// Numbas version: exam_results_page_options {"name": "Spesielle derivasjonsregler", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Spesielle derivasjonsregler", "tags": [], "metadata": {"description": "", "licence": "All rights reserved"}, "statement": "

Deriver funksjonene under.

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Hint: For å skrive inn f.eks. $\\frac{1}{x^{3a}}$ bør man skrive 1/(x^(3a)), for å skrive inn $\\cos^2x$ bør man skrive (cos(x))^2

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Bruk følgende spesielle derivasjonsregler

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$f(x)=a$ gir $f'(x)=0$ ($a$ er et konstant)

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$f(x)=ax$ gir $f'(x)=a$ ($a$ er et konstant)

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$f(x)=x^a$ gir $f'(x)=ax^{a-1}$ ($a\\neq0$ er et konstant)

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$f(x)=\\sin(kx)$ gir $f'(x)=k\\cos(kx)$ ($k$ er et konstant, $x$ er i radianer)

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$f(x)=\\cos(kx)$ gir $f'(x)=-k\\sin(kx)$ ($k$ er et konstant, $x$ er i radianer)

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$f(x)=\\tan(kx)$ gir $f'(x)=k\\dfrac{1}{\\cos^2(kx)}$ ($k$ er et konstant, $x$ er i radianer)

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$f(x)=e^{kx}$ gir $f'(x)=ke^{kx}$ ($k$ er et konstant)

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$f(x)=\\ln(kx)$ gir $f'(x)=\\frac{1}{x}$ ($k$ er et konstant)

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$\\simplify{{out1}}(\\simplify{{inp1}})=\\var{inp1}^\\var{k1}$

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$\\simplify{{out1}}'(\\simplify{{inp1}})=$ [[0]]

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$\\simplify{{out2}}(\\simplify{{inp2}})=\\var{k2}$

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$\\simplify{{out2}}'(\\simplify{{inp2}})=$ [[0]]

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$\\simplify{{out3}}(\\simplify{{inp3}})=\\sin(\\var{k3}\\var{inp3})$

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$\\simplify{{out3}}'(\\simplify{{inp3}})=$ [[0]]

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$\\simplify{{out4}}(\\simplify{{inp4}})=\\cos(\\var{k4}\\var{inp4})$

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$\\simplify{{out4}}'(\\simplify{{inp4}})=$ [[0]]

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$\\simplify{{out5}}(\\simplify{{inp5}})=\\tan(\\var{k5}\\var{inp5})$

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$\\simplify{{out5}}'(\\simplify{{inp5}})=$ [[0]]

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$\\simplify{{out6}}(\\simplify{{inp6}})=e^{\\var{k6}\\var{inp6}}$

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$\\simplify{{out6}}'(\\simplify{{inp6}})=$ [[0]]

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$\\simplify{{out7}}(\\simplify{{inp7}})=\\ln({\\var{k7}\\var{inp7}})$

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$\\simplify{{out7}}'(\\simplify{{inp7}})=$ [[0]]

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$\\simplify{{out8}}(\\simplify{{inp8}})=\\var{a}^{{\\var{k8}\\var{inp8}}}$

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$\\simplify{{out8}}'(\\simplify{{inp8}})=$ [[0]]

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