MATHHX B

MATHHX B

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10.2 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 10.1: 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 10.2.1
    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 10.2.1

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

  • a) \(f(x) = \ln (x)\)

  • b) \(f(x) = e^x\)

Løsning 10.2.1

  • a) \(f'(x) = \frac {1}{x}\)

  • b) \('f(x) = e^x\)

  • Eksempel 10.2.2
    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 10.2.3
    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 10.2.2

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 10.2.2

  • 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 10.2.4
    Vi vil differentiere funktionen givet ved udtrykket \(x^3\). Vi skriver:

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

Øvelse 10.2.3

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

Løsning 10.2.3

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

Differentialkvotient for opbyggede funktioner

Vi vil ofte møde funktioner som ikke direkte står i tabel 10.1 . I stedet vil de være bygget op af funktioner fra tabellen.

  • Eksempel 10.2.5
    Vi vil differentiere funktionen \(h(x)=x^2+3\). Vi ved fra tabel 10.1 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 10.2.6
    Vi vil differentiere funktionen \(5x^2\). Vi ved fra tabel 10.1 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 10.2: 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 10.2.7
    Vi vil differentiere funktionen \(h(x)=x^2+3\). Vi sammenligner med tabel 10.2 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 10.2.8
    Vi vil differentiere funktionen \(h(x)=5x^2\). Vi sammenligner med tabel 10.2 og ser at vores funktion har form som i formel (40), hvor

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

    Vi differentiere \(f\):

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

    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 10.2.4

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 10.2.4

  • 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 10.2.9
    Vi vil differentiere funktionen \(h(x)=x^2\cdot \ln (x)\). Vi sammenligner med tabel 10.2 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 10.2.5

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 10.2.5

  • 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 10.1, 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 10.1: 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 10.2.10
    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 10.2.6

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

  • a) \(f\circ g\)

  • b) \(g\circ f\)

Løsning 10.2.6

  • 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 10.2.11
    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 10.2.7

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 10.2.7

  • 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 10.2.12
    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 10.2.8

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 10.2.8

  • 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 10.2.

  • Eksempel 10.2.13
    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 differentiere \(g\) og \(f\):

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

    Ifølge formel (44) i ?? 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 10.2.9

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 10.2.9

  • a) \(h'(x)= \frac {2}{2x-1}.\)

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

  • c) \(f'(x)=8(2x+3)(x^2+3x-1)^7\)

  • 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 10.2.10

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 10.2.10

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