80 lines
1.4 KiB
Typst
80 lines
1.4 KiB
Typst
#import "@preview/grape-suite:1.0.0": exercise
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#import exercise: project, task, subtask
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#set text(lang: "de")
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#show: project.with(
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title: [Kettenregel],
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seminar: [Mathe Q2],
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show-outline: true,
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author: "Erik Grobecker",
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date: datetime(day: 18, month: 11, year: 2024),
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show-solutions: false,
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)
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#show heading.where(level: 1): it => {
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counter(math.equation).update(0)
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it
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}
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#show math.equation: set text(font: "New Computer Modern Math")
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= Kettenregel
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$
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f(x)&= underbrace((3/4 x^2 -3), u) dot underbrace(e^(1.4-x^2), v) \
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f'(x)&=u' dot v + u dot v'\
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u'&=3/4 dot 2x^1\ &
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= 3/2 dot x\
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v'&=e^(1.4-x^2)\
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&=e^(overbrace(1.4-x^2, "innere Ableitung")) \
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&=e^(1.4-x^2) dot overbrace((-2x), "innere Ableitung")\
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\
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"Einsetzen:"\
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f'(x)&=u' dot v + u dot v'\
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&=3/2x dot e^(1.4-x^2) + (3/4 x^2 -3) dot e^(1.4-x^2) dot (-2x) #h(2em)&| e "entfernen" \
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&=e^(1.4-x^2) dot (3/2x + (3/4x^2-3) dot (-2x))\
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&=e^(1.4-x^2) dot (3/2x -3/2x^3 + 6x)\
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&=e^(1.4-x^2) dot (-3/2x^3 + 15/2x)
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$
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Innere Ableitung:
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$
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u(x)&=1.4-x^2\
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u'(x)&=2x
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$
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#pagebreak()
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== Aufgabe
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Leite die folgenden Funktionen ab:
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*1)*
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$
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f(x)&=e^(2x)\
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f'(x)&=e^(2x)dot 2\
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&=2e^(2x)
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$
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*2)*
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$
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f(x)&=e^(3x)\
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f'(x)&=e^(3x) dot 3\
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&=3e^(3x)
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$
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*3)*
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$
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f(x)&=e^(-x)\
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f'(x)&=e^(-x) dot (-1)\
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&=-e^(-x)
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$
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*4)*
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$
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f(x)&=e^(0.5x)\
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f'(x)&=e^(0.5x) dot 1 / 2\
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&=1 / 2 e^(0.5x)
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$
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