{1}{2}*ln(\bruch{t}{3} [/mm] -1)
h ' (t) = [mm] \bruch{1}{2}*\bruch{1}{\bruch{t}{3} -1}*\bruch{1}{3}
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h ' (t) = [mm] \bruch{1}{2}*(t-3)^{-1}
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h '' (t) = - [mm] \bruch{1}{2}*\bruch{1}{(t-3)^2} [/mm]
7. f(x) = ln [mm] b*log_{b} [/mm] x
f(x) = ln [mm] b*b^x [/mm]
f ' (x) = ln b*ln [mm] b*b^x
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f '' (x) = ln b*ln b*ln [mm] b*b^x
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8. f(x) = [mm] e^{2*ln x} [/mm] + [mm] ln(e^{2x})
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könnte ich umschreiben...
f(x) = [mm] (e^{ln x})^2 [/mm] + [mm] ln^{e^{2x}}
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f(x) = [mm] x^2 [/mm] + 2x
f ' (x) = 2x + 2
f '' (x) = 2
9. f(t) = [mm] \bruch{1}{3}*sin(ln [/mm] t)
f ' (t) = [mm] \bruch{1}{3}*cos(ln [/mm] t) [mm] *\bruch{1}{t} [/mm]
f '' (t) = [mm] \bruch{1}{3}*(-sin(ln [/mm] t)) [mm] *\bruch{1}{t}*\bruch{1}{t} [/mm] + [mm] \bruch{1}{3}* [/mm] cos(ln t) [mm] *(-\bruch{1}{t^2})
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| Status: |
(Frage) beantwortet | | Datum: | 20:09 Fr 21.03.2008 | | Autor: | hase-hh |
3.
f(x) = [mm] log_{3} [/mm] x
hier nehem ich die ln-Variante, weil die Ableitung davon so einfach ist...
f(x) = [mm] \bruch{ln x}{ln 3}
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f(x) = [mm] \bruch{1}{ln 3}* [/mm] ln x
f ' (x) = [mm] \bruch{1}{ln 3} [/mm] * [mm] \bruch{1}{x}
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f '' (x) = - [mm] \bruch{1}{ln 3} [/mm] * [mm] \bruch{1}{x^2}
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7.
f(x) = ln b * [mm] log_{b} [/mm] x
f(x) = ln b * [mm] \bruch{ln x}{ln b}
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f(x) = ln x
f ' (x) = [mm] \bruch{1}{x}
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f '' (x) = - [mm] \bruch{1}{x^2}
[/mm]
Gruß
Wolfgang
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