// Numbas version: finer_feedback_settings
{"name": "Derivasjon av log og eksp", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Derivasjon av log og eksp", "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "<p>Deriver uttrykkene.</p>", "advice": "<p><strong>a)</strong></p>\n<p>Her er det lurt &aring; forenkle uttrykket&nbsp;<strong>f&oslash;r</strong> vi deriverer. Vi bruker logaritmesetningene:</p>\n<p>$y=\\simplify{ln(x^{a})+{b}ln({c}x)}=\\simplify{{a}ln(x)}+\\simplify{{b}(ln({c})+ln(x))}=\\simplify{({a}+{b})ln(x)+{b}ln({c})}$.</p>\n<p>Vi kan da derivere leddvis, og f&aring;r</p>\n<p>\\[ y'=(\\simplify{({a}+{b})ln(x)})'+(\\var{b}ln(\\var{c}))'=\\simplify{({a}+{b})/x}+0=\\underline{\\underline{\\simplify{({a}+{b})/x}}}\\]</p>\n<p></p>\n<p><strong>b)</strong></p>\n<p>Her b&oslash;r vi ogs&aring; forenkle f&oslash;rst:</p>\n<p>$y=\\simplify{ln(x^{a}/(sqrt(x)))}=\\ln(x^{\\var{a}}-\\simplify{ln(sqrt(x))}=\\var{a}\\ln(x)-\\ln(x^{\\frac{1}{2}})=\\var{a}\\ln(x)-\\frac{1}{2}\\ln(x)=\\simplify{((2*{a}-1)/2)}\\ln(x)$.</p>\n<p>Da f&aring;r vi:</p>\n<p>\\[y'=(\\simplify{((2*{a}-1)/2)}\\ln(x))'=\\underline{\\underline{\\simplify{((2*{a}-1)/(2x))}}}\\]</p>\n<p></p>\n<p><strong>c)</strong></p>\n<p>Det finnes ingen regler for &aring; forenkle logaritmen n&aring;r argumentet er en <strong>sum.</strong></p>\n<p>Her l&oslash;nner det seg &aring; huske \"snarveien\" $\\left(\\ln(u)\\right)=\\dfrac{u'}{u}$. Vi trenger da bare &aring; velge $u$ og regne ut $u'$:</p>\n<p>$u=\\simplify[unitFactor,!collectNumbers]{x^{d}+{e}x}, \\qquad u'=\\simplify{{d}x^{{d}-1}+{e}}$</p>\n<p>Vi f&aring;r da at $y'=\\dfrac{u'}{u}=\\underline{\\underline{\\dfrac{\\simplify{{d}x^{{d}-1}+{e}}}{\\simplify{x^{d}+{e}x}}}}$.</p>\n<p></p>\n<p>Ellers m&aring; vi bruke kjerneregelen fullt ut:</p>\n<p>$u=\\simplify[unitFactor,!collectNumbers]{x^{d}+{e}x}, \\qquad u'=\\simplify{{d}x^{{d}-1}+{e}}$</p>\n<p>$y(u)=\\ln(u), \\qquad y'(u)=\\dfrac{1}{u}$</p>\n<p>$y'=y'(u)\\cdot u' = \\dfrac{1}{u}\\cdot\\simplify{{d}x^{{d}-1}+{e}}=\\dfrac{\\simplify{{d}x^{{d}-1}+{e}}}{\\simplify{x^{d}+{e}x}}$</p>\n<p></p>\n<p><strong>d)</strong></p>\n<p>Her m&aring; vi bruke kjerneregelen:</p>\n<p>$u=\\simplify{x^{d}+x^{f}}, \\qquad u'=\\simplify{{d}x^{{d}-1}+{f}x^{{f}-1}}$</p>\n<p>$y(u)=e^{u}, \\qquad y'(u)=e^{u}$</p>\n<p>$y'=y'(u)\\cdot u'=e^{u}\\cdot 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