Differentiering
Find den afledte funktion f′(x) ved hjælp af differentieringsregler
Den afledte f′(x) fortæller med hvilken hastighed funktionen ændrer sig i hvert punkt — det er hældningen af grafen. Differentiering bruges til at finde ekstrema og bestemme tangentligninger.
På A-niveau forventes du at mestre tre regler for sammensatte udtryk: kædereglen (noget inde i noget), produktreglen (to udtryk ganget) og kvotientreglen (brøk med x i begge). Strategien: identificér typen, anvend reglen.
💡 Husk: brøker og rødder omskrives til potensform FØR differentiering — \(\tfrac{1}{x} = x^{-1}\) og \(\sqrt{x} = x^{0{,}5}\).
a = koefficient
n = eksponent
x = variabel
svar = resultat
Formler
Potensreglen
\(\dfrac{d}{dx}\bigl[\textcolor{#3b82f6}{a}\cdot x^{\textcolor{#f59e0b}{n}}\bigr] = \textcolor{#3b82f6}{a}\cdot\textcolor{#f59e0b}{n}\cdot x^{\textcolor{#f59e0b}{n}-1}\)
Kædereglen — differentier udefra ind
\(\dfrac{d}{dx}\bigl[f(\textcolor{#10b981}{g(x)})\bigr] = f'(\textcolor{#10b981}{g(x)})\cdot \textcolor{#10b981}{g'(x)}\)
De vigtigste kæderegel-typer
\(\bigl(e^{\textcolor{#10b981}{g(x)}}\bigr)' = e^{\textcolor{#10b981}{g(x)}}\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl(\ln(\textcolor{#10b981}{g(x)})\bigr)' = \dfrac{\textcolor{#10b981}{g'(x)}}{\textcolor{#10b981}{g(x)}}\)
\(\bigl(\sin(\textcolor{#10b981}{g(x)})\bigr)' = \cos(\textcolor{#10b981}{g(x)})\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl(\cos(\textcolor{#10b981}{g(x)})\bigr)' = -\sin(\textcolor{#10b981}{g(x)})\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl((\textcolor{#10b981}{g(x)})^n\bigr)' = n\cdot(\textcolor{#10b981}{g(x)})^{n-1}\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl(\ln(\textcolor{#10b981}{g(x)})\bigr)' = \dfrac{\textcolor{#10b981}{g'(x)}}{\textcolor{#10b981}{g(x)}}\)
\(\bigl(\sin(\textcolor{#10b981}{g(x)})\bigr)' = \cos(\textcolor{#10b981}{g(x)})\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl(\cos(\textcolor{#10b981}{g(x)})\bigr)' = -\sin(\textcolor{#10b981}{g(x)})\cdot\textcolor{#10b981}{g'(x)}\)
\(\bigl((\textcolor{#10b981}{g(x)})^n\bigr)' = n\cdot(\textcolor{#10b981}{g(x)})^{n-1}\cdot\textcolor{#10b981}{g'(x)}\)
Produktreglen
\((\textcolor{#3b82f6}{u} \cdot \textcolor{#10b981}{v})' = \textcolor{#3b82f6}{u}' \cdot \textcolor{#10b981}{v} + \textcolor{#3b82f6}{u} \cdot \textcolor{#10b981}{v}'\)
Kvotientreglen
\(\left(\dfrac{\textcolor{#3b82f6}{u}}{\textcolor{#10b981}{v}}\right)' = \dfrac{\textcolor{#3b82f6}{u}'\textcolor{#10b981}{v} - \textcolor{#3b82f6}{u}\textcolor{#10b981}{v}'}{\textcolor{#10b981}{v}^2}\)
Eksempler
Eksempel 1
Potensreglen: \(f(x) = 3x^4\)
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1
a = 3, n = 4
\(f'(x) = 3 \cdot 4 \cdot x^{4-1} = \textcolor{#ef4444}{12x^3}\)
\(f'(x) = 3 \cdot 4 \cdot x^{4-1} = \textcolor{#ef4444}{12x^3}\)
Eksempel 2
Kæderegel med sinus: \(f(x) = \sin(\textcolor{#10b981}{3x^2+2x})\)
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1
Identificér ydre og indre
Ydre: \(\sin(u)\) Indre: \(\textcolor{#10b981}{g(x) = 3x^2+2x}\)
Ydre: \(\sin(u)\) Indre: \(\textcolor{#10b981}{g(x) = 3x^2+2x}\)
2
Differentier indre funktion
\(\textcolor{#10b981}{g'(x) = 6x+2}\)
\(\textcolor{#10b981}{g'(x) = 6x+2}\)
3
Kæderegel: \((\sin u)' = \cos(u)\cdot u'\)
\(f'(x) = \cos(\textcolor{#10b981}{3x^2+2x})\cdot\textcolor{#10b981}{(6x+2)} = \textcolor{#ef4444}{(6x+2)\cos(3x^2+2x)}\)
\(f'(x) = \cos(\textcolor{#10b981}{3x^2+2x})\cdot\textcolor{#10b981}{(6x+2)} = \textcolor{#ef4444}{(6x+2)\cos(3x^2+2x)}\)
Eksempel 3
Kæderegel med potens: \(f(x) = (\textcolor{#10b981}{2x+3})^4\)
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1
Ydre: \(u^4\) Indre: \(\textcolor{#10b981}{g(x)=2x+3}\)
\(\textcolor{#10b981}{g'(x)=2}\)
\(\textcolor{#10b981}{g'(x)=2}\)
2
\((u^4)' = 4u^3\cdot u'\)
\(f'(x) = 4(\textcolor{#10b981}{2x+3})^3 \cdot \textcolor{#10b981}{2} = \textcolor{#ef4444}{8(2x+3)^3}\)
\(f'(x) = 4(\textcolor{#10b981}{2x+3})^3 \cdot \textcolor{#10b981}{2} = \textcolor{#ef4444}{8(2x+3)^3}\)
Eksempel 4
Kæderegel med e: \(f(x) = e^{\textcolor{#10b981}{2x^2-1}}\)
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1
Ydre: \(e^u\) Indre: \(\textcolor{#10b981}{g(x)=2x^2-1}\)
\(\textcolor{#10b981}{g'(x)=4x}\)
\(\textcolor{#10b981}{g'(x)=4x}\)
2
\((e^u)' = e^u\cdot u'\)
\(f'(x) = e^{\textcolor{#10b981}{2x^2-1}}\cdot\textcolor{#10b981}{4x} = \textcolor{#ef4444}{4x\cdot e^{2x^2-1}}\)
\(f'(x) = e^{\textcolor{#10b981}{2x^2-1}}\cdot\textcolor{#10b981}{4x} = \textcolor{#ef4444}{4x\cdot e^{2x^2-1}}\)
Eksempel 5
Produktregel: \(f(x) = \textcolor{#3b82f6}{\sin(2x)} \cdot \textcolor{#10b981}{e^{3x}}\)
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1
\(u = \textcolor{#3b82f6}{\sin(2x)}\), \(v = \textcolor{#10b981}{e^{3x}}\)
\(u' = \textcolor{#3b82f6}{2\cos(2x)}\), \(v' = \textcolor{#10b981}{3e^{3x}}\)
\(u' = \textcolor{#3b82f6}{2\cos(2x)}\), \(v' = \textcolor{#10b981}{3e^{3x}}\)
2
\((uv)' = u'v+uv'\)
\(f'(x) = \textcolor{#3b82f6}{2\cos(2x)}\cdot\textcolor{#10b981}{e^{3x}} + \textcolor{#3b82f6}{\sin(2x)}\cdot\textcolor{#10b981}{3e^{3x}} = \textcolor{#ef4444}{e^{3x}(2\cos(2x)+3\sin(2x))}\)
\(f'(x) = \textcolor{#3b82f6}{2\cos(2x)}\cdot\textcolor{#10b981}{e^{3x}} + \textcolor{#3b82f6}{\sin(2x)}\cdot\textcolor{#10b981}{3e^{3x}} = \textcolor{#ef4444}{e^{3x}(2\cos(2x)+3\sin(2x))}\)
Eksempel 6
Produktregel: \(f(x) = \textcolor{#3b82f6}{x^2}\cdot\textcolor{#10b981}{\ln(x)}\)
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1
\(u=\textcolor{#3b82f6}{x^2}\), \(v=\textcolor{#10b981}{\ln x}\)
\(u'=\textcolor{#3b82f6}{2x}\), \(v'=\textcolor{#10b981}{\tfrac{1}{x}}\)
\(u'=\textcolor{#3b82f6}{2x}\), \(v'=\textcolor{#10b981}{\tfrac{1}{x}}\)
2
\(f'(x) = \textcolor{#3b82f6}{2x}\cdot\textcolor{#10b981}{\ln x} + \textcolor{#3b82f6}{x^2}\cdot\textcolor{#10b981}{\tfrac{1}{x}} = \textcolor{#ef4444}{2x\ln x + x}\)
Eksempel 7
Kvotientreglen: \(f(x) = \dfrac{\textcolor{#3b82f6}{x^2}}{\textcolor{#10b981}{x+1}}\)
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1
Identificér u og v
\(\textcolor{#3b82f6}{u=x^2},\ u'=2x\) \(\textcolor{#10b981}{v=x+1},\ v'=1\)
\(\textcolor{#3b82f6}{u=x^2},\ u'=2x\) \(\textcolor{#10b981}{v=x+1},\ v'=1\)
2
\(f'(x) = \dfrac{\textcolor{#3b82f6}{2x}\cdot\textcolor{#10b981}{(x+1)} - \textcolor{#3b82f6}{x^2}\cdot\textcolor{#10b981}{1}}{\textcolor{#10b981}{(x+1)}^2} = \textcolor{#ef4444}{\dfrac{x^2+2x}{(x+1)^2}}\)
Eksempel 8
Omskrivning FØR differentiering: \(f(x) = \sqrt{x}\) og \(f(x) = \tfrac{3}{x^2}\)
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1
Omskriv til potensform
\(\sqrt{x} = x^{0{,}5}\) og \(\tfrac{3}{x^2} = 3x^{-2}\)
\(\sqrt{x} = x^{0{,}5}\) og \(\tfrac{3}{x^2} = 3x^{-2}\)
2
Anvend potensreglen
\((x^{0{,}5})' = 0{,}5\cdot x^{-0{,}5} = \textcolor{#ef4444}{\tfrac{1}{2\sqrt{x}}}\)
\((3x^{-2})' = 3\cdot(-2)\cdot x^{-3} = \textcolor{#ef4444}{-\tfrac{6}{x^3}}\)
\((x^{0{,}5})' = 0{,}5\cdot x^{-0{,}5} = \textcolor{#ef4444}{\tfrac{1}{2\sqrt{x}}}\)
\((3x^{-2})' = 3\cdot(-2)\cdot x^{-3} = \textcolor{#ef4444}{-\tfrac{6}{x^3}}\)
💡 Kædereglen huskeregel: Find den ydre funktion og den indre funktion. Differentier ydre (men hold indre uændret), gang med den afledte af indre. Sinus→cosinus→minus sinus→minus cosinus→sinus (cyklus).
Genkend opgavetypen
🔍 Sådan ser den ud til eksamen
- Let: potensregel, \(e^{ax}\), \(a\sin(bx)\) — direkte regler
- Mellem: \(\ln(g(x))\), \(\sin(g(x))\), \(e^{g(x)}\), \((g(x))^n\) — kæderegel, eller produkt af to funktioner
- Svær: dobbelt kæderegel, kvotientregel, kæde + produkt kombineret
- Nøgleord der afslører kæderegel: noget inde i noget andet — fx \(\sin(x^2)\), \(e^{x^2}\), \(\ln(3x+1)\)
Eksempel 3 — faseforskydning
Max = 5, min = −1, periode = \(\pi\), \(f(0) = 4\). Opstil \(f(t) = A\sin(\omega t + \varphi) + k\).
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1
Find A og k fra max/min
\(A = \dfrac{5-(-1)}{2} = 3\) \(k = \dfrac{5+(-1)}{2} = 2\)
\(A = \dfrac{5-(-1)}{2} = 3\) \(k = \dfrac{5+(-1)}{2} = 2\)
2
Find ω fra perioden
\(T = \pi \implies \omega = \dfrac{2\pi}{\pi} = 2\)
\(T = \pi \implies \omega = \dfrac{2\pi}{\pi} = 2\)
3
Find φ fra f(0)
\(f(0) = 3\sin(\varphi) + 2 = 4 \implies \sin(\varphi) = \dfrac{4-2}{3} = \dfrac{2}{3}\)... men her er \(A=3\), lad os bruge \(f(0)=4\):
\(3\sin(\varphi)+2=4 \implies \sin(\varphi)=\dfrac{2}{3}\)
Bemærk: dette er ikke en "pæn" vinkel — i generatorens opgaver er sinusværdien altid \(\frac{1}{2}\) eller \(1\) for at bruge enhedscirklen.
\(f(0) = 3\sin(\varphi) + 2 = 4 \implies \sin(\varphi) = \dfrac{4-2}{3} = \dfrac{2}{3}\)... men her er \(A=3\), lad os bruge \(f(0)=4\):
\(3\sin(\varphi)+2=4 \implies \sin(\varphi)=\dfrac{2}{3}\)
Bemærk: dette er ikke en "pæn" vinkel — i generatorens opgaver er sinusværdien altid \(\frac{1}{2}\) eller \(1\) for at bruge enhedscirklen.
4
Pænt eksempel: A=4, k=1, ω=2, f(0)=3
\(4\sin(\varphi)+1=3 \implies \sin(\varphi)=\dfrac{1}{2} \implies \varphi=\dfrac{\pi}{6}\) (fra enhedscirklen)
Facit: \(f(t) = 4\sin\!\left(2t+\dfrac{\pi}{6}\right)+1\)
\(4\sin(\varphi)+1=3 \implies \sin(\varphi)=\dfrac{1}{2} \implies \varphi=\dfrac{\pi}{6}\) (fra enhedscirklen)
Facit: \(f(t) = 4\sin\!\left(2t+\dfrac{\pi}{6}\right)+1\)
⚠️ Klassiske eksamensfejl
- Glemmer den indre afledte: \((\sin(3x))' = 3\cos(3x)\) — ikke bare \(\cos(3x)\)
- Cosinus → minus sinus: \((\cos(u))' = -\sin(u)\cdot u'\) — minustegnet glemmes
- Produktreglen: begge led differentieres — ikke kun ét
- ln differenteret: \((\ln x)' = \tfrac{1}{x}\) — ikke \(\tfrac{1}{\ln x}\)