MATHHX B

MATHHX B

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9.2 Differentialkvotienter ved tabelopslag

I sidste afsnit lærte vi at finde differentialkvotienter ved at aflæse hældninger. Det gav en god forståelse (håber jeg) for hvad en differentialkvotient er, men I praksis er det ikke sådan man bestemmer differentialkvotienter. Fremover vil vi slå differentialkvotienterne op i en tabel:

(image)

Tabel 9.1: Tabel med differentialkvotienter fra formelsamlingen. Kolonnen til højre skal vi ikke bruge i denne omgang

Jeg anbefaler, at I printer tabellen (eller formelsamlingen,) da vi skal bruge den hele tiden.

  • Eksempel 9.2.1
    Lad \(f(x)=\sqrt {x}\). Vil vil gerne bestemme \(f'(x)\).

    Vi finder \(\sqrt {x}\) i den midterste kolonne i tabellen og ser i kolonnen til venstre at \(f'(x)=\frac {1}{2\sqrt {x}}\).

Øvelse 9.2.1

Bestem ved tabelopslag \(f'(x)\) for følgende funktioner

  • a) \(f(x) = \ln (x)\)

  • b) \(f(x) = e^x\)

Løsning 9.2.1

  • a) \(f'(x) = \frac {1}{x}\)

  • b) \(f'(x) = e^x\)

  • Eksempel 9.2.2
    Vi vil bestemme den afledte funktion (differentialkvotienten) til funktionen \(f(x)=x^7\). Vi kan se at \(f\) har form som \(x^n\), som ifølge tabellen har differentialkvotient \(f'(x)=n\cdot x^{n-1}.\) Vi får

    \[f'(x)=7x^{7-1}=7x^6.\]

    Altså er \(f'(x)=7x^6\).

At bestemme en differentialkvotienten til en funktion \(f\) kaldes også at differentiere \(f\).

  • Eksempel 9.2.3
    Vi vil differentiere \(f(x)=3\). Vi ser at den har formen \(f(x)=a\) (\(a\) er en konstant). Ifølge tabellen får vi \(f'(x)=0\)

Øvelse 9.2.2

Bestem ved tabelopslag den afledte funktion for følgende funktioner:

  • a) \(f(x)=x^5\)

  • b) \(f(x)=2^x\)

  • c) \(f(x)=x^{-4}\)

  • d) \(f(x) = 5\)

  • e) \(f(x)=e^{2x}\)

  • f) \(f(x)=x^1\)

  • g) \(f(x)=x\)

  • h) \(f(x) = \pi \)

  • i) \(f(x) =2^3\)

Løsning 9.2.2

  • a) \(f'(x)=5x^4\)

  • b) \(f'(x)=\ln (2)\cdot 2^x\)

  • c) \(f'(x)=-4x^{-5}\)

  • d) \(f'(x) = 0\)

  • e) \(f'(x)=2e^{2x}\)

  • f) \(f'(x)=1\)

  • g) \(f'(x)=1\)

  • h) \(f'(x) = 0\)

  • i) \(f'(x)=0\)

Nogle gange får vi brug for at differentiere en funktion som ikke har noget navn. F.eks kan det være, at vi får at vide, at vi skal differentiere \(x^3\) (læg mærke til, at der ikke står ”\(f(x)=\)”).

  • Eksempel 9.2.4
    Vi vil differentiere funktionen givet ved udtrykket \(x^3\). Vi skriver:

    \[(x^3)'=3x^2\]

Øvelse 9.2.3

  • a) Bestem \(\big (\ln (x)\big )'\)

Løsning 9.2.3

  • a) \(\big (\ln (x)\big )'=\frac {1}{x}\)

Differentialkvotient for opbyggede funktioner

Vi vil ofte møde funktioner som ikke direkte står i tabel 9.1 . I stedet vil de være bygget op af funktioner fra tabellen.

  • Eksempel 9.2.5
    Vi vil differentiere funktionen \(h(x)=x^2+3\). Vi ved fra tabel 9.1 at \((x^2)'=2x^{2-1}=2x\) og \((3)'=0\). Så vi tænker at \(h'(x)=2x+0=2x\). Det er den også, men vi mangler faktisk et argument for, at det er ok differentiere de to led hver for sig.

  • Eksempel 9.2.6
    Vi vil differentiere funktionen \(5x^2\). Vi ved fra tabel 9.1 at \((5)'=0\) og \((x^2)'=2x\), så vi tænker at \(h'(x)=0\cdot 2x= 0\). Men det er…FORKERT.

For at differentiere funktionerne ovenstående eksempler korrekt har vi brug for nogle formler fra formelsamlingen:

(image)

Tabel 9.2: Udsnit fra formelsamlingen med formler for opbyggede funktioner. De to nederste linjer hører til på A-niveau

Jeg anbefaler at I printer tabellen (eller formelsamlingen) da vi skal bruge den hele tide.

  • Eksempel 9.2.7
    Vi vil differentiere funktionen \(h(x)=x^2+3\). Vi sammenligner med tabel 9.2 og ser at vores funktion har form som i formel (41), hvor:

    \[f(x)=x^2\quad \text {og} \quad g(x)=3\]

    Vi differentierer \(f\) og \(g\):

    \[f'(x)=2x\quad \text {og} \quad g'(x)=0.\]

    Ifølge tabellen er differentialkvotienten givet ved \(f'(x)+g'(x)\), så

    \[h'(x)=f'(x)+g'(x)=2x+0=2x.\]

    Altså er:

    \[h'(x)=2x.\]

  • Eksempel 9.2.8
    Vi vil differentiere funktionen \(h(x)=5x^2\). Vi sammenligner med tabel 9.2 og ser at vores funktion har form som i formel (40), hvor

    \[k=5 \quad \text {og} \quad f(x)=x^2.\]

    Vi differentierer \(f\):

    \[f'(x)=2x.\]

    Ifølge tabellen er differentialkvotienten givet ved \(k\cdot f'(x)\), så

    \[h'(x)=k\cdot f'(x)=5\cdot 2x=10x.\]

    Altså er

    \[h'(x)=10x.\]

Øvelse 9.2.4

Bestem differentialkvotienten for følgende funktioner:

  • a) \(h(x)=x^2+4\)

  • b) \(h(x)=e^{-2x}-x\)

  • c) \(f(x)=2x^3\)

  • d) \(f(x)=3x^2-2x+1\)

  • e) \(2\cdot 3^x +2\cdot e^{3x}\)

Løsning 9.2.4

  • a) \(h'(x)=2x\)

  • b) \(h'(x)=-2e^{-2x}-1\)

  • c) \(f'(x)=6\cdot x^2\)

  • d) \(f'(x)=6x-2\)

  • e) \((2\cdot 3^x +2\cdot e^{3x})'= 2\cdot \ln (3)\cdot 3^x+6e^{3x}\)

Produktfunktioner - A-niveau
  • Eksempel 9.2.9
    Vi vil differentiere funktionen \(h(x)=x^2\cdot \ln (x)\). Vi sammenligner med tabel 9.2 og ser at vores funktion har form som i formel (43), hvor:

    \[f(x)=x^2\quad \text {og} \quad g(x)=\ln (x)\]

    Vi differentierer \(f\) og \(g\):

    \[f'(x)=2x\quad \text {og} \quad g'(x)=\frac {1}{x}.\]

    Ifølge tabellen er differentialkvotienten givet ved \(f'(x)\cdot g(x) + f(x)\cdot g'(x)\), så

    \[h'(x)=2x \cdot \ln (x) + x^2 \cdot \frac {1}{x}=2x \cdot \ln (x) + x\]

    Altså er:

    \[h'(x)=2x \cdot \ln (x) + x.\]

Øvelse 9.2.5

Bestem den afledte funktion for følgende funktioner:

  • a) \(h(x)=2x\cdot \ln (x)\)

  • b) \(f(x)=\sqrt {x} \cdot 2^x\)

  • c) \(x^3\cdot e^x\)

Løsning 9.2.5

  • a) \(h'(x)=2 \cdot \ln (x) + 2x\cdot \frac {1}{x} = 2\cdot \ln (x)+2\)

  • b) \(f'(x)= \frac {1}{2\sqrt (x)}\cdot 2^x+\sqrt {x}\cdot \ln (2)\cdot 2^x = 2^x\big (\frac {1}{2\sqrt {x}}+ \ln (2) \cdot \sqrt {x}\big )\)

  • c) \((x^3\cdot e^x)'=3x^2\cdot e^x+ x^3\cdot e^x = x^2\cdot e^x (x + 3)\)

Sammensatte funktioner - A-niveau

For at forklare den sidste formel i tabel 9.2, har vi først brug for at introducere sammensatte funktioner.

Har man to funktioner \(f\) og \(g\) kan man sætte dem sammen til en ny funktion \(f\circ g\) (læses ”f bolle g”) som illustreret i følgende diagram:

(-tikz- diagram)

Figur 9.1: Funktionen \(f\circ g\) er sammensat af først \(g\) og så \(f\)

.

Vi regner forskriften for \((f\circ g )(x)\) ved at sætte forskriften for \(g\) ind i \(f\):

  • Eksempel 9.2.10
    Lad \(f(x)=\sqrt {x}\) og \(g(x)=x^2+x-1\). Vi regner forskriften for den sammensatte funktion \((f\circ g )(x)\) ved at sætte forskriften for \(g\) i stedet for \(x\) i forskriften for \(f\):

    \[(f\circ g )(x)=f\big ( g(x) \big )=\sqrt {x^2+x-1}\]

Øvelse 9.2.6

Lad \(f(x)=\ln (x)\) og \(g(x)=2x-1\). Regn forskrifterne for:

  • a) \(f\circ g\)

  • b) \(g\circ f\)

Løsning 9.2.6

  • a) \(f\circ g = \ln (2x-1)\)

  • b) \(g\circ f = 2\ln (x)-1\)

Selvom den sammensatte funktion formelt hedder \(f\circ g\), er der (HHX)-tradition for at betegne den med \(f\big ( g(x) \big )\), så det vil vi gøre fremover.

  • Eksempel 9.2.11
    Lad \(f(x)=x^2-9\) og \(g(x)=x-3\). Vi regner den forskriften for den sammensatte funktion:

    \[f\big (g(x)\big ) = (x-3)^2-9 = x^2+3^2-6x-9=x^2-6x.\]

    Altså \(f\big (g(x)\big ) =x^2-6x\).

Øvelse 9.2.7

Regn forskriften \(f\big ( g(x) \big )\) for den sammensatte funktion når:

  • a) \(f(x)=\sqrt {x}\) og \(g(x)=2x-1\)

  • b) \(f(x)=e^x\) og \(g(x)=x^2+2x\)

  • c) \(f(x)=x^2+3\) og \(g(x)=x+1\)

  • d) \(f(x) = x +1\) og \(g(x)=x^5\)

Løsning 9.2.7

  • a) \(f\big ( g(x) \big ) = \sqrt {2x-1}\)

  • b) \(f\big ( g(x) \big ) = e^{x^2+2x}\)

  • c) \(f\big ( g(x) \big ) = x^2 +2x+4\)

  • d) \(f\big ( g(x) \big ) = x^5+1\)

I forskrift \(h(x)\) som kan skrives på formen \(h(x)=f\big (g(x)\big )\) kaldes \(g\) den indre funktion og \(f\) den ydre funktion.

  • Eksempel 9.2.12
    I funktionen \(h(x)=\sqrt {x^4+x}\) er den indre funktion \(g(x)=x^4+x\) og den ydre funktion er \(f(x)=\sqrt {x}\).

Øvelse 9.2.8

Bestem den indre funktion \(g\) og den ydre funktion \(f\) i følgende sammensatte funktioner.

  • a) \(h(x)=\sqrt {x+2}\)

  • b) \(h(x)=3^{7x^2-2}\)

  • c) \((x^3-x^2+1)^7\)

Løsning 9.2.8

  • a) \(g(x)=x+2\) og \(f(x)=\sqrt {x}\)

  • b) \(g(x)=7x^2-2\) og \(f(x)=3^x\)

  • c) \(g(x)=x^3-x^2+1\) og \(f(x)=x^7\)

Vi er nu endelig klar til at demonstrere den sidste formel fra tabel 9.2.

  • Eksempel 9.2.13
    Vi vil differentiere funktion \(h(x)=\ln (x^2 + 1)\). Vi ser at den har form som en sammensat funktion \(f\big (g(x)\big )\).

    \[g(x)=x^2+1 \quad \text {og} \quad f(x)=\ln (x).\]

    Vi differentierer \(g\) og \(f\):

    \[g'(x)=2x \quad \text {og} \quad f'(x)=\frac {1}{x}.\]

    Ifølge formel (44) i tabel 9.2 er differentialkvotienten givet ved \(f'(g(x))\cdot g'(x)\), så

    \[h'(x)=f'(g(x))\cdot g'(x)=\frac {1}{x^2+1}\cdot 2x = \frac {2x}{x^2+1}.\]

    Altså

    \[h'(x)=\frac {2x}{x^2+1}.\]

Øvelse 9.2.9

Differentier følgende funktioner.

  • a) \(h(x)=\ln (2x-1)\)

  • b) \(h(x)=e^{x^3+3x}\)

  • c) \(f(x)=(x^2+3x-1)^8\)

  • d) \(\sqrt {3x^2+x}\)

Løsning 9.2.9

  • a) \(h'(x)= \frac {2}{2x-1}.\)

  • b) \(h'(x)=(3x^2+3)\cdot e^{x^3+3x}\)

  • c) \(f'(x)=8(x^2+3x-1)^7\cdot (2x+3)\)

  • d) \((\sqrt {3x^2+x})'=\frac {6x+1}{2\sqrt {3x^2+x}}\)

Vi slutter afsnittet af med en øvelse, hvor I får brug alle formlerne fra afsnittet.

Øvelse 9.2.10

Differentier følgende funktioner

  • a) \(f(x)=x^2+e^x\)

  • b) \(f(x)=5\cdot 6^x\)

  • c) \(f(x)=\ln (x)\cdot x\)

  • d) \(f(x)=\sqrt {x^2+3x}+x\)

  • e) \(f(x)=e^{4x}\cdot x^2 - \frac {1}{x}\)

  • f) \(f(x)=\frac {1}{x^7-x}\)

  • g) \(f(x) =\frac {2}{x}\)

  • h) \(f(x)=2\cdot 4^{x^2-4}\)

Løsning 9.2.10

  • a) \(f'(x)=2x+e^x\)

  • b) \(f'(x)=5\ln (6)\cdot 6^x\)

  • c) \(f'(x)=\ln (x)+1\)

  • d) \(f'(x)=\frac {2x+3}{2\sqrt {x^2+3x}}+1\)

  • e) \(f'(x)=4e^{4x}\cdot x^2 + e^{4x}\cdot 2x + \frac {1}{x^2}\)

  • f) \(f'(x)=-\frac {7x^6-1}{(x^7-x)^2}\)

  • g) \(f'(x) = - \frac {2}{x^2}\)

  • h) \(f'(x)=4x\cdot \ln (4)\cdot 4^{x^2-4}\)